10: Gases

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Conversion between Gas Pressure Units

Exercise \(\PageIndex{1}\)

A pressure of 1.00 atm has a metric equivalent of 1.01 × 10⁵ ________.

Answer: Pascals

Exercise \(\PageIndex{2}\)

Which of the following is the largest value of pressure?

a. 0.716 atm b. 18.3 in Hg c. 972 mm Hg d. 14.9 psi e. 86572 Pa

Answer

c. 972 mm Hg

Convert all to same units.

a. 0.716 atm

b. \(\large \frac{18.3 in Hg}{29.9213 (\frac{atm}{in Hg})} = 0.612\;atm\)

c. \(\large \frac{972 mm Hg}{760 (\frac{atm}{mm Hg})}=1.28\;atm\)

d. \(\large \frac{14.9 psi}{14.7\frac{atm}{psi}}=1.01\;atm\)

e. \(\large \frac{86572 Pa}{1.01*10^{5}\frac{atm}{Pa}}=0.857\;atm\)

Exercise \(\PageIndex{3}\)

A particular gas exerts a pressure of 356 mm Hg. What is this pressure in units of bar? (1 atm = 760 mm Hg = 101.3 kPa = 1.013 bar)

Answer: 0.473 bar

Exercise \(\PageIndex{4}\)

The local weather forecaster reports that the current barometric pressure is 15.9 inches of mercury. What is the current pressure in atmospheres?

Answer: 0.531 atm

Exercise \(\PageIndex{5}\)

It is possible to make a barometer using a liquid other than mercury. What would be the height (in meters) of a column of dichloromethane at a pressure of 0.790 atm, given that 0.790 atm is equal to a 0.758 m column of mercury and the densities of mercury and dichloromethane are 13.5 g/cm³ and 1.33 g/cm³, respectively.

Answer

7.69 m

\(\large d=\frac{P_{atm}}{hg}\) can be used to find the height in a column.

Use a ratio with the pressure in atm is equal to one.

\(\large \frac{P_{CH_{2}Cl_{2}}}{P_{Hg}}=\frac{d_{CH_{2}Cl_{2}}*g*h_{CH_{2}Cl_{2}}}{d_{Hg}*g*h_{Hg}}\)

\(\large \frac{0.790}{0.790}=\frac{1.33*981*h_{CH_{2}Cl_{2}}}{13.5*981*60}\) Note: Converted to cm from m.
\(\large h_{CH_{2}Cl_{2}}=609.02 cm\)

Gas Laws

Exercise \(\PageIndex{1}\)

Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant.

amount and volume
pressure and amount
temperature and volume
temperature and amount
pressure and temperature

Answer

P/T = constant
V/T = constant (Charles’ law)
P/n = constant
PV = constant (Boyle’s law)
V/n = constant (Avogadro’s law)

Ideal Gas Law

Exercise \(\PageIndex{2}\)

A sample of gaseous Cl₂ has a volume of 12.4 L at 500.0 K and 0.521 atm. How many moles are present?

Answer: \(n=\frac{PV}{RT}=\frac{0.521atm\left ( 12.4L \right )}{0.08206\frac{L\cdot atm}{mol\cdot K}\left ( 500K \right )}=0.157mol\)

Exercise \(\PageIndex{3}\)

What is the mass of chlorine gas in a 12.4L container at 500.0K and 0.521 atm?

Answer

\(n=\frac{PV}{RT}=\frac{0.521atm\left ( 12.4L \right )}{0.08206\frac{L\cdot atm}{mol\cdot K}\left ( 500K \right )}=0.157mol\)

\(0.157mol~Cl_{2}\left ( \frac{70.90g~Cl_{2}}{mol} \right )=11.1g\)

Exercise \(\PageIndex{4}\)

What volume is occupied by 1.06 mol of CO₂ gas at 299 K and a pressure of 0.89 atm?

Answer: \(V=\frac{nRT}{P}=\frac{1.06mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )299K}{0.89atm}=29L\)

Exercise \(\PageIndex{5}\)

What is the temperature of 1.41 mol of methane gas in a 5.0 L container at 1.00 atm?

