# 9: Gases

- Page ID
- 284427

## Gas Properties

Exercise \(\PageIndex{1a}\)

A pressure of 1.00 atm has a metric equivalent of 1.01 × 10^{5} ________.

**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/cm^{3} and 1.33 g/cm^{3}, 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 Cl_{2} 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 CO_{2} 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^{o}C?

**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 ^{o}C have if it had a pressure of 650.0 torr at 47.0 ^{o}C?

**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 V

_{1}= V_{2}and we can assume that no gas has been added or removed, so n_{1}=n_{2},\[ \frac{P_1\cancel{V_1}}{P_2 \cancel{V_2}}= \frac{\cancel{n_1}T_1}{\cancel{n_2}T_2}\]

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 T

_{1}= T_{2}and we can assume that no gas has been added or removed, so n_{1}=n_{2},\[ \frac{P_1V_1}{P_2 V_2}= \frac{\cancel{n_1}\cancel{T_1}}{\cancel{n_2}\cancel{T_2}}\]

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^{o}C 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^{o}C. 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 T

_{1}= T_{2}and we can assume that no gas has been added or removed, so n_{1}=n_{2},\[ \frac{P_1V_1}{P_2 V_2}= \frac{\cancel{n_1}\cancel{T_1}}{\cancel{n_2}\cancel{T_2}}\]

This reduces to Boyle's Law and

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

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

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Exercise \(\PageIndex{3c}\)

0.00240 mole of gas is injected into a sealed cylinder at 25.0^{o}C 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?
- P
_{2}S_{5} - N
_{2}O_{4} - P
_{4}S_{10}

**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

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## 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^{o}C. 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 \(\PageIndex{4b}\)

Nitrogen dioxide is formed in a closed container at 90^{o}C and 1.00 atm when 1.50 g NO and 2.00 mole of O_{2} are mixed. After the reaction is finished the pressure changes to 3.90 atm. What is the final temperature?

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