The Behavior of Gases—Review Guide
Matching
Column A
1. ideal gas constant (R) C
2. Boyle’s Law
H
3. Dalton’s Law of partial pressures
4. ideal gas law
G
5. combined gas law
6. Charles’s Law
7. Diffusion
8. partial pressure
Column B
a. the volume of a fixed mass of gas is directly
proportional to the Kelvin temperature if the
pressure is kept constant
b. at constant volume and temperature, the total
pressure exerted by a mixture of gases is equal
to the sum of the partial pressures of the
component gases
F
A
E
D
B
c.
0.0821
L x atm
mol x K
d. the contribution of each gas in a mixture
make to the total pressure
e. a gas tends to move to an area of lower
concentration until the concentration is uniform
throughout
f.
P1V 1 P 2V 2

T1
T2
g.
PV = nRT
h. For a given mass of gas at constant
temperature, the volume of gas varies inversely
with the pressure.
Multiple Choice:
9. As the temperature of a fixed volume of gas increases, the pressure will:
a) vary inversely
b) decrease
c) be unchanged
d) increase.
10. A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon
dioxide. What is the partial pressure of oxygen at 101.3 kPa total pressure if PHe = 84.0 kPa
and PCO2 = 0.10 kPa?
a) 10.3 kPa
b) 17.2 kPa
c) 34.4 kPa
d) 185.4 kPa
11. Increasing the volume of a given amount of gas at constant temperature caused the
pressure to decrease because
a) the molecules are striking a larger area with the same force
b) there are fewer molecules
c) the molcules are moving more slowly
d) there are more molecules
12. The volume of a gas is doubled while the temperature is held constant. The pressure of
the gas:
a) remains unchanged
b) is reduced by one half
c) is doubled
d) depends on the kind of gas
13. As the temperature of a gas in a balloon decreases
a) the volume increases
b) the pressure increases
c) the average kinetic energy of the gas particles decreases
d) all of the above are true
14. The volume of a gas is increased from 0.5 L to 4.0 L while the temperature is held
constant. The pressure of the gas
a) increases by a factor of four
b) decreases by a factor of eight
c) increases by a factor of eight
d) increases by a factor of two
15. A gas occupies 40.0 mL at -123 °C. What volume does it occupy at 27 °C, assuming
pressure is constant?
a) 182 mL
b) 8.80 mL
c) 80.0 mL
d) 20.0 mL
16. A gas occupies a volume of 0.2 L at 25 kPa. What volume will the gas occupy at 2.5 kPa?
a) 4 L
b) 20 L
c) 2 L
d) 0.02 L
17. Which of these changes would NOT cause an increase in the pressure of a contained
gas?
a) another gas is added to the container
b) additional amounts of the same gas are added to the container
c) the temperature is increased
d) the gas is moved to a larger container
18. If a balloon containing 1000. L of gas at 50. °C and 101.3 kPa rises to an altitude where
the pressure is 27.5 kPa and the temperature is 10. °C, its volume there is?
(1000.L) x(101.3kPa) (V 2) x(27.5kPa)

323K
283K
V2= 3227 L (sig figs ~ 3200 L)
19. Under the same conditions of temperature and pressure, what is the relative rate of
effusion of chlorine gas compared to helium gas? Give your answer with two significant
figures.
RateCl

RateHe
MMHe
4.0

 0.3
MMCl
35.45
20. A 10.0 g mass of krypton occupies 15.0 L at a pressure of 156 kPa. Find the volume of
the krypton when the pressure is increased to 215 kPa at the same temperature.
P1V1 = P2V2
(156kPa)(15.0L) = (215kPa)(V2) V2 = (156)(15) / (215) = 10.9 L
21. A gas has a volume of 550. mL at a temperature of –55.0 °C. What volume will the gas
occupy at 30.0 °C, assuming constant pressure?
V1 V 2

=
T1 T 2
550ml
V2

218 K 303K
V2= (550)(303) / (218) = 764 ml
22. A 250. mL sample of fluorine gas is collected over water at 25.0 °C and 760. torr
pressure. What is the pressure in torr of the dry fluorine gas alone? (Vapor pressure of
water at 25.0 °C = 23.8 torr)
PF = Patm – PH20 = 760 torr – 23.8 torr = 736 torr
23. What volume will 159.7g N2 gas occupy at 54.0 °C and 675 mm Hg?
P = 675 mmHg x
1atm
 0.89 atm
760mmHg
V= ?
1mole
= 5.70 moles
28.02 g
R = 0.0821 (L x atm / mol x K)
T = 54.0 C + 273 = 327 K
n= 159.7g x
PV= nRT
V=
nRT (5.70moles) x(0.0821atmL / mK ) x(327 K )
= 172L

P
(0.89atm)