Chemistry 11 (HL)
Unit 3 / IB Topic 1.4
Quantitative Chemistry 7
Gas Laws: Relationships for Fixed Masses of Gases
PAST PAPER GAS LAW QUESTIONS
Answers – Past Paper Questions
1.
A
2. A
3. B
4.
D
6.
C
7. D
8. B
9.
overall there will be no change to the pressure;
double absolute temperature and the pressure doubles;
double volume and the pressure halves;
5. B
OR
Use PV = nRT, Since n and R are constant;
V and T are both doubled;
P will remain unchanged;
MORE PRACTICE PROBLEMS
1. A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure
is increased to 60.0 mm Hg?
P1V1 = P2 V2
(40.0 mm Hg)(12.3 L) = (60.0 mmHg)V2
V2 = 8.20 L
2. If the pressure on a gas is decreased by one-half, how large will the volume change be?
Let P1 = 1 and V1 = 1
P1V1 = P2 V2
(1)(1) = (1/ 2)V2
V2 = 2
Therefore, the new volume will be TWICE the original volume.
3. Two hundred liters of helium at a certain pressure and 28.0 °C are compressed to 70.0 L and
stored in a tank with an internal pressure of 600.0 kPa at 28.0˚C. Find the original pressure of
the gas.
P1V1 = P2 V2
P1(200 L) = (600.0 kPa) (70.0 L)
P1 = 210 kPa
p. 1
Chemistry 11 (HL)
Unit 3 / IB Topic 1.4
4. You are now wearing scuba gear and swimming under water at a depth of 66.0 ft. You are
breathing air at 3.00 atm and your lung volume is 10.0 L. Your scuba gauge indicates that your
air supply is low so, to conserve air, you make a terrible and fatal mistake: you hold your
breath while you surface. What happens to your lungs? Why?
Assume the pressure at the surface is standard pressure (1.00 atm):
P1V1 = P2 V2
(3.00 atm) (10.0 L) = (1.00 atm) V2
V2 = 30.0 L
Lungs would inflate too rapidly and explode. (You should exhale while ascending to remove
air and reduce volume.)
5. Calculate the new temperature when 2.00 L at 20.0 °C is compressed to 1.00 L.
V1
V
= 2
T1
T2
2.00 L 1.00 L
=
293 K
T2
T2 = 147 K
T2 = -127ºC
6. Carbon dioxide is usually formed when gasoline is burned. If 30.0 L of CO2 is produced at a
3
temperature of 1.00 x 10 °C and allowed to reach room temperature (25.0 °C) without any
pressure changes, what is the new volume of the carbon dioxide?
V1
V
= 2
T1
T2
V2
30.0 L
=
1273 K
298 K
V2 = 7.02 L
7. A gas syringe contains 42.3 milliliters of a gas at 98.15 °C. Determine the final temperature
needed in order to reduce the volume to 29.00 mL.
V1
V
= 2
T1
T2
42.3 mL
29.00 mL
=
371.15 K
T2
T2 = 254.5 K
T2 = -18.5 ºC
p. 2
Chemistry 11 (HL)
Unit 3 / IB Topic 1.4
8. If the Kelvin temperature of a gas is doubled, the volume of the gas will increase by ____. Is
your answer the same if the question states “if the Celsius temperature of the gas is doubled,
the volume of the gas will increase by ____”? Explain.
Let V1 = 1 L and T1 (in K) = 1 K:
V1
V
= 2
T1
T2
V
1L
= 2
1K
2K
V2 = 2 L
The volume doubles when the absolute temperature (i.e. temp in degrees K) doubles.
If the Celsius temperature is doubled, consider:
Let V1 = 1 L and T1 (in ºC) = 1ºC
If the Celsius temperature doubles, T2 = 2ºC
V1
V
= 2
T1
T2
V2
1L
=
(1+ 273) K
(2 + 273) K
V2 = 1.007 L
The volume increase is negligible when the temperature in ºC increases.
9. 3.50 liters of a gas at 727.0 K will occupy how many liters at 153.0 K?
