1. Identify the two reversible processes from the diagram

For one mole of an ideal gas initially at $T_1=900\,\text{K}$:

• Process I is shown on a $U$ vs $S$ plot.
• Process II is shown on a $p$ vs $V$ plot.

We use the fact that for a reversible process, $dU = TdS - pdV$ so on a $U$-$S$ diagram, a vertical line means $S=\text{constant}$, i.e. a reversible adiabatic (isentropic) process.

Thus process I is reversible adiabatic from state 1 to state 2.

Process II is shown on a $p$-$V$ plot as a horizontal line, hence it is isobaric from state 2 to state 3.

2. Work done in Process I (reversible adiabatic)

For one mole of ideal gas, $U = nC_VT = C_VT$ with $C_V = \frac{5}{2}R$

In a reversible adiabatic process, $q=0 \implies W_I = -\Delta U = C_V(T_1-T_2)$

From the diagram, the final temperature after process I is $T_2=300\,\text{K}$. Hence, $W_I = \frac{5}{2}R(900-300)$ $W_I = \frac{5}{2}R\cdot 600 = 1500R$

3. Work done in Process II (isobaric)

For one mole of ideal gas in a reversible isobaric process, $W_{II} = p(V_3-V_2)$ Using $pV=RT$, $pV_2 = RT_2, \qquad pV_3 = RT_3$ So, $W_{II} = R(T_3-T_2)$

Given that the work done in the two processes is same, $W_I=W_{II}$ Therefore, $1500R = R(T_3-T_2)$ $T_3-T_2 = 1500$ Since $T_2=300\,\text{K}$, $T_3 = 1800\,\text{K}$

4. Find $\ln\left(\dfrac{V_3}{V_2}\right)$

Process II is isobaric, so for ideal gas, $\frac{V_3}{V_2} = \frac{T_3}{T_2}$ Thus, $\frac{V_3}{V_2} = \frac{1800}{300}=6$ Hence, $\ln\left(\frac{V_3}{V_2}\right)=\ln 6$

Now, $\ln 6 \approx 1.79$

This is not equal to 10.

5. Check consistency with the intended diagram information

A standard form of this problem usually has process I as an isothermal or some other path leading to a much larger work value, which could produce $\ln\left(\frac{V_3}{V_2}\right)=10$ But with the thermodynamic interpretation above, the result is clearly $\ln\left(\frac{V_3}{V_2}\right)=\ln 6$ not an integer.

Since the question is of integer type, it strongly suggests that either:

• some diagram information is missing from the prompt, or
• the stored correct answer corresponds to a different figure.

With the currently available information, the mathematically derived result is: $\ln\left(\frac{V_3}{V_2}\right)=\ln 6 \approx 1.79$