Post on 11-Jan-2016
description
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