9.11: Exercises
- Page ID
- 560844
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9.2: Gas Pressure
1. Why are sharp knives more effective than dull knives?
2. Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has?
Bridges have weight limits based on wheels or axles because more wheels distribute the vehicle’s weight over a larger area, reducing pressure on the bridge surface and minimizing structural stress.
3. Why should you roll or belly-crawl rather than walk across a thinly-frozen pond?
Rolling or belly-crawling across a thinly frozen pond spreads your body weight over a larger surface area, reducing pressure and lowering the risk of breaking the ice.
4. A typical barometric pressure in Redding, California, is about 750 mm Hg. Calculate this pressure in atm and kPa.
- Pressure in Redding, CA: 750 mm Hg
- In atm: (750 / 760 = 0.9868 atm)
- In kPa: (0.9868 × 101.325 = 99.96 kPa)
5. A typical barometric pressure in Denver, Colorado, is 615 mm Hg. What is this pressure in atmospheres and kilopascals?
- Pressure in Denver, CO: 615 mm Hg
- In atm: (615 / 760 = 0.8092 atm)
- In kPa: (0.8092 × 101.325 = 82.05 kPa)
6. A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?
- Pressure in Kansas City: 740 torr
- In atm: (740 / 760 = 0.9737 atm)
- In mm Hg: 740 mm Hg
- In kPa: (0.9737 × 101.325 = 98.64 kPa)
7. Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi?
- 32 psi to kPa: (32 × 6.89476 = 220.63 kPa)
8. During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars (1 bar = 0.987 atm). What is that pressure in torr
and kPa?
- Mars pressure: 6.50 millibars
- In atm: (6.50 × 0.000987 = 0.0064155 atm)
- In torr: (0.0064155 × 760 = 4.87 torr)
- In kPa: (0.0064155 × 101.325 = 0.65 kPa)
9. The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.
- Venus pressure: 88.8 atm
- In psi: (88.8 × 14.696 = 1305.13 psi)
- Earth sea level pressure: 14.696 psi
- Venus pressure is ~88.8 times greater than Earth’s
10. A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa. What is the pressure of this gas in atmospheres and torr?
- Pressure: 14.82 MPa
- In atm: (14.82 × 1000 / 101.325 = 146.28 atm)
- In torr: (146.28 × 760 = 11117.28 torr)
11. Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar.
a. What was the pressure in kPa?
b. The pressure near the seacoast in the northeastern United States is usually reported near 30.0 in. Hg. During a hurricane, the pressure may fall to near 28.0 in. Hg. Calculate the drop in pressure in torr.
a. torr
b.Pa
c.bar
a.torr
b.Pa
c.bar
a.mm Hg
b.atm
c.kPa
a.mm Hg
b.atm
c.kPa
17. How would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers?
9.3: Relating Pressure, Volume, Amount, and Temperature - The Ideal Gas Law
18. Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?
a.What is the meaning of the term “directly proportional?”
b.What are the “other things” that must be equal?
22.How would the graph in Figure 8.12 change if the number of moles of gas in the sample used to determine the curve were doubled?
a.the appropriate graph
b.Boyle’s law
27.A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can?
28. What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr?
29.A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
9.4: Stoichiometry of Gaseous Substances, Mixtures and Reactions
31.A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?
a. 0.100 L of CO2 at 307 torr and 26 °C
b. 8.75 L of C2H4, at 378.3 kPa and 483 K
c. 221 mL of Ar at 0.23 torr and –54 °C
36.A high altitude balloon is filled with 1.41 × 104 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
41. For a given amount of gas showing ideal behavior, draw labeled graphs of:
a. the variation of \(\text{P}\) with \(\text{V}\)
b. the variation of \(\text{V}\) with \(\text{T}\)
c. the variation of \(\text{P}\) with \(\text{T}\)
d. the variation of \(\frac{1}{\text{P}}\) with \(\text{V}\)
a. CCl2F2(g)
b. CH3CH2F(g)
a. Outline the steps necessary to answer the question.
b. Answer the question.
a. Outline the steps necessary to answer the following question: What volume of O2 at 23 °C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?
b. Answer the question.
\(\text{4H}_2\text{O}(g) + \text{3Fe}(s) \longrightarrow \text{Fe}_3\text{O}_4(s) + \text{4H}_2(g)\)
a. Outline the steps necessary to answer the following question: What volume of \(\text{H}_2\) at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.0 g of \(\text{H}_2\text{O}\)?
b. Answer the question.
67. The chlorofluorocarbon \(\text{CCl}_2\text{F}_2\) can be recycled into a different compound by reaction with hydrogen to produce \(\text{CH}_2\text{F}_2(g)\), a compound useful in chemical manufacturing: \(\text{CCl}_2\text{F}_2(g) + \text{4H}_2(g) \longrightarrow \text{CH}_2\text{F}_2(g) + \text{2HCl}(g)\)
a.Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875 K and 0.966 atm is produced by the decomposition of 1 ton (1.000 103 kg) of calcium carbonate?
b.Answer the question.
a.Outline the steps necessary to answer the following question: What volume of C2H2 at 1.005 atm and 12.2 °C is formed by the reaction of 15.48 g of CaC2 with water?
b.Answer the question.
a.What is the total volume of the CO2(g) and H2O(g) at 600 °C and 0.888 atm produced by the combustion of 1.00 L of C2H6(g) measured at STP?
b.What is the partial pressure of H2O in the product gases?
74. Methanol, \(\ce{CH3OH}\), is produced industrially by the following reaction:
\[\ce{CO(g) + 2H2(g) ->[copper\ catalyst][300^\circ\text{C},\ 300\text{ atm}] CH3OH(g)}\]
Find ratio of total volume of reactants to products in production of methanol
77. Ethanol, \(\text{C}_2\text{H}_5\text{OH}\), is produced industrially from ethylene, \(\text{C}_2\text{H}_4\), by the following sequence of reactions:
\(\text{3C}_2\text{H}_4 + \text{2H}_2\text{SO}_4 \longrightarrow \text{C}_2\text{H}_5\text{HSO}_4 + (\text{C}_2\text{H}_5)_2\text{SO}_4\) \(\text{C}_2\text{H}_5\text{HSO}_4 + (\text{C}_2\text{H}_5)_2\text{SO}_4 + \text{3H}_2\text{O} \longrightarrow \text{3C}_2\text{H}_5\text{OH} + \text{2H}_2\text{SO}_4\)
What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?
79.A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)
\(\ce{CH2(NH2)CO2H + HNO2 -> CH2(OH)CO2H + H2O + N2}\)
9.5: Effusion and Diffusion of Gases
81. A balloon filled with helium gas takes 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%?
83.Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses.
84.Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H2O and D2O.
85.Which of the following gases diffuse more slowly than oxygen? F2, Ne, N2O, C2H2, NO, Cl2, H2S
88.A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
89. When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of \(\ce{NH4Cl}\) forms where gaseous \(\ce{NH3}\) and gaseous \(\ce{HCl}\) first come into contact.
\[\ce{NH3(g) + HCl(g) -> NH4Cl(s)}\]
At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both \(\ce{NH3}\) and \(\ce{HCl}\), and find out how much faster \(\ce{NH3}\) diffuses than \(\ce{HCl}\).)
9.6: The Kinetic Molecular Theory
90.Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.
a.The pressure of the gas is increased by reducing the volume at constant temperature.
b.The pressure of the gas is increased by increasing the temperature at constant volume.
c.The average speed of the molecules is increased by a factor of 2.
a.What effect do these changes have on the pressure exerted by the gas?
b.What is the effect on the average kinetic energy of the molecules?
c.What is the effect on the root mean square speed of the molecules?
a.Is the pressure of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
b.Is the density of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
c.At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?
d.The average temperature of the gas in a hot-air balloon is 1.30 102 °F. Calculate its density, assuming the molar mass equals that of dry air.
e.The lifting capacity of a hot-air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?
f.An average balloon has a diameter of 60 feet and a volume of 1.1 105 ft3. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?
g.A balloon carries 40.0 gallons of liquid propane (density 0.5005 g/L). What volume of CO2 and H2O gas is produced by the combustion of this propane at STP?
h. A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?
9.7: Non-Ideal Gas Behavior
a.high pressure, small volume
b.high temperature, low pressure
c.low temperature, high pressure
CO, CO2, H2, He, NH3, SF6?
a.using the ideal gas law
b.using the van der Waals equation
c.Explain the reason for the difference.
d.Identify which correction (that for P or V) is dominant and why.
a.If XX behaved as an ideal gas, what would its graph of Z vs. P look like?
b.For most of this chapter, we performed calculations treating gases as ideal. Was this justified?
c.What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
d.What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
e.In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas?
Solutions
S9.2: Gas Pressure
S1. Sharp knives are more effective than dull knives
A sharp knife has a much smaller surface area along its cutting edge compared to a dull knife. Since pressure is defined as force divided by area (\(P = \frac{F}{A}\)), a smaller area concentrates the applied force, producing a significantly higher pressure that easily cuts through materials.
S2. Small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has
A vehicle's weight is distributed across its axles and wheels. Increasing the number of wheels or axles spreads the total force over a larger surface area, thereby reducing the pressure exerted on any single point of the bridge structure.
S3. Roll or belly-crawl rather than walk across a thinly-frozen pond
Walking concentrates your entire body weight onto the small surface area of your feet, creating high local pressure that can crack thin ice. Lying flat and crawling or rolling redistributes your weight over a much larger surface area, significantly lowering the pressure exerted on the ice surface.
S4. A typical barometric pressure in Redding, California, is about 750 mm Hg. It's pressure in atm and kPa:
* In atmospheres: \(750\text{ mm Hg} \times \frac{1\text{ atm}}{760\text{ mm Hg}} = 0.987\text{ atm}\)
* In kilopascals: \(0.987\text{ atm} \times \frac{101.325\text{ kPa}}{1\text{ atm}} = 100\text{ kPa}\)
S5. A typical barometric pressure in Denver, Colorado, is 615 mm Hg. Pressure in atmospheres and kilopascals:
* In atmospheres: \(615\text{ mm Hg} \times \frac{1\text{ atm}}{760\text{ mm Hg}} = 0.809\text{ atm}\)
* In kilopascals: \(0.809\text{ atm} \times \frac{101.325\text{ kPa}}{1\text{ atm}} = 82.0\text{ kPa}\)
S6. A typical barometric pressure in Kansas City is 740 torr. Pressure in atmospheres, in millimeters of mercury, and in kilopascals:
* In atmospheres: \(740\text{ torr} \times \frac{1\text{ atm}}{760\text{ torr}} = 0.974\text{ atm}\)
* In millimeters of mercury: Since \(1\text{ torr} = 1\text{ mm Hg}\), the pressure is \(740\text{ mm Hg}\).
* In kilopascals: \(0.974\text{ atm} \times \frac{101.325\text{ kPa}}{1\text{ atm}} = 98.7\text{ kPa}\)
S7. Canadian tire pressure gauges are marked in units of kilopascals. Reading on such a gauge that corresponds to 32 psi?