Answer: \(T=\frac{PV}{nR}=\frac{1.00atm\left ( 5.0L \right )}{1.41mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )}=43K\)

Exercise \(\PageIndex{6}\)

What is the pressure exerted by a 1.75 mol sample of water at 7.0L and 20^oC?

Answer: \(P=\frac{nRT}{V}=\frac{1.75mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )293.15K}{7.0L}=5.8atm\)

Two State

Exercise \(\PageIndex{7}\)

What pressure would 6.1 mole of a gas in a rigid container have at 20.0 ^oC have if it had a pressure of 0.45 atm at -45.0 ^oC?

Answer

\(P_{1}=P_{2}\frac{T_{1}}{T_{2}}=\left ( 0.45atm \right )\left ( \frac{293.15K}{228.15K} \right )=0.58atm\)

Exercise \(\PageIndex{8}\)

A pressure tank containing chlorine gas at a pressure of 2.00 atm and a temperature of 40^oC is set with a pressure relief valve set to open at a pressure of 10.0 atm. At what temperature will the relief valve open?

Answer: \(T_{2}=T_{1}\frac{P_{2}}{P_{1}}= \left ( 313.15K \right )\left ( \frac{10.0atm}{2.00atm} \right )=1570K\)

Exercise \(\PageIndex{9}\)

A gas is in a sealed cylinder at 25^oC with a piston which can expand or contract to change the volume. The initial volume and pressure are 75.0 L and 980.0 torr. A force is applied to the piston and the volume adjusts to a final pressure and temperature of 5.00 atm at 188^oC. What is the final volume?

Answer: \(V_{2}=V_{1}\left ( \frac{P_{1}}{P_{2}} \right )\left ( \frac{T_{2}}{T_{1}} \right )=75.0L\left ( \frac{980torr\left ( \frac{1atm}{760torr} \right )}{5.00atm} \right )\left ( \frac{461.15K}{298K} \right )=29.9L\)

Gas Phase Reactions

Exercise \(\PageIndex{1}\)

5.00 g of Mg is added to 50.0 mL of 0.800M HCl. A double displacement reaction occurs in a sealed container at 25.0^oC. What is the pressure of the hydrogen gas generated if the volume above the solution is 100.0 ml, and we ignore any pressure due to the evaporated water?

Answer: \(P_{H_{2}}=\frac{n_{H_{2}}RT}{V}=\frac{0.0200mol~H_{2}\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )298K}{0.100L}=4.89atm\)

Exercise \(\PageIndex{2}\)

Nitrogen dioxide is formed in a closed container at 90^oC and 1.00 atm when 1.50 g NO and 2.00 mole of O₂ are mixed. After the reaction is finished the pressure changes to 3.90 atm. What is the final temperature?

Answer: \(T_{2}=T_{1}\left ( \frac{n_{1}}{n_{2}} \right )\left ( \frac{P_{2}}{P_{1}} \right )=363.15K\left ( \frac{2.05mol}{2.025mol} \right )\left ( \frac{3.90atm}{1.00atm} \right )=1433.77K =1400K\)

Ideal Gas Law

Exercise \(\PageIndex{1}\)

What is the volume occupied by a mixture of 0.595 mol of N₂ gas and 0.685 mol of O₂ gases at 1.75 atm and 23.2°C?

Answer

17.9 L

\(\lV=\frac{nRT}{P}\)
n = 0.595 + 0.685 = 1.28 mole

T = 23.2°C + 273.15 = 296.35 K

P = 1.75 atm; R = 0.08205 L atm mol^-1 K^-1

\(\V = \frac{1.28*0.08205*296.35}{1.75} = 17.88\;L\)

Exercise \(\PageIndex{2}\)

A vessel with a volume of 20.5 L contains 79.9 g of nitrogen gas, 2.80 g of hydrogen gas, and 0.403 g of argon gas. At 25°C, what is the pressure in the vessel?

