GAS LAW PROBLEMS:
BOYLE’S LAW
1) 150.0 mL of a gas sample has a pressure of 745 torr. If the gas was transferred to a 250.0 mL container, what
would its pressure become? (assume constant T)
V1 = 150.0 mL
P1 = 745 torr
V2 = 250.0 mL
P2 = ?
P1V1 = P2V2
P2 =
P1 V1
(745 torr)(150.0 mL)
=
= 447 torr
V2
250.0 mL
2) The pressure inside a 3.00 L cylinder of gas is 1.080 atm. If the gas was compressed to a pressure of 400.0 kPa,
what would its volume become? (assume constant T)
V1 = 3.00 L
P1 = 1.080 atm
1 atm
P2 = 400.0 kPa(
) = 3.948 atm
101.325 kPa
V2 = ?
P1V1 = P2V2
V2 =
P1 V1
(1.080 atm)(3.00 L)
=
= 0.821 L
P2
3.948 atm
CHARLES’ LAW
1) The temperature of a 225 mL gas sample is 23.2°C. What is the temperature of the same gas sample if its
volume increased to 415 mL? (assume constant P)
V1 = 225 mL
T1 = 23.2°C + 273.15 = 296.4 K
T2 = ?
V2 = 415 mL
V1
V2
=
T1
T2
T2 =
V1 T2 = V2 T1
V2 T1
(415 mL)(296.4 K)
=
= 547 K
V1
225 mL
2) A balloon filled with 100.0 L of a gas has a temperature of 27°C inside a building. If the balloon was taken outside
on a winter day with the temperature at -15°C, what would the volume of the balloon become? (assume constant
P)
V1 = 100.0 L
T1 = 27°C = 300. K
T2 = -15°C = 258 K
V2 = ?
V1
V2
=
T1
T2
V2 =
V1 T2 = V2 T1
V1 T2
(100.0 L)(258 K)
=
= 86.0 L
T1
300. K
COMBINED GAS LAW
1) A 75.00 mL sample of gas has a temperature of 32.25°C and a pressure of 99.20 kPa. What would its volume be
at STP?
V1 = 75.00 mL
T1 = 32.25°C = 305.40 K
P1 = 99.20 kPa
V2 = ?
T2 = 273.15 K
P2 = 101.325 kPa
P1 V1 P2 V2
=
T1
T2
V2 =
P1 V1 T2 = P2 V2 T1
P1 V1 T2
(99.20 kPa)(75.00 mL)(273.15 K)
=
= 65.67 mL
P2 T1
(101.325 kPa)(305.40 K)
2) At 585.2 mmHg and 305 K, the volume of a gas sample is 3.33 L. If the volume became 3.85 L and the
temperature changed to 395 K, what will the pressure of the gas sample become?
P1 V1 P2 V2
=
T1
T2
P1 = 585.2 mmHg
T1 = 305 K
V1 = 3.33 L
V2 = 3.85 L
T2 = 395 K
P2 = ?
P2 =
P1 V1 T2 = P2 V2 T1
P1 V1 T2
(585.2 mmHg)(3.33 L)(395 K)
=
= 656 mmHg
V2 T1
(3.85 L)(305 K)
3) A gas sample has a volume of 1535 mL at 27.2°C and 800.0 torr. If the volume and pressure changed to 825 mL
and 732.5 torr, what will the temperature, in °C, of the gas become?
P1 V1 P2 V2
=
T1
T2
V1 = 1535 mL
T1 = 27.2°C = 300.4 K
P1 = 800.0 torr
V2 = 825 mL
P2 = 732.5 torr
T2 = ?
T2 =
P1 V1 T2 = P2 V2 T1
P2 V2 T1
(732.5 torr)(825 mL)(300.4 K)
=
= 148 K
P1 V1
(800.0 torr)(1535 mL)
T2 = 148 K – 273.15 = -125°C
GAS-LUSSAC’S LAW
1) The gas in a metal can has a pressure of 1.12 atm at 28.0°C. At what temperature, in °C, would the pressure
inside the can become 1.95 atm? (volume is constant since it is a metal can)
P1
P2
=
T1
T2
P1 = 1.12 atm
T1 = 28.0°C = 301.2 K
T2 = ?
P2 = 1.95 atm
T2 =
P1 T2 = P2 T1
P2 T1
(1.95 atm)(301.2 K)
=
= 524 K = 251°C
P1
1.12 atm
2) The pressure inside a football is 13.0 psi at 25.23°C. If the temperature outside during a game is 8.34°C, what
will the pressure inside the football become? (assume volume is constant)
P1
P2
=
T1
T2
P1 = 13.0 psi
T1 = 25.23°C = 298.38 K
T2 = 8.24°C = 281.49 K
P2 = ?
P2 =
P1 T2 = P2 T1
P1 T2
(13.0 psi)(281.49 K)
=
= 12.3 psi
T1
298.38 K
SAMPLE PROBLEMS
1) The pressure of a 450.0 mL sample of gas is 30.2 inHg. What will the pressure become if the volume changes to
525.0 mL? (temperature is constant)
V1 = 450.0 mL
P1 = 30.2 inHg
P2 = ?
V2 = 525.0 mL
P1V1 = P2V2
P2 =
P1 V1
(30.2 inHg)(450.0 mL)
=
= 25.9 inHg
V2
525 mL
2) At -25.2°C and 745 torr, the volume of a gas sample is 240.0 mL. What will the volume of the gas be at STP?
T1 = -25.2°C = 248.0 K
P1 = 745 torr
V1 = 240.0 mL
V2 = ?
T2 = 273.15 K
P2 = 760 torr
P1 V1 P2 V2
=
T1
T2
V2 =
P1 V1 T2 = P2 V2 T1
P1 V1 T2
(745 torr)(240.0 mL)(273.15 K)
=
= 259 mL
P2 T1
(760 torr)(248.0 K)