\(32\text{ psi} \times \frac{1\text{ atm}}{14.7\text{ psi}} \times \frac{101.325\text{ kPa}}{1\text{ atm}} = 2.2 \times 10^2\text{ kPa}\)
S8. During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars (1 bar = 0.987 atm). Pressure in torr and kPa?
* First convert millibars to bar: \(6.50\text{ mbar} = 0.00650\text{ bar}\)
* Convert to atmospheres: \(0.00650\text{ bar} \times \frac{0.987\text{ atm}}{1\text{ bar}} = 0.006416\text{ atm}\)
* In torr: \(0.006416\text{ atm} \times 760\text{ torr/atm} = 4.88\text{ torr}\)
* In kilopascals: \(0.006416\text{ atm} \times 101.325\text{ kPa/atm} = 0.650\text{ kPa}\)
S9. The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.
* Pressure on Venus in psi: \(88.8\text{ atm} \times 14.7\text{ psi/atm} = 1305\text{ psi}\)
* Pressure on Earth at sea level: \(14.7\text{ psi}\)
* Comparison: The atmospheric pressure on Venus is approximately \(88.8\) times greater than that of Earth.
S10. A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa. Pressure of this gas in atmospheres and torr:
* First convert MPa to Pa: \(14.82\text{ MPa} = 1.482 \times 10^7\text{ Pa}\)
* In atmospheres: \(1.482 \times 10^7\text{ Pa} \times \frac{1\text{ atm}}{101,325\text{ Pa}} = 146.3\text{ atm}\)
* In torr: \(146.3\text{ atm} \times 760\text{ torr/atm} = 1.112 \times 10^5\text{ torr}\)
S11. Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar.
a. Pressure in kPa: Using millibars: \(1013.9\text{ mbar} = 1.0139\text{ bar} \times \frac{100\text{ kPa}}{1\text{ bar}} = 101.39\text{ kPa}\)
b. The pressure near the seacoast in the northeastern United States is usually reported near 30.0 in. Hg. During a hurricane, the pressure may fall to near 28.0 in. Hg. Drop in pressure in torr.
* Pressure drop in inches of mercury: \(30.0\text{ in. Hg} - 28.0\text{ in. Hg} = 2.0\text{ in. Hg}\)
* Convert to torr: \(2.0\text{ in. Hg} \times \frac{25.4\text{ mm}}{1\text{ in.}} \times \frac{1\text{ torr}}{1\text{ mm Hg}} = 51\text{ torr}\)
S12. Why is it necessary to use a nonvolatile liquid in a barometer or manometer?
A volatile liquid would easily evaporate, creating vapor pressure in the sealed space above the liquid column. This unwanted gas pressure would push down on the liquid column, causing the instrument to read a lower measurement than the actual atmospheric or gas pressure.
S13. The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:
a. torr
For a closed-end manometer, \(P_{\text{gas}} = h\). Given \(h = 26.4\text{ cm} = 264\text{ mm}\), the pressure is \(264\text{ torr}\).
b. Pa
\(264\text{ torr} \times \frac{101,325\text{ Pa}}{760\text{ torr}} = 3.52 \times 10^4\text{ Pa}\)
c. bar
\(3.52 \times 10^4\text{ Pa} \times \frac{1\text{ bar}}{10^5\text{ Pa}} = 0.352\text{ bar}\)
S14. The pressure of a sample of gas is measured with an open-end manometer. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in. Hg, determine the pressure of the gas in:
a. torr
The mercury level is higher on the gas side, meaning \(P_{\text{gas}} < P_{\text{atm}}\).
\(P_{\text{atm}} = 29.92\text{ in. Hg} = 760\text{ mm Hg} = 760\text{ torr}\)
\(h = 6.00\text{ in. Hg} \times \frac{25.4\text{ mm}}{1\text{ in.}} = 152.4\text{ mm Hg} = 152.4\text{ torr}\)
\(P_{\text{gas}} = P_{\text{atm}} - h = 760\text{ torr} - 152.4\text{ torr} = 608\text{ torr}\)
b. Pa
\(608\text{ torr} \times \frac{101,325\text{ Pa}}{760\text{ torr}} = 8.11 \times 10^4\text{ Pa}\)
c. bar
\(8.11 \times 10^4\text{ Pa} \times \frac{1\text{ bar}}{10^5\text{ Pa}} = 0.811\text{ bar}\)
S15. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760.0 mm Hg, determine the pressure of the gas in:
a. mm Hg
The mercury level is lower on the gas side, meaning \(P_{\text{gas}} > P_{\text{atm}}\).
\(h = 13.7\text{ cm} = 137\text{ mm}\)
\(P_{\text{gas}} = P_{\text{atm}} + h = 760.0\text{ mm Hg} + 137\text{ mm Hg} = 897\text{ mm Hg}\)
b. atm
\(897\text{ mm Hg} \times \frac{1\text{ atm}}{760\text{ mm Hg}} = 1.18\text{ atm}\)
c. kPa
\(1.18\text{ atm} \times 101.325\text{ kPa/atm} = 120\text{ kPa}\)
S16. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760 mm Hg, determine the pressure of the gas in:
a. mm Hg
The mercury level is lower on the gas side, meaning \(P_{\text{gas}} > P_{\text{atm}}\).
\(h = 26.4\text{ cm} = 264\text{ mm}\)
\(P_{\text{gas}} = P_{\text{atm}} + h = 760\text{ mm Hg} + 264\text{ mm Hg} = 1024\text{ mm Hg}\)
b. atm
\(1024\text{ mm Hg} \times \frac{1\text{ atm}}{760\text{ mm Hg}} = 1.35\text{ atm}\)
c. kPa
\(1.35\text{ atm} \times 101.325\text{ kPa/atm} = 137\text{ kPa}\)
S17. Usese of a volatile liquid
* In a closed-end manometer, the volatile liquid evaporates into the vacuum space, generating an internal vapor pressure that exerts a downward force on the column, causing the measured gas pressure value to be erroneously low.
* In an open-end manometer, the volatile liquid evaporates into both the gas sample chamber and the open environment. The vaporized liquid escaping into the open side has negligible effect due to immediate diffusion, but evaporation inside the gas side adds to the total measured pressure, causing an erroneously high pressure calculation for the gas sample.
9.3: Relating Pressure, Volume, Amount, and Temperature - The Ideal Gas Law
S18. Effect of Solar Heating on Bicycle Tire Integrity
When a bicycle is left in the sun on a hot day, the thermal energy increases the temperature of the gas trapped inside the tire. According to the kinetic-molecular theory and Gay-Lussac's law, this temperature increase causes the gas molecules to move faster, striking the interior walls of the tire with greater force and frequency. This rapidly raises the internal pressure until it exceeds the structural limit of the tire material, resulting in a blowout.
S19. Volume Alteration of Scuba Bubbles During Ascent
As a bubble exhaled by a scuba diver rises toward the surface, the column of water above it decreases, which significantly drops the external hydrostatic pressure acting on the bubble. According to Boyle's law, the volume of a gas is inversely proportional to its pressure when temperature and amount remain constant. Therefore, as the ambient pressure decreases during the ascent, the volume of the intact gas bubbles must expand.
S20. Mathematical and Physical Interpretations of Inverse Proportionality
a. The term inversely proportional means that two interconnected variables change in opposite directions by an equivalent factor. For instance, if the volume of a gas sample is doubled, its pressure will be reduced by exactly half, provided all other systemic conditions remain constant.
b. The other things that must be held constant to observe Boyle's law are the absolute temperature (\(T\)) of the gas sample and the total amount of gas in moles (\(n\)).
S21. Mathematical and Physical Interpretations of Direct Proportionality
a. The term directly proportional means that two interconnected variables change in the exact same direction and by the identical factor. If the total number of molecules or moles of a gas is doubled, the volume occupied by that gas will also double.
b. The other things that must be held constant to satisfy this formulation of Avogadro's law are the absolute temperature (\(T\)) and the total external pressure (\(P\)) acting on the gas sample.
S22. Graphical Response of a Volume-Temperature Curve to Doubled Gas Quantity
Figure 8.12 plots volume as a function of temperature (Charles's law). According to the ideal gas law (\(V = \frac{nR}{P}T\)), the slope of this line is directly proportional to the number of moles (\(n\)). If the number of moles of gas in the sample used to determine the curve were doubled at a constant pressure, the slope of the linear plot would double, making the line twice as steep.
S23. Graphical Response of a Pressure-Volume Curve to Doubled Gas Quantity
Figure 8.13 plots pressure as a function of volume (Boyle's law), generating an inverse hyperbolic curve where \(P = \frac{nRT}{V}\). If the number of moles (\(n\)) is doubled while holding temperature constant, the value of the numerator doubles. Graphically, this shifts the entire hyperbolic curve outward away from the origin, doubling the pressure value measured at any given fixed volume.
S24. Necessary Thermodynamic Properties to Evaluate Mass from Empirical Gas Curves
To determine the absolute mass of the air sample from the pressure-volume data shown in Figure 8.13, we require the constant absolute temperature (\(T\)) at which the experiment was conducted, along with the average molar mass (\(M\)) of the air mixture to convert the calculated number of moles (\(n\)) into grams.
S25. Evaluation of Molar Volume from Graphic Reference Data
Using the underlying data coordinates of Figure 8.12 for an ideal gas at a temperature of \(150\text{ K}\) and a pressure of \(1\text{ atm}\), the volume occupied by \(1\text{ mol}\) of \(\ce{CH4}\) gas is determined via the ideal gas law:
\(V = \frac{nRT}{P} = \frac{1\text{ mol} \times 0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 150\text{ K}}{1\text{ atm}} = 12.3\text{ L}\)
S26. Graphic and Analytical Determination of Syringe Gas Pressure
a. Locating a volume of \(12.5\text{ mL}\) on the volume axis of the experimental syringe curve in Figure 8.13 and tracking horizontally to the corresponding pressure axis yields an empirical pressure of approximately \(2.4\text{ atm}\).
b. Using Boyle's law (\(P_1V_1 = P_2V_2\)) with standard initial reference parameters from the syringe system where \(P_1 = 1.0\text{ atm}\) at \(V_1 = 30.0\text{ mL}\):
\(P_2 = \frac{P_1V_1}{V_2} = \frac{1.0\text{ atm} \times 30.0\text{ mL}}{12.5\text{ mL}} = 2.40\text{ atm}\)
S27. Final Pressurization of an Incinerated Aerosol Container
This system behaves according to Gay-Lussac's law since volume and amount are rigid.
Convert initial and final temperatures to Kelvin:
\(T_1 = 23^\circ\text{C} + 273.15 = 296.15\text{ K}\)
\(T_2 = 475^\circ\text{C} + 273.15 = 748.15\text{ K}\)
Apply the pressure-temperature relationship:
\(P_2 = \frac{P_1 \times T_2}{T_1} = \frac{1344\text{ torr} \times 748.15\text{ K}}{296.15\text{ K}} = 3.40 \times 10^3\text{ torr}\)
S28. Identifying Initial Temperature Parameters for Carbon Monoxide Gas
Because the pressure of the system remains static at \(744\text{ torr}\) before and after the transition, this can be solved directly via Charles's law (\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)).
