Determination of Volume in Two Stage Reversible Gas Process

For one mole ($n = 1$) of an ideal monoatomic gas, the heat capacity at constant volume is: $C_{v,\text{m}} = C_{p,\text{m}} - R = \frac{5}{2}R - R = \frac{3}{2}R$

The adiabatic index (Poisson's ratio) $\gamma$ is given by: $\gamma = \frac{C_{p,\text{m}}}{C_{v,\text{m}}} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}$

Thus, $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.

Step 1: Process $\text{A} \rightarrow \text{B}$ (Reversible Adiabatic Expansion)

From the graph:

• Temperature at $\text{A}$, $T_1 = 600\text{ K}$
• Volume at $\text{A}$, $V_1 = 10\text{ m}^3$
• Temperature at $\text{B}$, $T_2 = 60\text{ K}$

Since $\text{A} \rightarrow \text{B}$ is a reversible adiabatic process: $q_{\text{A} \rightarrow \text{B}} = 0$

Using the relation $T V^{\gamma - 1} = \text{constant}$: $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

Substitute the known values: $600 \times 10^{2/3} = 60 \times V_2^{2/3}$

$\frac{600}{60} = \left(\frac{V_2}{10}\right)^{2/3}$

$10 = \left(\frac{V_2}{10}\right)^{2/3}$

$V_2 = 10 \times 10^{3/2} = 10^{5/2}\text{ m}^3$

Step 2: Process $\text{B} \rightarrow \text{C}$ (Reversible Isothermal Expansion)

Since the temperature remains constant at $T_2 = 60\text{ K}$ during $\text{B} \rightarrow \text{C}$, the heat absorbed in this reversible isothermal process is: $q_{\text{B} \rightarrow \text{C}} = n R T_2 \ln\left(\frac{V_3}{V_2}\right) = R T_2 \ln\left(\frac{V_3}{V_2}\right)$

The total heat absorbed in the entire process $\text{A} \rightarrow \text{B} \rightarrow \text{C}$ is: $q_{\text{total}} = q_{\text{A} \rightarrow \text{B}} + q_{\text{B} \rightarrow \text{C}} = 0 + R T_2 \ln\left(\frac{V_3}{V_2}\right)$

Given that $q_{\text{total}} = R T_2 \ln 10$: $R T_2 \ln\left(\frac{V_3}{V_2}\right) = R T_2 \ln 10$

$\ln\left(\frac{V_3}{V_2}\right) = \ln 10 \implies V_3 = 10 V_2$

Substitute $V_2 = 10^{5/2}\text{ m}^3$: $V_3 = 10 \times 10^{5/2} = 10^{7/2}\text{ m}^3$

Step 3: Calculation of $2 \log V_3$

Taking the base-10 logarithm of $V_3$: $\log_{10} V_3 = \log_{10}\left(10^{7/2}\right) = \frac{7}{2}$

Therefore: $2 \log V_3 = 2 \times \frac{7}{2} = 7$