Question

An ideal diatomic gas, with molecular rotation but without any
molecular oscillation, loses a certain amount of energy as heat Q.
Is the resulting decrease in the internal energy of the gas greater if the loss occurs in a constant-volume process or in a constant-pressure process?

Answer 1

Answer: constant volume process

Explanation: for constant volume process

ΔE=Q=nCνΔT

For constant pressure

ΔE=Q-W

ΔE=nCρΔT-PΔV

For an ideal diatomic gas,Cν=5/2,Cρ=7/2

since Q is the same for both processes

Qν=Qρ

∴nCνΔTν=nCρΔTρ

cancelling out n and R

5/2ΔTν=7/2ΔTρ

∴ΔTν=7/5ΔTρ

change in internal energy is directly proportional to ΔT

since ΔTν>ΔTρ

Decrease in the internal energy of the gas is greater for constant volume processes