Q: A scuba diver's tank contains 12.0 L of air at a pressure of 200 atm and a temperature of 27°C. If the diver releases air from the tank until the pressure inside is 1.00 atm at the same temperature, what volume of air at 1.00 atm and 27°C will the diver have available to breathe? (Assume ideal gas behavior.)

$$V_2$$ = \frac{$$P_1$$ $$V_1$$}{$$P_2$$} = \frac{200 \times 12.0}{1.00} = 2400\ \mathrm{L}$$

Q: A 0.500 mol sample of nitrogen gas is confined in a 10.0 L container at 300 K. What is the pressure exerted by the gas? (Use $$R = 0.08206\ \mathrm{L\ atm\ $$mol^{-1}$$\ K$$^{-1}$$})$$

$$P = \frac{nRT}{V} = \frac{0.500 \times 0.08206 \times 300}{10.0} = 1.23\ \mathrm{atm}$$

Q: A sample of helium gas occupies 2.00 L at 25°C and 1.00 atm. If the gas is heated to 100°C at constant pressure, what will be its new volume?

$$V_2$$ = $$V_1$$ \frac{$$T_2$$}{$$T_1$$} = 2.00 \times \frac{373}{298} = 2.50\ \mathrm{L}$$

Q: A mixture of gases contains 2.0 mol of O$$_2$$, 3.0 mol of N$$_2$$, and 1.0 mol of CO$$_2 in a 10.0 L $$container at 300 K. What is the partial pressure of N$$_2$$? ($$R = 0.08206\ \mathrm{L\ atm\ $$mol^{-1}$$\ K$$^{-1}$$})$$

$$P_{N_2$} = \frac{n$$_{N_2}$$RT}{V} = \frac{3.0 \times 0.08206 \times 300}{10.0} = 7.39\ \mathrm{atm}$$

Q: A 5.00 g sample of an unknown gas occupies 2.50 L at 1.00 atm and 25°C. What is the molar mass of the gas? ($$R = 0.08206\ \mathrm{L\ atm\ $$mol^{-1}$$\ K$$^{-1}$$})$$

$$M = \frac{mRT}{PV} = \frac{5.00 \times 0.08206 \times 298}{1.00 \times 2.50} = 48.9\ \mathrm{g\ $$mol^{-1}$$}$$

Question 1

A scuba diver's tank contains 12.0 L of air at a pressure of 200 atm and a temperature of 27°C. If the diver releases air from the tank until the pressure inside is 1.00 atm at the same temperature, what volume of air at 1.00 atm and 27°C will the diver have available to breathe? (Assume ideal gas behavior.)

Accepted Answer

$$V_2$$ = \frac{$$P_1$$ $$V_1$$}{$$P_2$$} = \frac{200 	imes 12.0}{1.00} = 2400\ \mathrm{L}$$

Question 2

Which of the following best explains why a balloon shrinks when placed in a freezer?

Accepted Answer

The average kinetic energy of the gas molecules decreases, causing the pressure and volume to decrease according to Charles's Law.

Question 3

A 0.500 mol sample of nitrogen gas is confined in a 10.0 L container at 300 K. What is the pressure exerted by the gas? (Use $$R = 0.08206\ \mathrm{L\ atm\ $$mol^{-1}$$\ K$$^{-1}$$})$$

Accepted Answer

$$P = \frac{nRT}{V} = \frac{0.500 	imes 0.08206 	imes 300}{10.0} = 1.23\ \mathrm{atm}$$

Question 4

A sample of helium gas occupies 2.00 L at 25°C and 1.00 atm. If the gas is heated to 100°C at constant pressure, what will be its new volume?

Accepted Answer

$$V_2$$ = $$V_1$$ \frac{$$T_2$$}{$$T_1$$} = 2.00 	imes \frac{373}{298} = 2.50\ \mathrm{L}$$

Question 5

Which of the following statements about the kinetic molecular theory is correct?

Accepted Answer

The average kinetic energy of gas particles is directly proportional to the absolute temperature.

Question 6

A mixture of gases contains 2.0 mol of O$$_2$$, 3.0 mol of N$$_2$$, and 1.0 mol of CO$$_2 in a 10.0 L $$container at 300 K. What is the partial pressure of N$$_2$$? ($$R = 0.08206\ \mathrm{L\ atm\ $$mol^{-1}$$\ K$$^{-1}$$})$$

Accepted Answer

$$P_{N_2$} = \frac{n$$_{N_2}$$RT}{V} = \frac{3.0 	imes 0.08206 	imes 300}{10.0} = 7.39\ \mathrm{atm}$$

Question 7

Which of the following best describes the effect of intermolecular forces on the pressure of a real gas compared to an ideal gas under the same conditions?

