Question

Charles's Law
A sample of helium gas at a pressure of 0.905 atm and a temperature of 230°C, occupies a volume of 615mL. If the gas is cooled at constant pressure until it's volume is 523mL, the temperature of the gas sample will be ____°C

A sample of oxygen gas at a pressure of 960 mm Hg and a temperature of 81°C, occupies a volume of 14.9 liters. I the gas is heated at constant pressure at a temperature of 123°C, the volume of the gas sample will be ____ L.

Answer 1

Charles's Law: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be directly related.

V/T = constant

where:

V is the volume of the gas

T is the temperature of the gas (measured in Kelvin).

k is a constant.

1.

V1/T1 = V2/T2

V1T2 = V2T1

T2 = V2T1 / V1

T1 = 230 °C = (230 + 273) K = 503 K

V1 = 615 mL

V2 = 523 mL

T2 = ?

Substituting the values, we get

T2 = V2T1 / V1

T2 = (523 mL) (503 K) / (615 mL)

T2 = 428 K

T2 = (428 - 273) °C

T2 = 155 °C

2.

1.

V1/T1 = V2/T2

V1T2 = V2T1

V2 = V1T2 / T1

T1 = 81 °C = (81 + 273) K = 354 K

V1 = 14.9 L

T2 = 123 °C = (123 + 273) K = 396 K

V2 = ?

Substituting the values, we get

V2 = V1T2 / T1

V2 = (14.9 L) (396 K) / (354 K)

V2 = 16.7 L

T2 = (428 - 273) °C

T2 = 155 °C