1. Initial state

Let the initial state be $A(P_0,V_0,T_0)$.

For an ideal gas, $P_0V_0=nRT_0$

The process consists of three steps:

• $A \to B$: isothermal expansion from $V_0$ to $4V_0$
• $B \to C$: isobaric compression to volume $V_0$
• $C \to A$: isochoric heating back to the initial state

We need the net heat exchanged in the whole cyclic process.

2. Step 1: Isothermal expansion $A \to B$

Since temperature is constant for an ideal gas, $Q_{AB}=W_{AB}=nRT_0\ln\left(\frac{4V_0}{V_0}\right)$ $Q_{AB}=P_0V_0\ln 4$ Using $\ln 4=2\ln 2$, $Q_{AB}=2P_0V_0\ln 2$

Also, at point $B$: $P_BV_B=P_0V_0$ with $V_B=4V_0$, so $P_B=\frac{P_0V_0}{4V_0}=\frac{P_0}{4}$ Thus state $B$ is $\left(\frac{P_0}{4},4V_0,T_0\right)$.

3. Step 2: Isobaric compression $B \to C$

Pressure remains constant at $P=\frac{P_0}{4}$ Volume changes from $4V_0$ to $V_0$.

Work done by gas: $W_{BC}=P\,(V_C-V_B)=\frac{P_0}{4}(V_0-4V_0)$ $W_{BC}=\frac{P_0}{4}(-3V_0)=-\frac{3}{4}P_0V_0$

Now find temperature at $C$: $T_C=\frac{P_CV_C}{nR}=\frac{(P_0/4)V_0}{nR}=\frac{T_0}{4}$ So, $\Delta T_{BC}=T_C-T_B=\frac{T_0}{4}-T_0=-\frac{3T_0}{4}$

For an isobaric process, the heat is $Q_{BC}=nC_P\Delta T$ But since options are independent of gas type, it is easier to use the first law over the full cycle. Still, for this step alone, internal energy change depends on $C_V$, so we avoid writing $Q_{BC}$ explicitly.

4. Step 3: Isochoric heating $C \to A$

Volume is constant at $V_0$, so $W_{CA}=0$ The system returns from $T_C=T_0/4$ to $T_0$. Again, $Q_{CA}=\Delta U_{CA}$ depends on the gas type if treated separately.

5. Use cyclic property

Since the gas returns to its initial state, over the complete cycle: $\Delta U_{\text{cycle}}=0$ Hence, $Q_{\text{net}}=W_{\text{net}}$

So we only need total work done by the gas.

Total work: $W_{\text{net}}=W_{AB}+W_{BC}+W_{CA}$ $W_{\text{net}}=2P_0V_0\ln 2-\frac{3}{4}P_0V_0+0$ $W_{\text{net}}=P_0V_0\left(2\ln 2-\frac{3}{4}\right)$

Therefore, $Q_{\text{net}}=P_0V_0\left(2\ln 2-0.75\right)$

6. Compare with options

This matches:

Option B: $P_0V_0(2\ln 2-0.75)$

7. Verification with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.