9: Gases

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Gas Properties

Exercise \(\PageIndex{1a}\)

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

Answer: Pascals

Exercise \(\PageIndex{1b}\)

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 \(\PageIndex{1c}\)

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 \(\PageIndex{1d}\)

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 \(\PageIndex{1e}\)

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 \(\PageIndex{2a}\)

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)

Exercise \(\PageIndex{2b}\)

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 \(\PageIndex{2c}\)

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 \(\PageIndex{2d}\)

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 \(\PageIndex{2e}\)

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 \(\PageIndex{2f}\)

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 Ideal Gas Problems

Exercise \(\PageIndex{3a}\)

What pressure would 2.1 mole of a gas in a rigid container have at 120.0 ^oC have if it had a pressure of 650.0 torr at 47.0 ^oC?

Answer

\[ \frac{P_1V_1}{P_2V_2}= \frac{n_1\cancel{R}T_1}{n_2\cancel{R}T_2}\]

the container is rigid so V₁= V₂ and we can assume that no gas has been added or removed, so n₁=n₂,

\[ \frac{P_1\cancel{V_1}}{P_2 \cancel{V_2}}= \frac{\cancel{n_1}T_1}{\cancel{n_2}T_2}\]

and this reduces to a Gay Lussac problem and as with all gas law problems, T must be in kelvin

\[ P_{1}=P_{2}\frac{T_{1}}{T_{2}}=\left ( 650.0 torr \right )\left ( \frac{393.15K}{310.15K} \right )=823.9torr\]

Exercise \(\PageIndex{3b}\)

A perfectly elastic ballon is placed in a vacuum table and has a volume of 22.0mL at 730.0 torr and room temperature. What is the final volume if the pressure is reduced to 240.0 torr at constant pressure??

Answer

\[ \frac{P_1V_1}{P_2V_2}= \frac{n_1\cancel{R}T_1}{n_2\cancel{R}T_2}\]

this is an isothermal process so T₁= T₂ and we can assume that no gas has been added or removed, so n₁=n₂,

\[ \frac{P_1V_1}{P_2 V_2}= \frac{\cancel{n_1}\cancel{T_1}}{\cancel{n_2}\cancel{T_2}}\]

This reduces to Boyle's Law and

\[\frac{P_1V_1}{P_2V_2}=1 \\ \; \\ or \\ \; \\ P_1V_1=P_2V_2\]

and \[V_1=V_2\left(\frac{P_2}{P_1}\right)V_1=22.0mL\left(\frac{730.0torr}{240.0torr}\right) = 66.9mL\]

Exercise \(\PageIndex{3c}\)

A gas is injected into a sealed cylinder at 25^oC with a frictionless piston which can expand or contract to change the volume. Initially there is 0.0890 mole with a volume of 35.0 mL at 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

\[ \frac{P_1V_1}{P_2V_2}= \frac{n_1\cancel{R}T_1}{n_2\cancel{R}T_2}\]

this is an isothermal process so T₁= T₂ and we can assume that no gas has been added or removed, so n₁=n₂,

\[ \frac{P_1V_1}{P_2 V_2}= \frac{\cancel{n_1}\cancel{T_1}}{\cancel{n_2}\cancel{T_2}}\]

This reduces to Boyle's Law and

\[\frac{P_1V_1}{P_2V_2}=1 \\ \; \\ or \\ \; \\ P_1V_1=P_2V_2\]

and \[V_1=V_2\left(\frac{P_2}{P_1}\right)V_1=22.0mL\left(\frac{730.0torr}{240.0torr}\right) = 66.9mL\]

Exercise \(\PageIndex{3c}\)

0.00240 mole of gas is injected into a sealed cylinder at 25.0^oC with a frictionless piston that can freely expand or contract to change the volume. Initially the volume was 35.0 mL at a pressure of 780.0 torr. After the gas was injected the pressure was raised to 812 torr while the temperature was maintained constant?

What were the initial number of moles gas in the container?
P₂S₅
N₂O₄
P₄S₁₀

Answer a: \[PV=nRT \;\; \therefore \;\; n=\frac{PV}{RT}=\frac{780torr\left ( \frac{1 atm}{760torrt} \right )0.0350L}{0.08206\left ( \frac{L\cdot atm}{mol \cdot K} \right )298K}=0.00147mol\]

Answer b: diphosphorus pentasulfide

Answer c: dinitrogen tetraoxide

Answer d: tetraphosphorus decasulfide

Gas Phase Stoichiometry

Exercise \(\PageIndex{4a}\)

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

Step 1: Write out and balance equation and write underneath it what is given:

\[\underset{m=5.00g \\fw= 24.3\frac{g}{mol}}{Mg(s)} \; + \; \underset{0.0500L \\ 0.800 M}{2HCl(aq)} \; \rightarrow \; Mg(Cl)_2(aq)+\underset{P=? atm \\ T=298K \\n=? \\ 0.100L??}{H_2(g)}\]

NOTE: We want \(P_{H_2}\), but have two unknowns (P and n), so we need to calculate n

Step 2: Calculate theoretical yield of n (based on complete consumption of limiting reagent)

\[5.00g \; Mg\left ( \frac{mol \; Mg}{24.3g} \right )\left ( \frac{1}{1mol \; Mg} \right )=0.206 \\ 0.0500L\left ( \frac{0.800mol \;HCl}{L} \right )\left ( \frac{1}{2mol \;HCl} \right )=0.0200
\\HCl =limiting \; reagent \\n_{H_2} =0.0200\left ( 1mol \; H_2(g) \right )=0.0200 mol\;H_2(g)\]

Step 3: Now that we have \(n_{H_2}\), use ideal gas law to solve for V at the conditions of the problem.

\[V=\frac{nRT}{P}=\frac{0.02500mol\left (0.08206\left ( \frac{l \cdot atm}{mol \cdot K} \right ) \right )298K}{760 \; torr\left ( \frac{1atm}{760 \;torr} \right )} = .6113L=611mL\]

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

Exercise \(\PageIndex{4b}\)

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