Convert the final temperature to Kelvin:
\(T_2 = 55^\circ\text{C} + 273.15 = 328.15\text{ K}\)
Isolate and solve for the initial absolute temperature \(T_1\):
\(T_1 = \frac{V_1 \times T_2}{V_2} = \frac{11.2\text{ L} \times 328.15\text{ K}}{13.3\text{ L}} = 276.3\text{ K}\)
Converting back to Celsius:
\(276.3 - 273.15 = 3.2^\circ\text{C}\)
S29. Isobaric Volume Expansion of Cryogenic Hydrogen Gas
The calculation relies on Charles's law because the pressure remains unaltered during thermal warming.
Convert temperatures to Kelvin:
\(T_1 = -196^\circ\text{C} + 273.15 = 77.15\text{ K}\)
\(T_2 = 100^\circ\text{C} + 273.15 = 373.15\text{ K}\)
Calculate the new expanded volume \(V_2\):
\(V_2 = \frac{V_1 \times T_2}{T_1} = \frac{2.50\text{ L} \times 373.15\text{ K}}{77.15\text{ K}} = 12.1\text{ L}\)
S30. Volumetric Adjustments to a Balloon via Incremental Lung Inflation
Avogadro's law dictates that gas volume is directly proportional to the molar amount of gas breaths when temperature and pressure are fixed. The initial state consists of \(3\text{ breaths}\) occupying \(1.7\text{ L}\). Adding \(5\text{ more breaths}\) brings the final total to \(3 + 5 = 8\text{ breaths}\).
\(V_2 = \frac{V_1 \times n_2}{n_1} = \frac{1.7\text{ L} \times 8\text{ breaths}}{3\text{ breaths}} = 4.5\text{ L}\)
9.4: Stoichiometry of Gaseous Substances, Mixtures, and Reactions
S31. Total Volume of a Standard Ground Level Weather Balloon
Convert the ambient ground temperature to Kelvin:
\(T = 25^\circ\text{C} + 273.15 = 298.15\text{ K}\)
Apply the ideal gas law to calculate the volume:
\(V = \frac{nRT}{P} = \frac{8.80\text{ mol} \times 0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 298.15\text{ K}}{0.992\text{ atm}} = 217\text{ L}\)
S32. Internal Pressure Build in an Inflated Automotive Airbag
First, find the moles of nitrogen gas using its molar mass (\(28.01\text{ g/mol}\)):
\(n = \frac{77.8\text{ g}}{28.01\text{ g/mol}} = 2.778\text{ mol}\)
Convert the working temperature to Kelvin:
\(T = 25^\circ\text{C} + 273.15 = 298.15\text{ K}\)
Calculate pressure inside the bag in atmospheres:
\(P = \frac{nRT}{V} = \frac{2.778\text{ mol} \times 0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 298.15\text{ K}}{66.8\text{ L}} = 1.0177\text{ atm}\)
Convert the atmospheric value to kilopascals:
\(1.0177\text{ atm} \times 101.325\text{ kPa/atm} = 103\text{ kPa}\)
S33. Molar and Mass Quantities of Boron Trifluoride in a Sealed Bulb
Calculate the chemical amount in moles using the ideal gas law:
\(n = \frac{PV}{RT} = \frac{1.220\text{ atm} \times 4.3410\text{ L}}{0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 788.0\text{ K}} = 0.08191\text{ mol}\)
Find the total mass utilizing the molar mass of \(\ce{BF3}\) (\(67.81\text{ g/mol}\)):
\(m = 0.08191\text{ mol} \times 67.81\text{ g/mol} = 5.554\text{ g}\)
S34. Sublimation Temperature Requirements for Iodine Vapor
Calculate the number of moles of \(\ce{I2}\) using its molecular molar mass (\(253.81\text{ g/mol}\)):
\(n = \frac{0.292\text{ g}}{253.81\text{ g/mol}} = 0.001150\text{ mol}\)
Convert the bulb volume to liters:
\(V = 73.3\text{ mL} = 0.0733\text{ L}\)
Isolate absolute temperature using the ideal gas law equation:
\(T = \frac{PV}{nR} = \frac{0.41284\text{ L}\cdot\text{atm}}{0.001150\text{ mol} \times 0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K}} = 359\text{ K}\)
Convert the thermodynamic value back to Celsius units:
\(359 - 273.15 = 86^\circ\text{C}\)
S35. Comprehensive Mass Extractions across Varied Gaseous Quantities
a. Convert parameters to standard analytical units: \(V = 0.100\text{ L}\), \(P = \frac{307}{760} = 0.4039\text{ atm}\), \(T = 26 + 273.15 = 299.15\text{ K}\).
\(n = \frac{0.4039 \times 0.100}{0.08206 \times 299.15} = 0.001645\text{ mol}\)
Mass of \(\ce{CO2}\) (\(44.01\text{ g/mol}\)): \(0.001645\text{ mol} \times 44.01\text{ g/mol} = 0.0724\text{ g}\)
b. Convert parameters to standard analytical units: \(V = 8.75\text{ L}\), \(P = \frac{378.3}{101.325} = 3.7335\text{ atm}\), \(T = 483\text{ K}\).
\(n = \frac{3.7335 \times 8.75}{0.08206 \times 483} = 0.08242\text{ mol}\)
Mass of \(\ce{C2H4}\) (\(28.05\text{ g/mol}\)): \(0.08242\text{ mol} \times 28.05\text{ g/mol} = 2.31\text{ g}\)
c. Convert parameters to standard analytical units: \(V = 0.221\text{ L}\), \(P = \frac{0.23}{760} = 0.0003026\text{ atm}\), \(T = -54 + 273.15 = 219.15\text{ K}\).
\(n = \frac{0.0003026 \times 0.221}{0.08206 \times 219.15} = 3.719 \times 10^{-6}\text{ mol}\)
Mass of \(\ce{Ar}\) (\(39.95\text{ g/mol}\)): \(3.719 \times 10^{-6}\text{ mol} \times 39.95\text{ g/mol} = 1.49 \times 10^{-4}\text{ g}\)
S36. Volumetric Expansion of a High-Altitude Hydrogen Balloon
First, determine the stable molar quantity (\(n\)) from the initial ground-level constraints:
\(P_1 = \frac{745\text{ torr}}{760\text{ torr/atm}} = 0.9803\text{ atm}\)
\(T_1 = 21^\circ\text{C} + 273.15 = 294.15\text{ K}\)
\(V_1 = 1.41 \times 10^4\text{ L}\)
Calculate the constant chemical amount:
\(n = \frac{P_1V_1}{RT_1} = \frac{0.9803 \times 1.41 \times 10^4}{0.08206 \times 294.15} = 572.82\text{ mol}\)
Next, compute the secondary volume (\(V_2\)) using the atmospheric conditions at an altitude of \(20\text{ km}\):
\(P_2 = \frac{63.1\text{ torr}}{760\text{ torr/atm}} = 0.08303\text{ atm}\)
\(T_2 = -48^\circ\text{C} + 273.15 = 225.15\text{ K}\)
Calculate the high-altitude volume:
\(V_2 = \frac{nRT_2}{P_2} = \frac{572.82 \times 0.08206 \times 225.15}{0.08303} = 1.27 \times 10^5\text{ L}\)
S37. Volumetric Equivalence of Stored Medical Oxygen at Normal Human Body Conditions
Determine the absolute moles of oxygen gas enclosed within the storage container:
\(n = \frac{PV}{RT} = \frac{151\text{ atm} \times 35.4\text{ L}}{0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 298.15\text{ K}} = 218.41\text{ mol}\)
Recalculate the corresponding gas volume under explicit physiological conditions (\(1\text{ atm}\) and \(37^\circ\text{C}\) or \(310.15\text{ K}\)):
\(V = \frac{nRT}{P} = \frac{218.41\text{ mol} \times 0.08206 \times 310.15\text{ K}}{1\text{ atm}} = 5.56 \times 10^3\text{ L}\)
S38. Verification of Volumetric Capacity in a High-Pressure Scuba Tank
Find the maximum moles of gas that the cylinder can hold when filled completely to capacity at its rated limit:
\(n = \frac{PV}{RT} = \frac{220\text{ bar} \times \left(\frac{1\text{ atm}}{1.01325\text{ bar}}\right) \times 18\text{ L}}{0.08206 \times 293.15\text{ K}} = 162.36\text{ mol}\)
Determine the volume that this specific molar quantity of air occupies at the diver's deployment depth conditions (\(2.37\text{ atm}\) and \(20^\circ\text{C}\)):
\(V = \frac{nRT}{P} = \frac{162.36\text{ mol} \times 0.08206 \times 293.15\text{ K}}{2.37\text{ atm}} = 1.65 \times 10^3\text{ L}\)
Because the tank can only supply \(1650\text{ L}\) of air under these specific environmental constraints, it was not filled to its full technical capacity, which requires a yield of \(1860\text{ L}\).
S39. Mass Retention of Liquid Butane in a Decompressed Cylinder
Calculate the initial moles of butane gas within the tank prior to opening:
\(n_{\text{initial}} = \frac{11.34 \times 10^3\text{ g}}{58.12\text{ g/mol}} = 195.11\text{ mol}\)
When opened, gas escapes until the internal cylinder pressure drops to equal the ambient atmosphere (\(0.983\text{ atm}\)). Calculate the moles remaining in the \(20.0\text{ L}\) space at \(300.15\text{ K}\):
\(n_{\text{remaining}} = \frac{PV}{RT} = \frac{0.983\text{ atm} \times 20.0\text{ L}}{0.08206 \times 300.15\text{ K}} = 0.7982\text{ mol}\)
Convert these residual moles back into standard mass units:
\(m_{\text{remaining}} = 0.7982\text{ mol} \times 58.12\text{ g/mol} = 46.4\text{ g}\)
S40. Molar Ingestion Rate of Pure Oxygen During Human Rest Cycles
Convert the atmospheric air pressure to atmospheres:
\(P = \frac{100\text{ kPa}}{101.325\text{ kPa/atm}} = 0.9869\text{ atm}\)
Convert the ambient room temperature into Kelvin units:
\(T = 25^\circ\text{C} + 273.15 = 298.15\text{ K}\)
Calculate the absolute moles of oxygen gas processed over a standard \(1.0\text{ h}\) cycle using the resting consumption volume (\(14\text{ L}\)):
\(n = \frac{PV}{RT} = \frac{0.9869\text{ atm} \times 14\text{ L}}{0.08206 \times 298.15\text{ K}} = 0.57\text{ mol}\)
S41. Labeled Graphical Geometries of Ideal Gas Equations
a. A plot of \(P\) versus \(V\) displays a downward curving, inverse hyperbolic trend line.
b. A plot of \(V\) versus \(T\) displays a straight, linear trend line with a positive slope originating from absolute zero.
c. A plot of \(P\) versus \(T\) displays a straight, linear trend line with a positive slope originating from absolute zero.
d. A plot of \(\frac{1}{P}\) versus \(V\) displays a straight, linear trend line with a positive slope passing through the coordinates of the origin.
