1 m3 of air at a temperature of 0 ° C is in a cylinder at a pressure of 2 • 10 ^ 5 Pa.

1 m3 of air at a temperature of 0 ° C is in a cylinder at a pressure of 2 • 10 ^ 5 Pa. What work will be done when it is heated isobaric at 10 ° C?

Given
V1 = 1m ^ 3
P = 2 * 10 ^ 5Pa
t1 = 0C
t2 = 10C
Let’s move on to the absolute temperature scale:
T1 = t1 + 273 = 0 + 273 = 273K
T2 = t2 + 273 = 10 + 273 = 283K
During the isobaric process, work is determined from the equation:
A = P * (V2 – V1), where
V2 – V1 – volume change.
To determine V2, let’s use Gay-Lussac’s law:
V1 / T1 = V2 / T2
V2 = V1 * (T2 / T1)
A = p * [V1 * (T2 / T1) – V1] = p * V1 * (T2 / T1 – 1) = 2 * 10 ^ 5 * 1 * (283/273 – 1) = 7326J
A = 7326J