Chapter 12 (Practice Test)

The Gas Laws

1. Identify the four factors that affect gases.

a. Pressure P

b. Volume V

c. Number of moles n

d. Temperature T (Kelvin)

2. PTotal = P1 + P2 + P3 + . . . Pn

A

B

B

A

m

m

rate

rate

P1V1 P2V2

T2T1

=

V1 V2

T2T1

=

3.

4. P1V1 P2V2=

PV = n RT5.

1

2

2

1

m

m

or6.

P1 P2

T2T1

=

7.

8.

Dalton’s Law of Partial Pressure

(F)Combined Gas Law

(D)Boyle’s Law (A)

Ideal Gas Law (E)

Graham’s Law (G)

Charles’s Law (B)

Gay-Lussac’s Law (C)

9. Identify STP:

Standard temperature _____________________

Standard pressure ________________________

0°C

101.3 kPa 760 mmHg 1 atm.

273 K

10.At what point during the process of cooling do the gas laws become ineffective?

When the gas condenses into a liquid.

Gas laws don’t work for substances that are not gases.

Gas laws deviate from ideal behavior at high pressures and low temperatures.

11.Use the Kinetic Theory to describe why a decrease in the volume of a gas will cause and increase in pressure as stated by Boyle’s Law.

Particles moving at the same speed (same temp.) in a smaller space will hit the walls more often. Since pressure is the result of collisions with the walls, the pressure will increase with more collisions.

12. Which law shows an inverse relationship between pressure and volume?

13. Which law compares the rates of two gases?

14. Identify the Gas Law used if the pressure was held constant.

15. What would happen to the pressure exerted by a gas if the volume is decreased and temperature is held constant?

(a) Boyle’s Law: P1V1 = P2V2

(c) Graham’s Law:A

B

B

A

m

m

rate

rate

1

2

2

1

m

m

or

(b) Charles’s Law:P1 P2

T2T1

=V1 V2

(a) Pressure would increase.PV PV

16. In order to solve problems using Charles’s Law or the Combined Gas Law the temperature must be measured in:

17. Which of these changes would NOT cause an increase in the pressure of a gaseous system?

a) The container is made larger.

b) Additional amounts of the same gas are added to the container.

c) The temperature is increase.

d) Another gas is added to the container.

(c) Kelvin

Pressure would decrease not increase.

PVPV

18. What happens to the temperature of a gas when it is expanded quickly?

19. What happens to the pressure of a gas inside a container if the temperature of the gas is lowered?

20. As the temperature of the gas in a balloon decreases ____.

(c) The temperature decreases.

(c) The pressure decreases.P1 P2

T2T1

=

Direct relationship so… P and V should decrease also.

This is how they make liquid air. (Joule-Thomson expansion engine)

T2T1

=V1 V2P1 P2

T2T1

=

Temp. measures KEave

(b) Lower temp. = lower KEave.

21. If a balloon is squeezed, what happens to the air pressure within the balloon.

22. Which of the following molecules would have the greatest velocity if each gas had the same kinetic energy?

a. Br2 b. Cl2 c. NH3 d. H2 e. Ar

(a) Pressure increases.

The smallest molecule travels the fastest.

P V

A

B

B

A

m

m

rate

rate

1

2

2

1

m

m

or

79.90x 2

159.80

35.45x 2

70.90

14.01x 1

14.01

1.01 x 3+ 3.03 = 17.04

1.01x 22.02 39.95

Chapter 12 (Practice Test)

The Gas Laws(Problem Section)

1. A gas sample has a volume of 300.0 L when under a pressure of 55.0 kPa. What is the new volume if the pressure is increased to 165 kPa while the temperature is held constant.

V1 = 300.0 L

55.0 kPa

165 kPa

V2

(55.0 kPa)

165 kPa

Calculator 55 x 300 ÷ 165 = 100 L

P1V1 = P2V2

(300.0 L)

Answer w/ 3 Sig. Figs. V2 = 100. L or 1.00 x 102 L

P1 =

V2 =

P2 =Divide both sides by P2

P2P2

V2 =

______ L

2. A quantity of gas occupies a volume of 804 mL at a temperature of 27 °C. At what temperature will the volume of the gas be 402 mL, assuming there is no change in pressure?

V1 = 804 mL

27 °C

402 mL

V2

Calculator 402 x 300 ÷ 804 = 150 K

V1 =

Answer w/ 3 Sig. Figs. T2 = 150. K or 1.50 x 102 K

T1 =

V2 =

T2 =

T2T1

Solve for T2 so cross multiply then divide

______ K

Convert to KELVINS!

+ 273 = 300. K

(402) 804=

T2(300)804 804

3. The gas in a closed container has a pressure of 3.00 x 102 kPa at 30C. What will the pressure be if the temperature is lowered to –172.0C ?

P1 = 3 x 102 kPa

30 °CP2

Calculator 101 x 300 ÷ 303 = 100. kPa

P1 =

Answer w/ 3 Sig. Figs.

= 100. kPa or 1.00 x 102 kPa

T1 =

P2 =

T2 =

T2T1

Solve for P2 ______ kPa

Convert to KELVINS!

+ 273 = 303. K

P2 300=

101303

(101) (101)-172.0 °C + 273 = 101.0 KP2

=

(300)

303

(101)

= 300. kPa

4. A 100 L sample of gas is at a pressure of 80 kPa and a temperature of 200 K. What volume does the same sample of gas occupy at STP ?

V1 =P2

Calculator 273 x 80 x 100 ÷ 101.3 ÷ 200 = 107.79862 L

P1 =

Answer w/ 1 Sig. Fig.