S42. Avogadro-Based Atom Density Comparisons Between Gaseous Methane and Hydrogen
According to Avogadro's law, equal volumes of gases at identical temperature and pressure contain an identical number of total gas molecules. Therefore, \(1\text{ L}\) of methane gas (\(\ce{CH4}\)) contains the exact same number of molecules as \(1\text{ L}\) of hydrogen gas (\(\ce{H2}\)). However, because each individual molecule of methane contains \(4\) hydrogen atoms, whereas each molecule of hydrogen gas contains only \(2\) hydrogen atoms, the liter of methane contains twice as many total atoms of hydrogen.
S43. Comparative Volumetric Footprints of Chlorofluorocarbon Substitutes at STP
At standard temperature and pressure conditions, \(1\text{ mol}\) of any ideal gas occupies a fixed volume of \(22.414\text{ L}\).
a. Find the moles of \(\ce{CCl2F2}\) using its specific molar mass (\(120.91\text{ g/mol}\)):
\(n = \frac{10.0\text{ g}}{120.91\text{ g/mol}} = 0.08271\text{ mol}\)
\(V = 0.08271\text{ mol} \times 22.414\text{ L/mol} = 1.85\text{ L}\)
b. Find the moles of \(\ce{CH3CH2F}\) using its specific molar mass (\(48.06\text{ g/mol}\)):
\(n = \frac{10.0\text{ g}}{48.06\text{ g/mol}} = 0.20807\text{ mol}\)
\(V = 0.20807\text{ mol} \times 22.414\text{ L/mol} = 4.66\text{ L}\)
S44. Manometric Pressure Rise from Radioactive Alpha Particle Decay
Because each individual alpha particle captures electrons to transform into a neutral helium atom, the total number of gaseous helium atoms generated equals the number of decayed particles (\(1.16 \times 10^{18}\)).
Convert particles to moles:
\(n = \frac{1.16 \times 10^{18}}{6.022 \times 10^{23}\text{ atoms/mol}} = 1.926 \times 10^{-6}\text{ mol}\)
Convert volume and temperature parameters to standard units: \(V = 0.125\text{ L}\), \(T = 298.15\text{ K}\).
Calculate the pressure in atmospheres:
\(P = \frac{nRT}{V} = \frac{1.926 \times 10^{-6} \times 0.08206 \times 298.15}{0.125} = 3.771 \times 10^{-4}\text{ atm}\)
Convert the result to pascals:
\(3.771 \times 10^{-4}\text{ atm} \times 101,325\text{ Pa/atm} = 38.2\text{ Pa}\)
S45. High-Altitude Peak Pressure Computations for an Expanding Weather Balloon
Because the total moles of gas contained within the sealed balloon remain constant, use the combined gas law (\(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)).
Convert operational temperatures to Kelvin:
\(T_1 = 21^\circ\text{C} + 273.15 = 294.15\text{ K}\)
\(T_2 = 5.24^\circ\text{C} + 273.15 = 278.39\text{ K}\)
Isolate and solve for the final peak pressure (\(P_2\)):
\(P_2 = \frac{P_1 \times V_1 \times T_2}{V_2 \times T_1} = \frac{0.981\text{ atm} \times 100.21\text{ L} \times 278.39\text{ K}}{144.53\text{ L} \times 294.15\text{ K}} = 0.644\text{ atm}\)
S46. Constant Volume Pressure Response to Thermal Doubling
According to Gay-Lussac's law, when the total volume of a fixed gas sample is held constant, the absolute pressure is directly proportional to its thermodynamic temperature in Kelvin. Therefore, if the absolute temperature is doubled, the pressure of the gas will double.
S47. Isothermal Pressure Response to Volumetric Tripling
According to Boyle's law, when the temperature of a fixed gas sample is held constant, the internal pressure is inversely proportional to the volume it occupies. Therefore, if the volume of the gas is tripled, the final pressure will drop to exactly one-third of its initial value.
S48. Ideal Density Evaluations of Gaseous Nitrous Oxide
Convert the target pressure into atmospheres:
\(P = \frac{113.0\text{ kPa}}{101.325\text{ kPa/atm}} = 1.1152\text{ atm}\)
Apply the gas density equation using the molar mass of \(\ce{N2O}\) (\(44.01\text{ g/mol}\)):
\(d = \frac{PM}{RT} = \frac{1.1152\text{ atm} \times 44.01\text{ g/mol}}{0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 325\text{ K}} = 1.84\text{ g/L}\)
S49. Ideal Density Evaluations of Gaseous Freon-12
Convert the temperature to Kelvin:
\(T = 30.0^\circ\text{C} + 273.15 = 303.15\text{ K}\)
Apply the gas density equation using the molecular molar mass of Freon-12, \(\ce{CF2Cl2}\) (\(120.91\text{ g/mol}\)):
\(d = \frac{PM}{RT} = \frac{0.954\text{ atm} \times 120.91\text{ g/mol}}{0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 303.15\text{ K}} = 4.64\text{ g/L}\)
S50. Mass Density Discrepancies Between Dry and Water-Vapor Saturated Air Systems
Dry air is denser than air saturated with water vapor at identical temperature and pressure. Saturated air is a mixture where a portion of air molecules is replaced by water vapor (\(\ce{H2O}\)) molecules. Because the molar mass of water vapor (\(18.02\text{ g/mol}\)) is significantly lower than the average molar mass of dry air (\(28.97\text{ g/mol}\)), Avogadro's law implies that replacing heavier nitrogen and oxygen molecules with lighter water molecules reduces the total mass per unit volume, making humid air less dense.
S51. Total Capacity Mass of an Oxygen Ventilation Cylinder
Convert the temperature to Kelvin:
\(T = 28.0^\circ\text{C} + 273.15 = 301.15\text{ K}\)
Determine the moles of oxygen gas using the ideal gas law:
\(n = \frac{PV}{RT} = \frac{10.0\text{ atm} \times 3.00\text{ L}}{0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 301.15\text{ K}} = 1.214\text{ mol}\)
Calculate the mass using the molecular mass of \(\ce{O2}\) (\(32.00\text{ g/mol}\)):
\(m = 1.214\text{ mol} \times 32.00\text{ g/mol} = 38.8\text{ g}\)
S52. Molar Mass Analytical Extrapolations of an Isolated Gas Sample
Convert the pressure and temperature values to standard units:
\(P = \frac{307\text{ torr}}{760\text{ torr/atm}} = 0.40395\text{ atm}\)
\(T = 26^\circ\text{C} + 273.15 = 299.15\text{ K}\)
Determine the moles of gas present in the \(0.100\text{ L}\) system:
\(n = \frac{PV}{RT} = \frac{0.40395\text{ atm} \times 0.100\text{ L}}{0.08206 \times 299.15\text{ K}} = 0.0016454\text{ mol}\)
Calculate the molar mass (\(M\)):
\(M = \frac{\text{mass}}{n} = \frac{0.0494\text{ g}}{0.0016454\text{ mol}} = 30.0\text{ g/mol}\)
S53. Molar Mass Analytical Extrapolations of a High-Temperature Gas Sample
Convert the pressure and temperature parameters to standard units:
\(P = \frac{777\text{ torr}}{760\text{ torr/atm}} = 1.0224\text{ atm}\)
\(T = 126^\circ\text{C} + 273.15 = 399.15\text{ K}\)
\(V = 125\text{ mL} = 0.125\text{ L}\)
Determine the moles of gas present:
\(n = \frac{PV}{RT} = \frac{1.0224\text{ atm} \times 0.125\text{ L}}{0.08206 \times 399.15\text{ K}} = 0.0039017\text{ mol}\)
Calculate the molar mass (\(M\)):
\(M = \frac{\text{mass}}{n} = \frac{0.281\text{ g}}{0.0039017\text{ mol}} = 72.0\text{ g/mol}\)
S54. Experimental Strategies to Distinguish Propene Polymeric Formulas
To prove that the formula is \(\ce{C3H6}\) and not \(\ce{CH2}\), vaporize a known mass of the pure propene sample inside a sealed container of fixed volume at a specific temperature, then measure the resulting pressure. Use the ideal gas law to calculate the experimental molar mass (\(M = \frac{mRT}{PV}\)). If the formula is propene (\(\ce{C3H6}\)), the empirical molar mass will be approximately \(42.08\text{ g/mol}\), whereas a monomeric \(\ce{CH2}\) structure would yield a mass of only \(14.03\text{ g/mol}\).
S55. Characterization of a Gaseous Phosphorus Fluoride Compound
At standard temperature and pressure (STP), \(1\text{ mol}\) of gas occupies exactly \(22.414\text{ L}\). Calculate the molar mass from the given density:
\(M = \text{density} \times \text{molar volume} = 3.93\text{ g/L} \times 22.414\text{ L/mol} = 88.1\text{ g/mol}\)
The empirical formula unit for phosphorus fluoride must contain a single phosphorus atom (\(30.97\text{ g/mol}\)) combined with fluorine atoms (\(19.00\text{ g/mol}\)):
Mass of fluorine = \(88.1 - 30.97 = 57.13\text{ g}\)
Number of fluorine atoms = \(\frac{57.13\text{ g}}{19.00\text{ g/mol}} \approx 3\)
The molecular formula is determined to be phosphorus trifluoride, \(\ce{PF3}\).
S56. Stepwise Deductions and Formula Proofs for an Unknown Organic Gas
a. Step 1: Use the mass percentages to find the empirical formula of the compound. Step 2: Convert the given experimental pressure, volume, and temperature parameters into standard ideal gas units. Step 3: Calculate the total moles of gas using the ideal gas law. Step 4: Divide the sample mass by the calculated moles to find the molar mass. Step 5: Compare the true molar mass to the empirical formula mass to find the molecular multiplier.
b. Find the empirical formula from a \(100\text{ g}\) reference base:
Moles of \(\ce{C}\) = \(39 / 12.01 = 3.247\text{ mol}\)
Moles of \(\ce{N}\) = \(45 / 14.01 = 3.212\text{ mol}\)
Moles of \(\ce{H}\) = \(16 / 1.008 = 15.873\text{ mol}\)
Dividing by the smallest value (\(3.212\)) yields a ratio of \(\ce{CH5N}\), giving an empirical mass of \(31.06\text{ g/mol}\).
Convert gas properties: \(P = \frac{99.5}{101.325} = 0.9820\text{ atm}\), \(V = 0.125\text{ L}\), \(T = 22 + 273.15 = 295.15\text{ K}\).
\(n = \frac{PV}{RT} = \frac{0.9820 \times 0.125}{0.08206 \times 295.15} = 0.005068\text{ mol}\)
\(M = \frac{0.157\text{ g}}{0.005068\text{ mol}} = 31.0\text{ g/mol}\)
Because the calculated molecular mass matches the empirical mass, the molecular formula is \(\ce{CH5N}\).
