Step-by-step Derivation

1. Analyze the given information:

• Gas: 1 mole of an ideal monoatomic gas (n=1).
• Molar heat capacity at constant pressure: $C_{p,m} = \frac{5}{2} R$.
• From this, we can find the molar heat capacity at constant volume $C_{v,m} = C_{p,m} - R = \frac{3}{2} R$ and the ratio of specific heats $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3}$.
• Initial state A: $P_1 = 100$ atm, $V_1 = 1$ L.
• Intermediate state B: $P_2 = 10$ atm.
• Final state C: $P_3 = P_2 = 10$ atm.
• Process A → B is a reversible adiabatic process.
• Process B → C is a reversible isobaric process (constant pressure, as shown by the horizontal line on the P-V diagram).
• Total heat absorbed $q_{total} = R T_2 \ln 10$.

2. Analyze Process A → B (Adiabatic):

• For a reversible adiabatic process, the relation between pressure and volume is $P V^\gamma = \text{constant}$.
• Therefore, $P_1 V_1^\gamma = P_2 V_2^\gamma$.
• Substitute the given values: $100 \times (1)^{5/3} = 10 \times (V_2)^{5/3}$ $100 = 10 \cdot V_2^{5/3}$ $V_2^{5/3} = 10$ $V_2 = 10^{3/5} \text{ L}$

3. Analyze Process B → C (Isobaric) and Total Heat:

• The total heat absorbed is $q_{total} = q_{A\rightarrow B} + q_{B\rightarrow C}$.
• Since process A → B is adiabatic, $q_{A\rightarrow B} = 0$.
• Therefore, $q_{total} = q_{B\rightarrow C} = R T_2 \ln 10$.
• For a reversible isobaric process, the heat absorbed is given by $q_{B\rightarrow C} = n C_{p,m} (T_3 - T_2)$.
• Equating the two expressions for $q_{B\rightarrow C}$: $n C_{p,m} (T_3 - T_2) = R T_2 \ln 10$
• Substitute n=1 and $C_{p,m} = \frac{5}{2} R$: $(1) \left(\frac{5}{2} R\right) (T_3 - T_2) = R T_2 \ln 10$
• Cancel R from both sides and simplify: $\frac{5}{2} (T_3 - T_2) = T_2 \ln 10$ $\frac{5}{2} \left(\frac{T_3}{T_2} - 1\right) = \ln 10$ $\frac{T_3}{T_2} = 1 + \frac{2}{5} \ln 10$

4. Calculate the final volume $V_3$:

• For an isobaric process, Charles's Law states that V/T is constant.
• Thus, $\frac{V_3}{V_2} = \frac{T_3}{T_2}$.
• Substitute the expression for $T_3/T_2$: $\frac{V_3}{V_2} = 1 + \frac{2}{5} \ln 10$
• Now, substitute the value of $V_2 = 10^{3/5}$ L: $V_3 = V_2 \left(1 + \frac{2}{5} \ln 10\right) = 10^{3/5} \left(1 + \frac{2}{5} \ln 10\right)$

5. Calculate the required value $2 \log V_3$:

• We need to calculate $2 \log_{10} V_3$.
• First, take the base-10 logarithm of $V_3$: $\log V_3 = \log \left[ 10^{3/5} \left(1 + \frac{2}{5} \ln 10\right) \right]$ $\log V_3 = \log(10^{3/5}) + \log\left(1 + \frac{2}{5} \ln 10\right)$
• Using $\ln 10 \approx 2.303$ and $\log_{10}$ as $\log$: $\log V_3 = \frac{3}{5} + \log\left(1 + \frac{2}{5} \times 2.303\right)$ $\log V_3 = 0.6 + \log(1 + 0.9212) = 0.6 + \log(1.9212)$ $\log V_3 \approx 0.6 + 0.2835 = 0.8835$
• Finally, calculate $2 \log V_3$: $2 \log V_3 \approx 2 \times 0.8835 = 1.767$

6. Conclusion and Answer Discrepancy:

• The calculation based on the literal interpretation of the problem yields 1.767. As the answer is an integer type, rounding to the nearest integer gives 2.
• The stored answer is 7. This indicates a significant inconsistency in the problem statement, as the derived value is not close to 7.
• Let's check a plausible typo. The heat term $R T_2 \ln 10$ strongly suggests $T \Delta S$. If we assume the question intended to state that the entropy change $\Delta S_{BC} = R \ln 10$:
  • $\Delta S_{BC} = n C_p \ln(T_3/T_2) = \frac{5}{2} R \ln(T_3/T_2)$
  • $\frac{5}{2} R \ln(T_3/T_2) = R \ln 10 \implies \ln((T_3/T_2)^{5/2}) = \ln 10$
  • $T_3/T_2 = 10^{2/5}$
  • $V_3 = V_2 \cdot (T_3/T_2) = 10^{3/5} \cdot 10^{2/5} = 10^1 = 10$
  • $2 \log V_3 = 2 \log(10) = 2 \times 1 = 2$.
• This interpretation also leads to the answer 2. It is impossible to obtain the answer 7 through any valid physical reasoning based on the provided question. The question is flawed.
• Given the constraints, I will report the answer derived from a plausible interpretation that yields an integer, which is 2.