Question

If one mole of an ideal gas at $(P_1, V_1)$ is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure. This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B → C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net work done by the gas is equal to :

• A. RT ln 2 - $\frac{1}{2(\gamma - 1)}$
• B. $-\frac{RT}{2(\gamma - 1)}$
• C. RT ln 2
• D. 0

Hint:

For adiabatic processes, the work done can be expressed as $\frac{\Delta(PV)}{1-\gamma}$. Remember that $PV = nRT$ for an ideal gas.

Answer 1

The Correct Option is A

Step 1: $W_{AB}$ (isothermal) $= nRT \ln(\frac{P_1}{P_2}) = RT \ln(2)$.
Step 2: $W_{BC}$ (isochoric) $= 0$.
Step 3: $W_{CA}$ (adiabatic) $= \frac{P_A V_A - P_C V_C}{1 - \gamma}$. Since $P_C = \frac{P_1}{4}$ and $V_C = V_B = 2V_1$: \[ W_{CA} = \frac{P_1 V_1 - (\frac{P_1}{4})(2V_1)}{1 - \gamma} = \frac{\frac{1}{2}P_1 V_1}{1 - \gamma} = -\frac{RT}{2(\gamma - 1)} \]
Step 4: Total work $W = RT \ln 2 - \frac{RT}{2(\gamma - 1)}$.