S57. Total Boundary Pressure of a Multi-Component Calibration Mixture
Find the individual molar counts for each component gas:
* Moles of \(\ce{CO2}\): \(350\text{ g} / 44.01\text{ g/mol} = 7.953\text{ mol}\)
* Moles of \(\ce{O2}\): \(805\text{ g} / 32.00\text{ g/mol} = 25.156\text{ mol}\)
* Moles of \(\ce{N2}\): \(4880\text{ g} / 28.01\text{ g/mol} = 174.223\text{ mol}\)
Total moles (\(n_{\text{total}}\)) = \(7.953 + 25.156 + 174.223 = 207.332\text{ mol}\)
Convert the systemic calibration temperature to Kelvin:
\(T = 25^\circ\text{C} + 273.15 = 298.15\text{ K}\)
Calculate the total pressure in atmospheres inside the \(36.0\text{ L}\) cylinder:
\(P_{\text{total}} = \frac{n_{\text{total}}RT}{V} = \frac{207.332 \times 0.08206 \times 298.15}{36.0\text{ L}} = 141\text{ atm}\)
S58. Fractional Partial Pressures within a High-Pressure Diagnostic Mix
According to Dalton's law, the partial pressure of an individual gas component in a mixture is equal to the total pressure multiplied by its percentage contribution:
* \(P_{\ce{CO2}} = 146\text{ atm} \times 0.050 = 7.3\text{ atm}\)
* \(P_{\ce{O2}} = 146\text{ atm} \times 0.120 = 17.5\text{ atm}\)
* \(P_{\ce{N2}} = 146\text{ atm} \times (1.00 - 0.050 - 0.120) = 146\text{ atm} \times 0.830 = 121.2\text{ atm}\)
S59. Partial Pressure Profiles of Gaseous Petroleum Distillates
The partial pressure exerted by each individual gas components is directly calculated from its specified percentage of the total pressure (\(307.2\text{ kPa}\)):
* \(P_{\ce{CH4}} = 307.2\text{ kPa} \times 0.900 = 276.5\text{ kPa}\)
* \(P_{\ce{C2H6}} = 307.2\text{ kPa} \times 0.089 = 27.3\text{ kPa}\)
* \(P_{\ce{C3H8}} = 307.2\text{ kPa} \times 0.011 = 3.4\text{ kPa}\)
S60. Total Physical Volume of a Mixed Ideal Gas Sample at STP
First, evaluate the total molar quantity within the container:
* Moles of \(\ce{H2}\): \(0.200\text{ g} / 2.016\text{ g/mol} = 0.09921\text{ mol}\)
* Moles of \(\ce{N2}\): \(1.00\text{ g} / 28.01\text{ g/mol} = 0.03570\text{ mol}\)
* Moles of \(\ce{Ar}\): \(0.820\text{ g} / 39.95\text{ g/mol} = 0.02053\text{ mol}\)
Total moles (\(n_{\text{total}}\)) = \(0.09921 + 0.03570 + 0.02053 = 0.15544\text{ mol}\)
Using the standard molar volume at STP (\(22.414\text{ L/mol}\)):
\(V = 0.15544\text{ mol} \times 22.414\text{ L/mol} = 3.48\text{ L}\)
S61. Explosive Danger Thresholds of Hydrogen-Oxygen Content Mixtures
Determine the partial pressure exerted by hydrogen gas using Dalton's law:
\(P_{\ce{H2}} = 34.5\text{ atm} - 33.2\text{ atm} = 1.3\text{ atm}\)
Calculate the percentage of oxygen gas present in the total final system:
\(\text{Percentage of }\ce{O2} = \left(\frac{33.2\text{ atm}}{34.5\text{ atm}}\right) \times 100\% = 96.2\%\)
Because the percentage of oxygen in this gas mixture (\(96.2\%\)) far exceeds the maximum non-explosive safety limit (which requires oxygen to remain below \(3.0\%\)), this mixture is highly explosive.
S62. Partial Vapor Pressure Measurements of Trace Elemental Mercury Contamination
First, find the concentration of gaseous mercury in grams per liter:
\(2 \times 10^{-6}\text{ mg/L} = 2 \times 10^{-9}\text{ g/L}\)
Convert this concentration into a molarity value using the atomic mass of \(\ce{Hg}\) (\(200.59\text{ g/mol}\)):
\(\text{Moles per liter} = \frac{2 \times 10^{-9}\text{ g/L}}{200.59\text{ g/mol}} = 9.97 \times 10^{-12}\text{ mol/L}\)
Using a standard volume baseline of \(1\text{ L}\) at a temperature of \(26^\circ\text{C}\) (\(299.15\text{ K}\)), calculate the partial pressure:
\(P = \frac{nRT}{V} = \left(9.97 \times 10^{-12}\text{ mol}\right) \times 0.08206 \times 299.15\text{ K} = 2.45 \times 10^{-10}\text{ atm}\)
Convert this value to torr units:
\(2.45 \times 10^{-10}\text{ atm} \times 760\text{ torr/atm} = 1.9 \times 10^{-7}\text{ torr}\)
S63. Vapor Correction Adjustments for Carbon Monoxide Collected Over Water
According to Dalton's law of partial pressures, when a gas is collected over water, the measured total pressure equals the pressure of the target dry gas plus the vapor pressure of the water at that temperature. Looking up the standard vapor pressure of water at \(18^\circ\text{C}\) yields a correction value of \(15.5\text{ torr}\).
\(P_{\ce{CO}} = P_{\text{total}} - P_{\ce{H2O}} = 756\text{ torr} - 15.5\text{ torr} = 741\text{ torr}\)
S64. Molar Mass Determination of a Water-Displaced Gas Synthesis
Subtract the vapor pressure of water at \(27^\circ\text{C}\) (\(26.7\text{ torr}\)) from the total experimental pressure to find the pressure of the dry gas:
\(P_{\text{gas}} = 753\text{ torr} - 26.7\text{ torr} = 726.3\text{ torr} = \frac{726.3}{760} = 0.95566\text{ atm}\)
Convert the volume and temperature parameters to standard values:
\(V = 265\text{ mL} = 0.265\text{ L}\)
\(T = 27^\circ\text{C} + 273.15 = 300.15\text{ K}\)
Determine the moles of dry gas collected:
\(n = \frac{PV}{RT} = \frac{0.95566\text{ atm} \times 0.265\text{ L}}{0.08206 \times 300.15\text{ K}} = 0.010287\text{ mol}\)
Calculate the molar mass (\(M\)):
\(M = \frac{\text{mass}}{n} = \frac{0.472\text{ g}}{0.010287\text{ mol}} = 45.9\text{ g/mol}\)
S65. Stepwise Synthesis Scaling for the Thermal Decomposition of Mercuric Oxide
a. Step 1: Write down the balanced chemical decomposition equation. Step 2: Convert the starting mass of solid mercuric oxide (\(\ce{HgO}\)) into moles. Step 3: Apply the stoichiometric molar ratio from the balanced equation to find the moles of oxygen gas (\(\ce{O2}\)) produced. Step 4: Convert the experimental reaction temperature into Kelvin. Step 5: Calculate the final volume of oxygen gas using the ideal gas law with the given pressure.
b. The reaction stoichiometry is given as: \(\ce{2HgO(s) -> 2Hg(l) + O2(g)}\).
Convert the reactant mass to moles using the formula mass of \(\ce{HgO}\) (\(216.59\text{ g/mol}\)):
\(5.36\text{ g HgO} \times \frac{1\text{ mol}}{216.59\text{ g}} = 0.024747\text{ mol}\)
Find the moles of \(\ce{O2}\) produced:
\(0.024747\text{ mol HgO} \times \frac{1\text{ mol O}_2}{2\text{ mol HgO}} = 0.012374\text{ mol O}_2\)
Convert the temperature to Kelvin: \(T = 23 + 273.15 = 296.15\text{ K}\).
Calculate the volume of oxygen gas produced at \(0.975\text{ atm}\):
\(V = \frac{nRT}{P} = \frac{0.012374\text{ mol} \times 0.08206 \times 296.15\text{ K}}{0.975\text{ atm}} = 0.309\text{ L}\)
S66. Stepwise Synthesis Scaling for the Steam Combustion of Iron
a. Step 1: Convert the initial mass of the water reactant into moles. Step 2: Use the stoichiometric coefficients from the balanced chemical equation to find the moles of hydrogen gas (\(\ce{H2}\)) generated. Step 3: Convert the operational pressure and temperature parameters into standard units. Step 4: Compute the final gas volume using the ideal gas law.
b. The balanced chemical reaction equation is: \(\ce{4H2O(g) + 3Fe(s) -> Fe3O4(s) + 4H2(g)}\).
Convert the mass of \(\ce{H2O}\) to moles using its molar mass (\(18.015\text{ g/mol}\)):
\(15.0\text{ g H}_2\text{O} \times \frac{1\text{ mol}}{18.015\text{ g}} = 0.83264\text{ mol}\)
The molar ratio between \(\ce{H2O}\) and \(\ce{H2}\) is \(4:4\) (or \(1:1\)), so the reaction generates \(0.83264\text{ mol of H2}\).
Convert the pressure and temperature parameters to standard units:
\(P = \frac{745\text{ torr}}{760\text{ torr/atm}} = 0.98026\text{ atm}\)
\(T = 20^\circ\text{C} + 273.15 = 293.15\text{ K}\)
Calculate the final volume of hydrogen gas generated:
\(V = \frac{nRT}{P} = \frac{0.83264\text{ mol} \times 0.08206 \times 293.15\text{ K}}{0.98026\text{ atm}} = 20.4\text{ L}\)
S67. Balanced Stoichiometric Recycling Equations for Chlorofluorocarbon Management
To balance the chemical recycling reaction transforming \(\ce{CCl2F2}\) into \(\ce{CH2F2}\) and \(\ce{HCl}\):
\[\ce{CCl2F2(g) + 2H2(g) -> CH2F2(g) + 2HCl(g)}\]
S68. Volumetric Yields of Nitrogen Gas from the Decomposition of Sodium Azide
The balanced chemical equation for airbag inflation is:
\[\ce{2NaN3(s) -> 2Na(s) + 3N2(g)}\]
Convert the mass of sodium azide to moles using its molar mass (\(65.01\text{ g/mol}\)):
\(125\text{ g NaN}_3 \times \frac{1\text{ mol}}{65.01\text{ g}} = 1.9228\text{ mol}\)
Determine the moles of nitrogen gas produced based on the stoichiometric ratio:
\(1.9228\text{ mol NaN}_3 \times \frac{3\text{ mol N}_2}{2\text{ mol NaN}_3} = 2.8842\text{ mol N}_2\)
Convert the environment's pressure and temperature parameters to standard units:
\(P = \frac{756\text{ torr}}{760\text{ torr/atm}} = 0.99474\text{ atm}\)
\(T = 27^\circ\text{C} + 273.15 = 300.15\text{ K}\)
Calculate the final volume of nitrogen gas generated:
\(V = \frac{nRT}{P} = \frac{2.8842\text{ mol} \times 0.08206 \times 300.15\text{ K}}{0.99474\text{ atm}} = 71.4\text{ L}\)
S69. Stepwise Synthesis Scaling for the Calcination Decomposition of Lime
a. Step 1: Write down the balanced chemical equation for the thermal decomposition of calcium carbonate (\(\ce{CaCO3}\)). Step 2: Convert the metric ton mass of the calcium carbonate reactant into grams, and then into moles. Step 3: Use the balanced reaction ratio to find the moles of carbon dioxide (\(\ce{CO2}\)) produced. Step 4: Calculate the volume of the carbon dioxide gas product using the ideal gas law with the given temperature and pressure.
b. The balanced decomposition equation is: \(\ce{CaCO3(s) -> CaO(s) + CO2(g)}\).
