Question

An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change ΔS of the gas is

Accepted Answer

The correct answer is (B) mR ln(P1/P2). For a reversible isothermal process on an ideal gas, ΔS = mR ln(V2/V1). Since PV=constant at fixed T, V2/V1=P1/P2, so ΔS = mR ln(P1/P2). (This is fully consistent with ΔS=Q/T1 for a reversible process — the 'heat loss Q' given in the problem doesn't add new information, it's a consequence, not an independent input; option (D) is wrong because entropy of the GAS itself does change even though the universe's net entropy change is zero for a reversible process.)

Explanation