Discussion
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
Answer: B
The entropy change is \(mR\ln(P_{1}/P_{2})\).
For an ideal gas:
\[ \Delta S = mc_p\ln\frac{T_2}{T_1} - mR\ln\frac{P_2}{P_1} \]
The process is isothermal, so the temperature term vanishes and \(\Delta S = mR\ln(P_{1}/P_{2})\). Since heat is lost, the gas is being compressed (P2 > P1) and \(\Delta S\) is negative, as it must be for a reversible process with heat leaving.
\(mR\ln(P_{2}/P_{1})\) has the wrong sign. Subtracting Q/T1 counts the heat transfer twice, because for a reversible process the expression already equals the heat transfer over T. Zero applies to a reversible adiabatic process, not to an isothermal one.
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