Convert mass to moles using the molar mass of \(\ce{CaCO3}\) (\(100.09\text{ g/mol}\)):
\(1.000 \times 10^6\text{ g CaCO}_3 \times \frac{1\text{ mol}}{100.09\text{ g}} = 9991\text{ mol}\)
Because the stoichiometric ratio is \(1:1\), the reaction produces \(9991\text{ mol of CO2}\).
Calculate the final volume of carbon dioxide gas produced at \(875\text{ K}\) and \(0.966\text{ atm}\):
\(V = \frac{nRT}{P} = \frac{9991\text{ mol} \times 0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 875\text{ K}}{0.966\text{ atm}} = 7.42 \times 10^5\text{ L}\)
S70. Stepwise Synthesis Scaling for the Aqueous Hydrolysis of Calcium Carbide
a. Step 1: Write down the balanced chemical equation for the reaction of solid calcium carbide (\(\ce{CaC2}\)) with liquid water to produce acetylene gas (\(\ce{C2H2}\)) and calcium hydroxide. Step 2: Convert the starting mass of the calcium carbide reactant into moles. Step 3: Use the stoichiometric coefficients to determine the moles of acetylene gas generated. Step 4: Convert the system temperature to Kelvin. Step 5: Use the ideal gas law to calculate the final volume of acetylene gas produced.
b. The balanced chemical equation is: \(\ce{CaC2(s) + 2H2O(l) -> C2H2(g) + Ca(OH)2(aq)}\).
Convert the mass of \(\ce{CaC2}\) to moles using its molar mass (\(64.10\text{ g/mol}\)):
\(15.48\text{ g CaC}_2 \times \frac{1\text{ mol}}{64.10\text{ g}} = 0.2415\text{ mol}\)
Because the reaction ratio is \(1:1\), it yields \(0.2415\text{ mol of C2H2}\).
Convert the reaction temperature to Kelvin: \(T = 12.2^\circ\text{C} + 273.15 = 285.35\text{ K}\).
Calculate the volume of acetylene gas produced at \(1.005\text{ atm}\):
\(V = \frac{nRT}{P} = \frac{0.2415\text{ mol} \times 0.08206 \times 285.35\text{ K}}{1.005\text{ atm}} = 5.63\text{ L}\)
S71. Volumetric Requirements for the Stoichiometric Combustion of Ethane Gas
The balanced chemical equation for the combustion of ethane gas is:
\[\ce{2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)}\]
According to Avogadro's law, when different gases are measured under identical temperature and pressure conditions, their reaction volumes are directly proportional to their stoichiometric coefficients.
Calculate the volume of oxygen gas required to completely combust \(12.00\text{ L}\) of ethane:
\(V_{\ce{O2}} = 12.00\text{ L C}_2\text{H}_6 \times \frac{7\text{ L O}_2}{2\text{ L C}_2\text{H}_6} = 42.00\text{ L}\)
9.4: Stoichiometry of Gaseous Substances, Mixtures, and Reactions
S72. Volumetric Stoichiometry of Nitric Oxide Oxidation
According to Avogadro's law, gas volumes at the same temperature and pressure are directly proportional to their stoichiometric coefficients. The balanced chemical equation is:
\[\ce{2NO(g) + O2(g) -> 2NO2(g)}\]
The required volume of oxygen is:
\[V_{\ce{O2}} = 8.0\text{ L NO} \times \frac{1\text{ L O}_2}{2\text{ L NO}} = 4.0\text{ L}\]
The produced volume of nitrogen dioxide is:
\[V_{\ce{NO2}} = 8.0\text{ L NO} \times \frac{2\text{ L NO}_2}{2\text{ L NO}} = 8.0\text{ L}\]
S73. High-Temperature Volumetric and Partial Pressure Analysis of Ethane Combustion
The balanced chemical equation for the combustion of ethane gas is:
\[\ce{2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)}\]
a. First, determine the number of moles of ethane reactant initially present at STP conditions:
\[n_{\ce{C2H6}} = \frac{1.00\text{ L}}{22.414\text{ L/mol}} = 0.044615\text{ mol}\]
Using the reaction ratios, calculate the moles of gaseous products generated:
\[n_{\ce{CO2}} = 0.044615\text{ mol} \times \frac{4}{2} = 0.08923\text{ mol}\]
\[n_{\ce{H2O}} = 0.044615\text{ mol} \times \frac{6}{2} = 0.13385\text{ mol}\]
Total gaseous product moles:
\[n_{\text{total}} = 0.08923 + 0.13385 = 0.22308\text{ mol}\]
Convert the reaction environment temperature to Kelvin:
\[T = 600^\circ\text{C} + 273.15 = 873.15\text{ K}\]
Calculate the combined volume of the products at $0.888\text{ atm}$:
\[V = \frac{n_{\text{total}}RT}{P} = \frac{0.22308\text{ mol} \times 0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 873.15\text{ K}}{0.888\text{ atm}} = 18.0\text{ L}\]
b. The mole fraction of water vapor in the total product gas stream is:
\[X_{\ce{H2O}} = \frac{6}{4 + 6} = 0.600\]
Apply Dalton's law to find the partial pressure of water vapor:
\[P_{\ce{H2O}} = X_{\ce{H2O}} \times P_{\text{total}} = 0.600 \times 0.888\text{ atm} = 0.533\text{ atm}\]
S74. Volumetric Ratios for Industrial Methanol Gas Synthesis
The balanced chemical reaction is:
\[\ce{CO(g) + 2H2(g) -> CH3OH(g)}\]
The total stoichiometric volume of the gas phase reactants represents $1 + 2 = 3\text{ volumes}$, while the product gas represents $1\text{ volume}$. Therefore, the direct ratio of the total volume of reactants to products is $3:1$.
\subsubsection*{S75. Gas Volumetric Yield from Barium Peroxide Thermal Decomposition}
The balanced decomposition equation is:
\[\ce{2BaO2(s) -> 2BaO(s) + O2(g)}\]
Find the chemical amount of barium peroxide decomposed using its molar mass ($169.33\text{ g/mol}$):
\[n_{\ce{BaO2}} = \frac{129.7\text{ g}}{169.33\text{ g/mol}} = 0.76596\text{ mol}\]
Determine the moles of oxygen gas produced:
\[n_{\ce{O2}} = 0.76596\text{ mol BaO}_2 \times \frac{1\text{ mol O}_2}{2\text{ mol BaO}_2} = 0.38298\text{ mol O}_2\]
Convert the system pressure to atmospheres:
\[P = \frac{127.4\text{ kPa}}{101.325\text{ kPa/atm}} = 1.2573\text{ atm}\]
Calculate the volume of oxygen gas using the ideal gas law:
\[V = \frac{nRT}{P} = \frac{0.38298\text{ mol} \times 0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 423.0\text{ K}}{1.2573\text{ atm}} = 10.6\text{ L}\]
S76. Empirical Identification of an Unknown Gaseous Oxide
According to Avogadro's law, gas volume ratios at equal temperature and pressure correspond to the direct mole ratios of the substances. The volumetric ratio of the constituent products is:
\[2.50\text{ L N}_2 : 1.25\text{ L O}_2 = 2\text{ mol N}_2 : 1\text{ mol O}_2\]
This gives a simplified element atomic ratio of $4\text{ N}$ atoms to $2\text{ O}$ atoms. Since the volume of the original gas sample decomposed ($2.50\text{ L}$) is identical to the volume of the nitrogen gas produced ($2.50\text{ L}$), each molecule of the original unknown gas must contain two nitrogen atoms. Therefore, the formula of the colorless gas is dinitrogen monoxide, \(\ce{N2O}\).
S77. Mass-to-Volume Industrial Synthesis Scaling for Ethylene Hydration
The sequential steps show a net overall stoichiometric conversion matching a $1:1$ molar ratio:
\[\ce{C2H4 -> C2H5OH}\]
Find the theoretical mass yield required before accounting for the $90.1\%$ operational loss:
\[\text{Theoretical Mass} = \frac{1000\text{ kg}}{0.901} = 1109.88\text{ kg} = 1.10988 \times 10^6\text{ g}\]
Convert this theoretical mass of ethanol into moles using its molar mass ($46.07\text{ g/mol}$):
\[n_{\ce{C2H5OH}} = \frac{1.10988 \times 10^6\text{ g}}{46.07\text{ g/mol}} = 24091\text{ mol}\]
Because of the $1:1$ relationship, $24091\text{ mol of C2H4}$ is required. Using the standard molar volume at STP ($22.414\text{ L/mol}$), calculate the total volume:
\[V = 24091\text{ mol} \times 22.414\text{ L/mol} = 5.40 \times 10^5\text{ L}\]
S78. Molar Mass Analytical Characterization of Macromolecular Hemoglobin
Convert the operational environment parameters to standard units:
\[P = \frac{743\text{ torr}}{760\text{ torr/atm}} = 0.97763\text{ atm}\]
\[V = 1.53\text{ mL} = 0.00153\text{ L}\]
\[T = 37^\circ\text{C} + 273.15 = 310.15\text{ K}\]
Calculate the moles of oxygen gas combined with the sample:
\[n_{\ce{O2}} = \frac{PV}{RT} = \frac{0.97763\text{ atm} \times 0.00153\text{ L}}{0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 310.15\text{ K}} = 5.8753 \times 10^{-5}\text{ mol}\]
Since one molecule of hemoglobin binds four molecules of oxygen, the molar ratio is $1:4$:
\[n_{\text{hemoglobin}} = \frac{5.8753 \times 10^{-5}\text{ mol}}{4} = 1.4688 \times 10^{-5}\text{ mol}\]
Determine the molar mass of hemoglobin:
\[M = \frac{1.0\text{ g}}{1.4688 \times 10^{-5}\text{ mol}} = 6.8 \times 10^4\text{ g/mol}\]
S79. Empirical Synthesis Formula of a Gaseous Xenon Fluoride Compound
Let the initial pressure of the xenon fluoride gas sample be $P_{\ce{XeF_x}} = 18\text{ torr}$. When hydrogen is added, the total initial pressure increases to $72\text{ torr}$, meaning the partial pressure of added hydrogen is:
\[P_{\ce{H2, initial}} = 72 - 18 = 54\text{ torr}\]
The general balanced reaction equation with hydrogen gas to form xenon gas and hydrogen fluoride gas is:
\[\ce{XeF_x(g) + \frac{x}{2} H2(g) -> Xe(g) + x HF(g)}\]
Solid $\ce{KOH}$ selectively absorbs and removes all of the generated $\ce{HF}$ gas. The final pressure of $36\text{ torr}$ is due only to the remaining unreacted hydrogen gas and the newly produced xenon gas. Since each molecule of the original compound contains exactly one xenon atom, the partial pressure of xenon gas produced must equal the initial pressure of the compound:
\[P_{\ce{Xe}} = 18\text{ torr}\]
The partial pressure of the unreacted residual hydrogen gas inside the bulb is:
\[P_{\ce{H2, residual}} = 36 - 18 = 18\text{ torr}\]
Calculate the partial pressure of the hydrogen gas consumed during the spark reaction:
\[P_{\ce{H2, consumed}} = P_{\ce{H2, initial}} - P_{\ce{H2, residual}} = 54 - 18 = 36\text{ torr}\]
The molar ratio of consumed hydrogen gas to the initial xenon fluoride compound is proportional to their partial pressures:
\[\frac{\text{Moles of }\ce{H2}}{\text{Moles of compound}} = \frac{36\text{ torr}}{18\text{ torr}} = 2\]
Thus, $2\text{ moles of H2}$ are required per mole of the compound, meaning each mole of the compound contains $4\text{ moles of F}$ atoms. The empirical formula is \(\ce{XeF4}\).
