How many degrees is it necessary to cool a gas with a temperature of 273 K and a volume of 21 liters

How many degrees is it necessary to cool a gas with a temperature of 273 K and a volume of 21 liters, so that its volume at the same pressure becomes equal to 3 liters?

Given:

T1 = 273 ° Kelvin – initial gas temperature;

V1 = 21 liters = 0.021 m3 (cubic meter) – initial gas volume;

V2 = 3 liters = 0.003 m3 – final gas volume.

It is required to determine dt (degree Kelvin) – how many degrees the gas must be cooled.

According to the condition of the problem, the process takes place at constant pressure. Then:

V1 / T1 = V2 / T2;

T2 = V2 * T1 / V1 = 0.003 * 273 / 0.021 = 273/7 = 39 ° Kelvin.

Then:

dt = T1 – T2 = 273 – 39 = 234 ° Kelvin.

Answer: the gas must be cooled by 234 ° Kelvin.