Gas at a pressure of 32 kPa and a temperature of 290 degrees K occupies a volume of 87 liters.

Gas at a pressure of 32 kPa and a temperature of 290 degrees K occupies a volume of 87 liters. Find the volume of gas under normal conditions.

Given:

P1 = 32 kPa = 32 * 10 ^ 3 Pascal – initial gas pressure;

T1 = 290 degrees Celsius = 563 degrees Kelvin – the initial gas temperature;

V1 = 87 liters = 0.087 m ^ 3 – initial gas volume;

P2 = 10 ^ 5 Pascals – normal pressure;

T2 = 0 degrees Celsius = 273 degrees Kelvin – normal temperature.

It is required to determine the volume of gas under normal conditions V2 (liter).

P1 * V1 / T1 = P2 * V2 / T2, hence:

V2 = P1 * V1 * T2 / (T1 * P2) = 32 * 10 ^ 3 * 0.087 * 273 / (10 ^ 5 * 563) =

= 760/56300 = 0.013 m ^ 3 = 13 liters.

Answer: under normal conditions, the gas will occupy a volume equal to 13 liters.