Question

An ideal gas exists in a state with pressure $P_0$, volume $V_0$. It is isothermally expanded to 4 times of its initial volume($V_0$), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is

• A. $P_0V_0$ (2ln 2 - 0.75)
• B. $P_0V_0$ (2ln 2 - 0.25)
• C. $P_0V_0$ (In 2 - 0.75)
• D. $P_0V_0$ (In 2 - 0.25)

Answer 1

Answer: (A)

$P_0V_0(2\ln2 - 0.75)$

Answer 2

We have a three‐step cyclic process for an ideal gas:

1. Isothermal Expansion from $(P_0,V_0)$ to $(P_0/4,4V_0)$:

   For an isothermal process, $\Delta U=0$ so

   $$Q_1 = W_1 = P_0V_0 \ln\frac{4V_0}{V_0} = P_0V_0 \ln 4 = P_0V_0 (2\ln2).$$

2. Isobaric Compression at $P = P_0/4$ from $4V_0$ to $V_0$:

   Temperatures:

   Initial: $T_2 = \frac{P_0/4\times 4V_0}{R} = \frac{P_0V_0}{R}$

   Final: $T_3 = \frac{P_0/4\times V_0}{R} = \frac{P_0V_0}{4R}$

   $\Delta T = T_3 - T_2 = -\frac{3P_0V_0}{4R}$.

   First law: $Q_2 = \Delta U + W_2$.

   Work done:

   $$W_2 = P\Delta V = \frac{P_0}{4}(V_0 - 4V_0) = -\frac{3P_0V_0}{4}.$$

   Change in internal energy:

   $$\Delta U = nC_V\Delta T = -\frac{3P_0V_0}{4}\frac{C_V}{R}.$$

   Thus,

   $$Q_2 = -\frac{3P_0V_0}{4}\left(\frac{C_V}{R}+1\right).$$

   Note: We will see that the contribution due to $C_V$ cancels with the next step.

3. Isochoric Heating at $V_0$ from $T_3 = \frac{P_0V_0}{4R}$ back to the initial temperature $T_1 = \frac{P_0V_0}{R}$:

   Since $W_3=0$,

   $$Q_3 = \Delta U = n C_V \Delta T = n C_V \left(\frac{P_0V_0}{R} - \frac{P_0V_0}{4R}\right) = \frac{3P_0V_0}{4}\frac{C_V}{R}.$$

Adding the contributions:

$$Q_{\text{total}} = Q_1 + Q_2 + Q_3 = P_0V_0 (2\ln2) -\frac{3P_0V_0}{4}\left(\frac{C_V}{R}+1\right)+\frac{3P_0V_0}{4}\frac{C_V}{R}.$$

Notice that the terms with $\frac{C_V}{R}$ cancel:

$$Q_{\text{total}} = P_0V_0 (2\ln2) -\frac{3P_0V_0}{4}.$$

Thus,

$$Q_{\text{total}} = P_0V_0\left(2\ln2 -0.75\right).$$