V1
V
= 2
T1
T2
V2
3.50 L
=
727.0 K 153.0 K
V2 = 0.737 L
10. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from
20.0 °C to 30.0 °C.
P1
P
= 2
T1
T2
P2
1 atm
=
293 K
303 K
P2 = 1.03 atm
Therefore the pressure CHANGE = 1.03 atm – 1 atm = 0.03 atm
p. 3
Chemistry 11 (HL)
Unit 3 / IB Topic 1.4
11. A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard temperature?
P1
P
= 2
T1
T2
P2
0.370 atm
=
323 K
273 K
P2 = 0.313 atm
12. The pressure in a scuba tank at a certain temperature is 130.0 atm. When it cools to 25.0˚C,
the pressure is 3059 kPa. What was the original temperature of the tank?
Convert P1 from atm to kPa:
P1 = 130.0 atm x
101 kPa
= 13130 kPa
1 atm
P1
P
= 2
T1
T2
I think this is an unrealistic scenario!
13130 kPa
3059 kPa
=
T1
298 K
T1 = 1279 K
T1 = 1006 ºC
13. 500.0 liters of a gas are prepared at 700.0 mm Hg and 200.0 °C. The gas is placed into a tank
under high pressure. When the tank cools to 20.0 °C, the pressure of the gas is 30.0 atm.
What is the volume of the gas?
Convert P1 from mm Hg to atm:
P1 = 700.0 mm Hg x
1 atm
= 0.921 atm
760 mm Hg
P1V1
PV
= 2 2
T1
T2
(0.921 atm) (500.0 L) (30.0 atm) V2
=
473 K
293 K
V2 = 9.51 L
14. What is the final pressure of a gas sample at 760 mm Hg that is subjected to a temperature
change from 22.0 °C to 30.0 °C and a volume change from 400.0 mL to 867.0 mL?
p. 4
Chemistry 11 (HL)
Unit 3 / IB Topic 1.4
P1V1
PV
= 2 2
T1
T2
(760 mm Hg) (400.0 mL) P2 (867.0 mL)
=
295 K
303 K
P2 = 360 mm Hg
15. A gas sample occupies 3.25 liters at 24.5 °C and 1825 mm Hg. Determine the temperature at
which the gas will occupy 4250 mL at 1.50 atm.
Convert P1 from mm Hg to atm:
P1 = 1825 mm Hg x
1 atm
= 2.40 atm
760 mm Hg
P1V1
PV
= 2 2
T1
T2
(2.40 atm) (3.25 L) (1.50 atm) (4.25 L)
=
297.5 K
T2
T2 = 243 K
T2 = -30ºC
16. If the absolute temperature of a given quantity of gas is doubled and the pressure tripled, what
happens to the volume of the gas?
Let P1 = 1 atm, V1 = 1 L and T1 = 1 (in Kelvin)
Note: “Absolute temperature” refers to temperature in Kelvin degrees.
P1V1
PV
= 2 2
T1
T2
(1 atm) (1 L) (3 atm) V2
=
1K
2K
V2 = 2/3
The new volume of the gas will be 2/3 of the original volume.
17. The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume of its
container is increased by moving a piston from 85.0 mL to 350.0 mL. What would the final
temperature be if the original temperature was 90.0 °C?
p. 5
Chemistry 11 (HL)
Unit 3 / IB Topic 1.4
P1V1
PV
= 2 2
T1
T2
(1200.0 mm Hg) (85.0 mL) (850.0 mm Hg) (350.0 mL)
=
363 K
T2
T2 = 1058 K
T2 = 786ºC
Answers – More Practice Problems
1.
2.
volume doubles
3.
210 kPa
4.
30 L
8.20 L
5.
-127˚C
6.
7.02 L
7.
-18.5˚C
8.
2x; no
9.
0.737 L
10. 0.03 atm
11.
0.313 atm
12.
1.00 x 10 ˚C
13. 9.51 L
14.
360 mm Hg
15.
243 K = -30˚C
16. 2/3 less
17.
786 ˚C
3
p. 6