= 100 L or 1 x 102 L

P1 =

V2 =

T2 =

T2T1

Solve for V2

______ L

200 K

80 kPa

101.3 (100)= 273200

(80) V2

273 K=

100 L

T1 =

P2 = 101.3 kPa

V1 V2

(273)(273)

(101.3)(101.3)

V2 (100)

(200)

(80)(273)

(101.3)

5. What is the pressure of a 5.0 L container which contains 3.0 moles of a gas at a temperature of 0C? Note: Use Ideal gas law. R=8.31 (kPa x L)/(mol x K)

5. What is the pressure of a 5.0 L container which contains 3.0 moles of a gas at a temperature of 0C? Note: Use Ideal gas law. R=8.31 (kPa x L)/(mol x K)

P =n

Calculator 3 x 8.31 x 273 ÷ 5 = 1,361.178 kPa

P =

Answer w/ 2 Sig. Figs.

1,400 kPa or 1.4 x 103 kPa

V =

T =

T

Solve for P

0°C

3.0 mol

5.0 L(5) =

____

n =

R = 8.31 kPa•Lmol•K

V R

= 273 K

kPa

P (3) (8.31)(273)

(5) =P (3) (8.31)(273)

(5) (5)=P

P =

6. What volume does 14.0 g of Hydrogen gas (H2) take up at a temperature of 30C and a pressure of 120 kPa? Note: Use Ideal gas law. R=8.31 (kPa x L)/(mol x K)

6. What volume does 14.0 g of Hydrogen gas (H2) take up at a temperature of 30C and a pressure of 120 kPa? Note: Use Ideal gas law. R=8.31 (kPa x L)/(mol x K)

P =n

Calculator 6.93 x 8.31 x 303 ÷ 120 = 145.4104575 L

P =

Answer w/ 2 Sig. Figs. 150 L = 1.5 x 102 L

V =

T =

T

Solve for V

30°C

14.0 g H2

____ V =

120 kPa

n =

R = 8.31 kPa•Lmol•K

V R

+ 273

L

(120) (6.93) (8.31)(303)

V =(120) (6.93) (8.31)(303)

(120) (120)=

V =

xg H2

mol H2

2.02

1= 6.93 mol

= 303 K

1.01x 22.02

1

HHydrogen

1.01

7. Two gases CH4 and SO2 are released at the same time from opposite ends of the room. You are in the center of the room. Which gas will reach you first? (hint: mass of S=32; C=12; O=16; and H=1.0)

Part A:Circle one: Methane (CH4) or Sulfur dioxide (SO2)

Part B: How much quicker will the gas reach you? v1 =

CH4 is 2 times faster than SO2

m1 =

m2 =

Lighter gases travel faster

SO2

16 g/mol

CH4

v2 =CH4

16

64

SO2= 64 g/mol

1x 4 4 = 16

12x 112 +

16x 2 32 = 64

32x 132 +

A

B

B

A

m

m

rate

rate

1

2

2

1

m

m

or

=1

4=2

8. A balloon contains 3 gases (hydrogen, oxygen, and carbon dioxide). If the balloon occupies a volume of 1.0 L , hydrogen gas has a pressure of 100 kPa, oxygen has a pressure of 85 kPa and carbon dioxide has a pressure of 115 kPa. What is the total pressure of the gases exerted on the balloon?

P =PT =

300 kPa

P =H2

115 kPa

85 kPa=

100 kPa

P =

PT =

100 kPa + 85 kPa+ 115 kPa

PT =

____ kPa

O2

CO2

P +H2

P +O2PCO2

PT

9. Given a 32.0 g sample of methane gas (CH4) , determine the pressure that would be exerted on a container with a volume of 850 cm3 at a temperature of 30 C.

Note: Carbon = 12.0 g/mol and Hydrogen = 1.0 g/mol

P =n

Calculator 2 x 8.31 x 303 ÷ 0.850 = 5,924.54 kPa

P =

Answer w/ 3 Sig. Figs.

5,920 kPa = 5.92 x 103 kPa

V =

T =

T

Solve for P

30°C

32.0 g CH4

850 cm3

(0.850)=

____

n =

R =8.31 kPa•Lmol•K

V R

+ 273

kPa

P (2) (8.31)(303)

P =(0.850) (2) (8.31)(303)

(0.850) (0.850)

=

P =

xg CH4

mol CH4

16

1 = 2 mol

= 303 K

= 850 mL= 0.850

Use Ideal Gas Law when given grams.

L

1000 mL = 1 L so…

6

CCarbon12.01

1

HHydrogen

1.01

12.0x 1

12.0

CH4

1.0x 4 + 4.0 = 16.0 g

10. Given the following balanced equation.N2(g) + 3 H2(g) 2 NH3(g)

What volume of ammonia gas (NH3) would be produced at 95.0 kPa and 70 C if you were given a 28.0 g sample of nitrogen gas (N2)?

P =n

Calculator 2 x 8.31 x 343 ÷ 95.0 = 60.0069 L

P =

Answer w/ 3 Sig. Figs. 60.0 L

V =

T =

T

Solve for V

70°C

28.0 g N2

______V=95.0 kPa

n =

R =8.31 kPa•Lmol•K

V R

+ 273

(95.0) (2) (8.31)(343)

V =(95.0) (2) (8.31)(343)(95.0) (95.0)

=

V =

xg N2

mol N2

28.0

1= 2 mol NH3

= 343 K

Use Ideal Gas Law when given grams.

L

xmol N2

mol NH3

1 2

7

NNitrogen

14.01

N2

14.01x 2

28.02g