Being under a pressure of 3 * 10 to the fifth degree Pa, the air expanded isobarically and at the same time a work

Being under a pressure of 3 * 10 to the fifth degree Pa, the air expanded isobarically and at the same time a work of 165 kJ was performed. determine the increase in gas volume

P = 3 * 10 ^ 5 Pa.

A = 165 kJ = 165000 J.

ΔV -?

Isobaric is a gas process in which the gas pressure P does not change, that is, throughout the entire process it remains unchanged.

In this process, the work of gas A is determined by the formula: A = P * ΔV, where ΔV is the change in gas volume.

ΔV = A / P.

ΔV = 165000 J / 3 * 10 ^ 5 Pa = 0.55 m3.

Answer: with isobaric expansion, the air volume increased by ΔV = 0.55 m3.