Question

A gas is expanded from volume V₀ to 2V₀ under three different processes. Process 1 is isobaric, process 2 is isothermal and process 3 is isothermal and process 3 is adiabatic. Let DU₁, DU₂ and DU₃ be the change in internal energy of the gas in these three processes. Then –

• A. DU₁>DU₂>DU₃
• B. DU₁<DU₂<DU₃
• C. DU₂<DU₁<DU₃
• D. DU₂<DU₃<DU₁

Answer 1

Answer: (A)

DU₁>DU₂>DU₃

Answer 2

Process 2 is an isothermal process

Hence, DU₂ = 0

Process 1 is an isobaric (P = constant) expansion.

Hence, temperature of the gas will increase

or DU₁ = positive

Process 3 is an adiabatic expansion. Hence, temperature will decrease

or DU₃ = negative

Therefore, DU₁ > DU₂ > DU₃ is the correct option.