Accepted Answer

The pressure of a real gas is lower than predicted by the ideal gas law due to attractive intermolecular forces.

Question 8

A 1.00 mol sample of argon gas is confined in a 2.30 L container at 273 K. Calculate the pressure using the van der Waals equation. ($$a = 1.336\ \mathrm{L^2$\ atm\ $$mol^{-2}$$}$$, $$b = 0.0318\ \mathrm{L\ $$mol^{-1}$$}$$, $$R = 0.08206\ \mathrm{L\ atm\ $$mol^{-1}$$\ K$$^{-1}$$})$$

Accepted Answer

$$P = \frac{nRT}{V - nb} - \frac{a n^2$}{V^2$} = \frac{1.00 	imes 0.08206 	imes 273}{2.30 - 1.00 	imes 0.0318} - \frac{1.336 	imes 1.00^2$}{2.30^2$} = 9.63\ \mathrm{atm}$$

Question 9

Which of the following conditions would cause a gas to deviate most from ideal behavior?

Accepted Answer

Low temperature and high pressure

Question 10

A 5.00 g sample of an unknown gas occupies 2.50 L at 1.00 atm and 25°C. What is the molar mass of the gas? ($$R = 0.08206\ \mathrm{L\ atm\ $$mol^{-1}$$\ K$$^{-1}$$})$$

Accepted Answer

$$M = \frac{mRT}{PV} = \frac{5.00 	imes 0.08206 	imes 298}{1.00 	imes 2.50} = 48.9\ \mathrm{g\ $$mol^{-1}$$}$$

Question 11

Hydrogen gas is collected over water at 25°C and a total pressure of 758.2 Torr. The vapor pressure of water at 25°C is 23.8 Torr. What is the partial pressure of the hydrogen gas?

Accepted Answer

$$P_{H_2$} = P_{total} - P_{H_2O} = 758.2 - 23.8 = 734.4\ \mathrm{Torr}$$

Question 12

Which of the following best describes the process of effusion?

Accepted Answer

The process by which a gas escapes from a container through a small hole into a vacuum.

Question 13

Calculate the ratio of the rates of effusion of helium (He, $$M = 4.00\ \mathrm{g\ $$mol^{-1}$$}) to $$oxygen (O$$_2$$, $$M = 32.00\ \mathrm{g\ $$mol^{-1}$$}) at $$the same temperature.

Accepted Answer

$$	ext{Rate}_{He}/	ext{Rate}_{O_2$} = \sqrt{\frac{M_{O_2$}}{M_{He}}} = \sqrt{\frac{32.00}{4.00}} = 2.83$$

Question 14

A 30.0 L sample of hydrogen gas at STP reacts with excess oxygen to form water according to the reaction: $$2H_2(g) + O_2$(g) \rightarrow 2H_2O(l). $$What mass of water is produced?

Accepted Answer

$$n_{H_2$} = \frac{30.0}{22.4} = 1.34\ \mathrm{mol}$$; \quad m_{H_2O} = 1.34\ \mathrm{mol} 	imes 18.0\ \mathrm{g\ $$mol^{-1}$$} = 24.1\ \mathrm{g}$

Question 15

Which of the following is the correct expression for the density of a gas using the ideal gas law?

Accepted Answer

$$d = \frac{MP}{RT}$$

Question 16

A hospital stores oxygen gas at 300 K in a 50.0 L tank at a pressure of 150 atm. How many moles of oxygen are present? ($$R = 0.08206\ \mathrm{L\ atm\ $$mol^{-1}$$\ K$$^{-1}$$})$$

Accepted Answer

$$n = \frac{PV}{RT} = \frac{150 	imes 50.0}{0.08206 	imes 300} = 305\ \mathrm{mol}$$

Question 17

Which of the following best explains why the molar volume of a real gas is greater than that predicted by the ideal gas law at high pressures?

Accepted Answer

The finite volume of gas particles becomes significant at high pressures.

Question 18

A gas mixture contains 0.50 mol He, 0.30 mol Ne, and 0.20 mol Ar. If the total pressure is 1.00 atm, what is the partial pressure of Ne?

Accepted Answer

$$P_{Ne} = X_{Ne} P_{total} = \frac{0.30}{1.00} 	imes 1.00 = 0.30\ \mathrm{atm}$$

Chapter 5: Gases – Properties, Laws, and Molecular Theory

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