Thermodynamic Reversible Processes on Internal Energy versus Entropy Plot

To find the value of $\ln\frac{V_3}{V_2}$, we analyze the two reversible processes depicted in the $\frac{U}{R}$ versus $S$ plot:

Process $\mathbf{I}$:

• This process is represented by a vertical line, meaning the entropy remains constant ($\Delta S = 0$).
• Therefore, Process $\mathbf{I}$ is a reversible adiabatic process.
• From the graph, the initial internal energy is: $\frac{U_1}{R} = 2250\text{ K} \implies U_1 = 2250R$
• The final internal energy for Process $\mathbf{I}$ is: $\frac{U_2}{R} = 450\text{ K} \implies U_2 = 450R$
• The temperature at state 2 is given by: $T_2 = \frac{U_2}{n C_{V,m}} = \frac{450R}{1 \times \frac{5}{2}R} = 180\text{ K}$
• For a reversible adiabatic process ($q = 0$), the work done by the gas ($w_{\mathbf{I}}$) according to the First Law of Thermodynamics is: $w_{\mathbf{I}} = -\Delta U = -(U_2 - U_1) = -(450R - 2250R) = 1800R$

Process $\mathbf{II}$:

• This process is represented by a horizontal line, meaning the internal energy is constant ($\Delta U = 0$).
• For an ideal gas, constant internal energy implies constant temperature ($T = T_2 = 180\text{ K}$). Therefore, Process $\mathbf{II}$ is a reversible isothermal process.
• The work done by the gas during this reversible isothermal expansion from volume $V_2$ to $V_3$ is: $w_{\mathbf{II}} = nRT_2 \ln\left(\frac{V_3}{V_2}\right) = 1 \times R \times 180 \times \ln\left(\frac{V_3}{V_2}\right) = 180R \ln\left(\frac{V_3}{V_2}\right)$

Equating the work done in both processes: Since the magnitude of the work done by the gas in both processes is identical: $w_{\mathbf{I}} = w_{\mathbf{II}}$ $1800R = 180R \ln\left(\frac{V_3}{V_2}\right)$ $\ln\left(\frac{V_3}{V_2}\right) = \frac{1800R}{180R} = 10$

Thus, the value of $\ln\frac{V_3}{V_2}$ is 10.