S80. Quantitative van Slyke Nitrogen Assessment of Glycine Purity
First, subtract the water vapor pressure at $29^\circ\text{C}$ ($30.0\text{ torr}$) from the total measured pressure to find the pressure of the dry nitrogen gas:
\[P_{\ce{N2}} = 735\text{ torr} - 30.0\text{ torr} = 705\text{ torr} = \frac{705}{760} = 0.92763\text{ atm}\]
Convert the volume and temperature parameters to standard units:
\[V = 3.70\text{ mL} = 0.00370\text{ L}\]
\[T = 29^\circ\text{C} + 273.15 = 302.15\text{ K}\]
Calculate the moles of nitrogen gas collected:
\[n_{\ce{N2}} = \frac{PV}{RT} = \frac{0.92763\text{ atm} \times 0.00370\text{ L}}{0.08206 \times 302.15\text{ K}} = 1.3837 \times 10^{-4}\text{ mol}\]
The balanced reaction equation shows a $1:1$ stoichiometric ratio between glycine and nitrogen gas, meaning the sample contains $1.3837 \times 10^{-4}\text{ mol of glycine}$. Convert this amount to mass using the molar mass of glycine ($75.07\text{ g/mol}$):
\[\text{Mass of glycine} = 1.3837 \times 10^{-4}\text{ mol} \times 75.07\text{ g/mol} = 0.010387\text{ g}\]
Calculate the weight percentage of glycine in the sample:
\[\text{Percentage} = \left(\frac{0.010387\text{ g}}{0.0604\text{ g}}\right) \times 100\% = 17.2\%\]
9.5: Effusion and Diffusion of Gases
S81. Comparative Effusion Decompression Lifetimes of Hydrogen and Helium
According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Since the time required to complete effusion is inversely proportional to the rate:
\[\frac{\text{Time}_{\ce{H2}}}{\text{Time}_{\ce{He}}} = \sqrt{\frac{M_{\ce{H2}}}{M_{\ce{He}}}}\]
Substitute the molecular molar masses ($M_{\ce{H2}} = 2.016\text{ g/mol}$ and $M_{\ce{He}} = 4.003\text{ g/mol}$):
\[\text{Time}_{\ce{H2}} = 6\text{ hours} \times \sqrt{\frac{2.016}{4.003}} = 4.24\text{ hours}\]
S82. Structural Factors Governing Molecular Densities at Equilibrium
The center illustration of Figure 8.27 shows two bulbs at equal pressure and temperature but containing different gases. If one bulb contains a gas with a larger molecular size or significant intermolecular forces, its behavior will deviate from that of an ideal gas. At identical pressures and temperatures, real gases occupy different molar volumes depending on their specific van der Waals constants, meaning the total number of molecules per unit volume will not be perfectly identical.
S83. Analytical Derivation of Graham's Law of Effusion
The kinetic-molecular theory states that at a given temperature, the average kinetic energy of two different gases is equal:
\[\text{KE}_1 = \text{KE}_2 \implies \frac{1}{2} M_1 (u_{\text{rms}, 1})^2 = \frac{1}{2} M_2 (u_{\text{rms}, 2})^2\]
Rearranging this equation to find the ratio of their root-mean-square velocities yields:
\[\frac{u_{\text{rms}, 1}}{u_{\text{rms}, 2}} = \sqrt{\frac{M_2}{M_1}}\]
Since the rate of effusion ($R$) is directly proportional to the molecular speed of the gas particles, substituting $R$ for $u_{\text{rms}}$ gives Graham's law equation:
\[\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}}\]
S84. Gas-Phase Diffusion Fractionation of Deuterium Oxide
Apply Graham's law using the given molar masses ($M_{\ce{H2O}} = 18.01\text{ g/mol}$ and $M_{\ce{D2O}} = 20.03\text{ g/mol}$):
\[\frac{\text{Rate}_{\ce{H2O}}}{\text{Rate}_{\ce{D2O}}} = \sqrt{\frac{20.03}{18.01}} = 1.054\]
The relative rate of diffusion shows that light water vapor molecules diffuse $1.054$ times faster than heavy water vapor molecules.
S85. Kinetic Sorting of Retarded Diffusion Profiles Relative to Oxygen
According to Graham's law, any gas with a molar mass greater than that of molecular oxygen ($M_{\ce{O2}} = 32.00\text{ g/mol}$) will diffuse more slowly. Evaluating the molar masses of the choices:
\begin{itemize}
\item \(\ce{F2}\) ($38.00\text{ g/mol}$): Slower
\item \(\ce{Ne}\) ($20.18\text{ g/mol}$): Faster
\item \(\ce{N2O}\) ($44.01\text{ g/mol}$): Slower
\item \(\ce{C2H2}\) ($26.04\text{ g/mol}$): Faster
\item \(\ce{NO}\) ($30.01\text{ g/mol}$): Faster
\item \(\ce{Cl2}\) ($70.90\text{ g/mol}$): Slower
\item \(\ce{H2S}\) ($34.08\text{ g/mol}$): Slower
\end{itemize}
The gases that diffuse more slowly than oxygen are \(\ce{F2}\), \(\ce{N2O}\), \(\ce{Cl2}\), and \(\ce{H2S}\).
S86. Validation Calculations for the Uranium Hexafluoride Enrichment Process
Apply Graham's law to the two isotopic forms of uranium hexafluoride gas:
\[\frac{\text{Rate}(^{235}\ce{UF6})}{\text{Rate}(^{236}\ce{UF6})} = \sqrt{\frac{352.041206\text{ g/mol}}{349.034348\text{ g/mol}}} = \sqrt{1.008615} = 1.00429\]
Subtracting $1$ and converting to a percentage shows that $^{235}\ce{UF6}$ diffuses $0.43\%$ (approximately $0.4\%$) faster than $^{236}\ce{UF6}$, verifying the value.
S87. Relative Isotopic and Allotropic Diffusion Ratios of Hydrogen and Ozone
Apply Graham's law to each pair:
Isotopic hydrogen ratio:
\[\frac{\text{Rate}(^1\ce{H2})}{\text{Rate}(^2\ce{H2})} = \sqrt{\frac{4.0}{2.0}} = 1.4\]
Oxygen-ozone allotropic ratio:
\[\frac{\text{Rate}(\ce{O2})}{\text{Rate}(\ce{O3})} = \sqrt{\frac{48.0}{32.0}} = 1.22\]
S88. Molecular Mass Identification of an Unknown Diffusing Gas Component
Apply Graham's law to the unknown gas and the carbon dioxide reference gas:
\[\frac{\text{Rate}_{\text{unknown}}}{\text{Rate}_{\ce{CO2}}} = \sqrt{\frac{M_{\ce{CO2}}}{M_{\text{unknown}}}}\]
Substitute the given rates and the molar mass of $\ce{CO2}$ ($44.01\text{ g/mol}$):
\[\frac{83.3\text{ mL/s}}{102\text{ mL/s}} = \sqrt{\frac{44.01}{M_{\text{unknown}}}}\]
\[0.81667 = \sqrt{\frac{44.01}{M_{\text{unknown}}}}\]
Squaring both sides and solving for the unknown mass yields:
\[0.66694 = \frac{44.01}{M_{\text{unknown}}} \implies M_{\text{unknown}} = 66.0\text{ g/mol}\]
S89. Multi-Component Diffusion Interception Distances in a Linear Tube Setup
Find the relative diffusion rates of ammonia gas ($17.03\text{ g/mol}$) and hydrochloric acid gas ($36.46\text{ g/mol}$):
\[\frac{\text{Rate}_{\ce{NH3}}}{\text{Rate}_{\ce{HCl}}} = \sqrt{\frac{36.46}{17.03}} = 1.463\]
This means ammonia gas travels $1.463$ times farther than hydrochloric acid gas in the same amount of time. Let $d$ be the distance traveled by the $\ce{HCl}$ gas molecules. The distance traveled by the $\ce{NH3}$ molecules is $1.463d$. Set their sum equal to the total length of the tube:
\[d + 1.463d = 87.0\text{ cm}\]
\[2.463d = 87.0\text{ cm} \implies d = 35.32\text{ cm}\]
Calculate the position from the ammonia source plug:
\[\text{Distance} = 87.0 - 35.32 = 51.7\text{ cm}\]
9.6: The Kinetic-Molecular Theory
S90. Kinetic-Molecular Interpretation of Container Shape Filling
According to the postulates of the kinetic-molecular theory, gas molecules are in continuous, rapid, and random motion in all directions. They travel in straight lines until they collide elastically with other molecules or the walls of the container. Because there are no significant attractive or repulsive forces between ideal gas molecules, they move freely to fill the entire available volume, taking the shape of any container.
S91. Thermal Fluctuations of Individual Molecular Velocities
Yes, the speed of any single molecule can double at a constant temperature. Temperature is a measure of the average kinetic energy of all the molecules in a gas sample. Individual molecules continuously collide with each other, exchanging energy. Through these random collisions, a single molecule can gain enough kinetic energy to double its speed, while other molecules slow down, keeping the total average energy of the system constant.
S92. Kinetic Energy Responses to Systemic Property Adjustments
a. The average kinetic energy remains unchanged because the temperature of the system is held constant.
b. The average kinetic energy increases because the temperature of the system increases, and kinetic energy is directly proportional to absolute temperature.
c. The average kinetic energy increases by a factor of 4 because kinetic energy is proportional to the square of the molecular speed ($\text{KE} = \frac{1}{2}mu^2$).