Answer: 5.07 atm

Exercise \(\PageIndex{3}\)

Water can be decomposed by electrolysis into hydrogen gas and oxygen gas. What mass of water must decompose to fill a 3.00 L flask to a total pressure of 2.00 atm at 298 K with a mixture hydrogen and oxygen? (R = 0.08206 L⋅atm/mol⋅K)

2 H₂O(l) → 2 H₂(g) + O₂(g)

Answer: 2.95 g

Exercise \(\PageIndex{4}\)

Carbon monoxide reacts with oxygen to form carbon dioxide.

2 CO(g) + O₂(g) → 2 CO₂(g)

In a 1.00 L flask, 4.30 atm of CO reacts with 2.50 atm of O2. Assuming that the temperature remains constant, what is the final pressure in the flask?

Answer: 4.95 atm

Mole Fraction and Partial Pressure

Exercise \(\PageIndex{5}\)

The partial pressures of CH₄, N₂, and O₂ in a sample of a gas mixture were found to be 191 mm Hg, 451 mm Hg, and 583 mm Hg, respectively. Calculate the mole fraction of each gas.

Answer

\(\large \chi _{CH_{4}} = 0.156\;,\;\chi _{N_{2}} = 0.368\;,\;\chi _{O_{2}} = 0.476\)

Total Pressure = 191 + 451 + 583 = 1225 mm Hg

\(\large \chi _{CH_{4}} = \frac{\rho _{CH_{4}}}{P}=\frac{191}{1225} = 0.1559\)

\(\large \chi _{N_{2}} = \frac{\rho _{N_{2}}}{P}=\frac{451}{1225} = 0.3681\)
\(\large \chi _{O_{2}} = \frac{\rho _{O_{2}}}{P}=\frac{583}{1225} = 0.4759\)

Exercise \(\PageIndex{6}\)

A gaseous mixture containing 7.60 mol Ar and 7.40 mol CO₂ has a total pressure of 3.50 atm. What is the partial pressure of CO₂?

Answer

\(\large \rho _{CO_{2} = 1.73 atm\)

\(\large \chi _{Ar} = \frac{7.60}{7.60+7.40}=0.507\)

\(\large \chi _{CO_{2}} = \frac{7.40}{7.60+7.40}=0.493\)

\(\large \rho _{CO_{2}= P_{total}\;*\;\chi _{CO_{2}}=3.50\;*\;0.493=1.726\; atm\]

Exercise \(\PageIndex{7}\)

A 10.0 L flask at 318 K contains a mixture of Ar and CH₄ with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715, what is the mass percent of Ar?

Answer: 86.2%

Kinetic Molecular Theory

Exercise \(\PageIndex{1}\)

Which of the following statements is/are postulates of the kinetic–molecular theory of gases?

Gas particles are in constant, random motion.
The distance between gas particles is large in comparison to their size.
The average kinetic energy of gas particles is proportional to the kelvin temperature of the gas.

a. 1 only b. 2 only c. 3 only d. 1 and 2 e. 1, 2, and 3

Answer: e. 1, 2, and 3

Exercise \(\PageIndex{2}\)

Calculate the root-mean-square velocity for the O₂ molecules in a sample of O₂ gas at 35.9°C. (R = 8.3145 J/K⋅mol)

Answer

2.41*10⁵ J*kg^-1

\(v = \sqrt{\frac{3RT}{M}}\)

R = 8.314 J/K*mol; T = 35.9 + 273.15 = 309.05 K; M = (O₂) = 32*10^-3 kg/mol

\(v = \sqrt{\frac{3*8.3145*309.05}{32*10-3}}=2.408*10^{5}\;J*kg^{-1}\)

Exercise \(\PageIndex{3}\)

Place the following gases in order of increasing average velocity at 300 K: Ar, CH₄, N₂, and N₂O.

Answer: N₂O < Ar < N₂ < CH₄

Exercise \(\PageIndex{4}\)

At STP, as the molar mass of the molecules that make up a pure gas increases, the

root mean square speed of the molecules increases.
root mean square speed of the molecules decreases.
root mean square speed of the molecules remains constant.
root mean square speed increases to a maximum, then decreases.
none of the above.