S93. Velocity Profile Shifts in Cooled Helium Samples
When a sample of helium gas is cooled, its average kinetic energy decreases, shifting its molecular speed distribution curve toward lower velocities. This causes the peak of the curve to become narrower and taller. This shape looks more like the velocity distribution curve of water vapor (\(\ce{H2O}\)) than that of hydrogen gas (\(\ce{H2}\)) at room temperature because water has a higher molar mass and moves with a lower, more concentrated speed distribution.
S94. Kinetic Energy and Speed Ratios of Sulfur Dioxide and Oxygen Mixtures
Because both gases exist together in the same mixture, they share the same temperature. Therefore, the ratio of the average kinetic energy of an $\ce{SO2}$ molecule to that of an $\ce{O2}$ molecule is exactly $1:1$.
The ratio of their root-mean-square speeds depends on their molar masses ($M_{\ce{SO2}} = 64.06\text{ g/mol}$ and $M_{\ce{O2}} = 32.00\text{ g/mol}$):
\[\frac{u_{\text{rms}}(\ce{SO2})}{u_{\text{rms}}(\ce{O2})} = \sqrt{\frac{M_{\ce{O2}}}{M_{\ce{SO2}}}} = \sqrt{\frac{32.00}{64.06}} = 0.707\]
S95. Comprehensive Thermodynamic Adjustments to a Carbon Monoxide Sample
a. The initial parameters at STP are $T_1 = 273.15\text{ K}\text{ and }V_1 = 1\text{ L}$. The final conditions are $T_2 = 546.3\text{ K}$ (doubled temperature) and $V_2 = 2\text{ L}$ (doubled volume). According to the ideal gas law, doubling the absolute temperature tends to double the pressure, while doubling the volume cuts the pressure in half. Therefore, these changes cancel out, and there is no net effect on the pressure.
b. The average kinetic energy doubles because the absolute temperature of the gas system is doubled.
c. The root-mean-square speed increases by a factor of $\sqrt{2}$ (approximately $1.41$) because the velocity is proportional to the square root of the absolute temperature.
S96. Root-Mean-Square Velocity Evaluations for Molecular Nitrogen
The root-mean-square speed of a gas is inversely proportional to the square root of its molar mass at a constant temperature. Using the speeds and masses of molecular hydrogen ($2.016\text{ g/mol}$) and nitrogen ($28.01\text{ g/mol}$):
\[\frac{u_{\text{rms}}(\ce{N2})}{u_{\text{rms}}(\ce{H2})} = \sqrt{\frac{M_{\ce{H2}}}{M_{\ce{N2}}}}\]
\[u_{\text{rms}}(\ce{N2}) = 1.6\text{ km/s} \times \sqrt{\frac{2.016}{28.01}} = 0.43\text{ km/s}\]
9.7: Non-Ideal Gas Behavior
S97. Thermodynamic and Lift Capacity Analyses of Hot Air Balloons
a. The pressure of the gas inside the hot-air balloon is equal to the atmospheric pressure outside because the bottom of the balloon is open to the atmosphere, allowing the pressures to balance.
b. The density of the gas inside the balloon is less than that of the outside air because the air inside is hotter, causing the gas to expand and reducing its mass per unit volume.
c. Calculate the average molar mass of dry air using the ideal gas density equation where $T = 20^\circ\text{C} + 273.15 = 293.15\text{ K}$:
\[M = \frac{dRT}{P} = \frac{1.2256\text{ g/L} \times 0.08206\text{ L}\cdot\text{atm/mol}\cdot\text{K} \times 293.15\text{ K}}{1\text{ atm}} = 29.48\text{ g/mol}\]
d. Convert the balloon internal temperature to Kelvin:
\[T = \frac{1.30 \times 10^2 - 32}{1.8} + 273.15 = 54.44^\circ\text{C} + 273.15 = 327.59\text{ K}\]
Calculate the internal gas density:
\[d = \frac{PM}{RT} = \frac{1\text{ atm} \times 29.48\text{ g/mol}}{0.08206 \times 327.59\text{ K}} = 1.097\text{ g/L}\]
e. The difference in mass between $1.00\text{ L}$ of cool air and hot air is:
\[\Delta m = 1.2256\text{ g} - 1.097\text{ g} = 0.129\text{ g}\]
f. Convert the volume of the balloon into liters:
\[1.1 \times 10^5\text{ ft}^3 \times 28.3168\text{ L/ft}^3 = 3.1148 \times 10^6\text{ L}\]
Calculate total mass of displaced cool external air:
\[m_{\text{cool}} = 3.1148 \times 10^6\text{ L} \times 1.2256\text{ g/L} = 3.8175 \times 10^6\text{ g} = 8416\text{ lbs}\]
Calculate total mass of the internal hot gas:
\[m_{\text{hot}} = 3.1148 \times 10^6\text{ L} \times 1.097\text{ g/L} = 3.4169 \times 10^6\text{ g} = 7533\text{ lbs}\]
The total lifting power is the difference between these masses:
\[\text{Lifting Power} = 8416 - 7533 = 883\text{ lbs}\]
Subtracting the weight of the rigging gives the net cargo capacity:
\[\text{Capacity} = 883 - 500 = 383\text{ lbs}\]
g. Convert liquid propane volume to mass, then to moles using its density and mass ($44.1\text{ g/mol}$):
\[40.0\text{ gal} \times 3.78541\text{ L/gal} = 151.42\text{ L}\]
\[\text{Mass} = 151.42\text{ L} \times 500.5\text{ g/L} = 75784\text{ g}\]
\[n_{\ce{C3H8}} = \frac{75784\text{ g}}{44.1\text{ g/mol}} = 1718.5\text{ mol}\]
The balanced combustion equation is: $\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)}$.
Calculate the volumes of the products produced at STP:
\[V_{\ce{CO2}} = (1718.5 \times 3) \times 22.414\text{ L/mol} = 1.16 \times 10^5\text{ L}\]
\[V_{\ce{H2O}} = (1718.5 \times 4) \times 22.414\text{ L/mol} = 1.54 \times 10^5\text{ L}\]
h. Using a standard heat of combustion for propane of approximately $2220\text{ kJ/mol}$:
\[\text{Total Heat} = 1718.5\text{ mol} \times 2220\text{ kJ/mol} = 3.815 \times 10^6\text{ kJ}\]
Calculate the approximate rate of heat loss over the $90\text{ minute}$ flight:
\[\text{Rate} = \frac{3.815 \times 10^6\text{ kJ}}{90\text{ min}} = 4.24 \times 10^4\text{ kJ/min}\]
S98. Invariance of Graham's Law Across Variable Thermal Baselines
According to Graham's law, the ratio of the diffusion rates of two gases depends only on the inverse ratio of the square roots of their molar masses:
\[\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}}\]
Because this formula contains no temperature variables, the ratio remains constant regardless of whether the system is at $0^\circ\text{C}$ or $100^\circ\text{C}$.
S99. Graphical Identification of Real Gas Behavior Deviations
In the graphs shown in the referenced layout, Gas C and Gas E exhibit behavior that is significantly different from that expected for ideal gases. Ideal gases maintain a constant compressibility factor ($Z = 1$) and follow linear volume-temperature tracks. The non-linear curves for Gas C and Gas E reveal real-gas behavior due to intermolecular attractions and molecular volume constraints.
S100. Intermolecular and Structural Compressibility Deviations in Carbon Dioxide
As shown in the graphical analysis plots, the plot of $PV$ for $\ce{CO2}$ drops below the ideal gas line at low pressures because of intermolecular attractive forces. These forces pull the molecules together, reducing the force of their collisions with the container walls and lowering the pressure below ideal expectations. At higher pressures, the curve rises sharply above the ideal line because the physical volume of the gas molecules becomes significant, making the gas harder to compress.
S101. Ideal Compatibility Conditions for Real Gas Systems
Real gases behave most like ideal gases under condition b (high temperature, low pressure) because the molecules move too fast and are too far apart to interact. Real gases deviate most from ideal behavior under condition c (low temperature, high pressure) because the molecules move slowly and are forced close together, making intermolecular attractions and molecular volume significant.
S102. Molecular Root Factors Governing Real Gas Deviations
The two primary factors responsible for real gases deviating from ideal behavior are the significant attractive forces between gas molecules, which reduce collision force, and the actual physical volume occupied by the gas molecules themselves, which limits the free space available for motion at high pressures.
S103. Volumetric Factor Corrections for High-Mass Actinide Hexafluorides
The correction for molecular volume (the van der Waals constant $b$) is largest for the largest molecule in the group. Comparing the choices, sulfur hexafluoride, $\ce{SF6}$, has the largest molecular size and electron cloud, so it requires the largest volume correction factor.
S104. Comparative Pressure Calculations for Carbon Dioxide Gas
a. Apply the ideal gas law where $V = 0.245\text{ L}$, $n = 0.467\text{ mol}$, and $T = 159^\circ\text{C} + 273.15 = 432.15\text{ K}$:
\[P = \frac{nRT}{V} = \frac{0.467\text{ mol} \times 0.08206 \times 432.15\text{ K}}{0.245\text{ L}} = 67.6\text{ atm}\]
b. Apply the van der Waals equation using the constants for $\ce{CO2}$ ($a = 3.59\text{ L}^2\cdot\text{atm/mol}^2$, $b = 0.0427\text{ L/mol}$):
\[P = \frac{nRT}{V - nb} - \frac{an^2}{V^2} = \frac{0.467 \times 0.08206 \times 432.15}{0.245 - (0.467 \times 0.0427)} - \frac{3.59 \times (0.467)^2}{(0.245)^2}\]
\[P = \frac{16.560}{0.22506} - \frac{0.7829}{0.060025} = 73.58 - 13.04 = 60.5\text{ atm}\]
c. The ideal gas law assumes that gas molecules have no volume and experience no intermolecular forces. The van der Waals equation corrects for these assumptions, which is why it yields a lower, more accurate pressure.
d. The pressure correction factor ($\frac{an^2}{V^2} = 13.04\text{ atm}$) is dominant because the strong attractive forces between the polarizable $\ce{CO2}$ molecules significantly reduce the pressure exerted on the container walls.
S105. Compressibility Profiles and Temperature Responses of Real Gases
a. If a gas behaved ideally, its graph of Z versus P would be a perfectly horizontal straight line at Z=1
b. Yes, treating gases as ideal throughout the chapter was justified because most calculations were performed near standard room temperatures and atmospheric pressures, where real gas deviations are small enough to be negligible.
c. The actual volume of gas molecules increases the value of Z, pushing it above 1. This effect is small at low pressures and large volumes where molecules are far apart. It becomes large at high pressures where the molecules are crowded together.
d. Intermolecular attractions lower the value of Z, pulling it below 1. This effect is small at high temperatures where fast-moving molecules break free of attractions. It becomes large at low temperatures where slower molecules are pulled together.
e. Real gases show the largest deviations from ideal behavior at very low temperatures, where molecular motion slows down enough for intermolecular attractions to take effect.