Answer: b. root mean square speed of the molecules decreases.

Effusion

Exercise \(\PageIndex{1}\)

Methane gas, CH₄, effuses through a barrier at a rate of 0.961 mL/minute. If an unknown gas effuses through the same barrier at a rate of 0.343 mL/minute, what is the molar mass of the gas?

Answer

126 g/mol

\(R_{E} = \frac{1}{\sqrt{mm}}\)

\(\frac{0.961}{0.343}=\sqrt{\frac{m}{16}}\)

m = 125.5 g/mol

Exercise \(\PageIndex{2}\)

How long will it take 10.0 mL of Ne gas to effuse through a porous barrier if it has been observed that 175 minutes are required for 10.0 mL of Ar gas to effuse through the same barrier?

Answer: 124 min

Exercise \(\PageIndex{3}\)

The molar mass of an unknown gas was measured by an effusion experiment. It was found that the unknown gas effused at a rate of 9.3 × 10⁻⁶ mol/hr, whereas the nitrogen gas effused at a rate of 3.9 × 10⁻⁶ mol/hr. The molar mass of the gas is _____.

Answer: 4.9 g/mol

Exercise \(\PageIndex{4}\)

The rate of effusion of an unknown gas was measured and found to be 10.7 mL/min. Under identical conditions, the rate of effusion of pure oxygen (O₂) gas is 12.6 mL/min. Based on this information, the identity of the unknown gas could be:

a. N₂CH₂OF₂ b. CO₂N₂O c. NO d. C₂H₆ e. none of these

Answer: b. CO₂N₂O

Theory

Exercise \(\PageIndex{1}\)

Non-ideal behavior for a gas is most likely to be observed under conditions of

high temperature and low pressure.
high temperature and high pressure.
low temperature and low pressure.
standard temperature and pressure.
low temperature and high pressure.

Answer: e. low temperature and high pressure.

Exercise \(\PageIndex{2}\)

Which of the following statements concerning real gases is/are CORRECT?

Real gases are always liquids or solids at temperatures below 273.15 K.
The pressure of a real gas is higher than predicted by the ideal gas law.
The molecules in a real gas are attracted to each other.

a. 1 only b. 2 only c. 3 only d. 2 and 3 e. 1, 2, and 3

Answer: c. 3 only

Exercise \(\PageIndex{3}\)

Real gases are those that

only behave ideally at high pressures or low temperatures
deviate from ideal behavior
are only available naturally in the earth's atmosphere
are called real gases because their behavior can easily be modeled
have an even number of protons

Answer: b. deviate from ideal behavior

Exercise \(\PageIndex{4}\)

Real gases deviate from ideal behavior because of intermolecular forces and _____.

pressures within the chemical bonds
nonzero molecular volumes
ionization energies
molecular vibrations
a wide distribution of molecular speeds

Answer: b. nonzero molecular volumes

Exercise \(\PageIndex{5}\)

The ideal gas law can be modified to correct for the errors arising from nonideality. The modified equation is known as the ________ equation of state for a real gas.

Answer: van der Waals

van der Waals

Exercise \(\PageIndex{6}\)

The van der Waals constants for xenon are a = 4.19 (L²·atm)/mol² and b = 0.0510 L/mol. If a 0.250 mol sample of xenon in a container with a volume of 3.65 L is cooled to −90°C, what is the pressure of the sample assuming ideal gas behavior? What would be the actual pressure under these conditions?

Answer

Ideal Pressure: 1.03 atm; Actual Pressure: 1.01 atm

\(P = \frac{nRT}{V}\)

\(P = \frac{(0.250 mol)(0.08206 L*atm*mol^{-1}*K^{-1})(183.15 K)}{3.65 L}=1.029\;atm\)

\(P = \frac{nRT}{V-nb}-\frac{an^{2}}{V^{2}}\)

\(P = \frac{0.250*0.08206*183.15}{3.65-(0.250*0.0510)}-\frac{4.19*0.0250^{2}}{3.65^{2}}=1.0330-0.0196=1.0133\;atm\)