4 mole of an ideal gas at `27^@ C` is isothermally expanded to 7 times its volume. Then it is heated at constant volume so that the pressure comes to the original value. The total heat absorbed in the two process is 110.85 kJ. Now `C_V` for the gas (in calm `"oI"^(-1)K^(-1)` ) _____ . [Given that ln 7 = 1.95]

Explanation

To solve the problem, we need to analyze the two processes that the gas undergoes: isothermal expansion and constant volume heating. We will calculate the heat absorbed in each process and then use the total heat absorbed to find the value of \( C_V \).

### Step-by-Step Solution:

1. **Identify Given Data:**
- Number of moles, \( n = 4 \) moles
- Initial temperature, \( T = 27^\circ C = 300 \, K \)
- Final volume after isothermal expansion, \( V_2 = 7V_1 \) (where \( V_1 \) is the initial volume)
- Total heat absorbed, \( Q_{total} = 110.85 \, kJ \)
- Given \( \ln(7) = 1.95 \)

2. **Calculate Heat Absorbed During Isothermal Expansion (Q_AB):**
- For an isothermal process, the change in internal energy \( \Delta U = 0 \).
- The heat absorbed \( Q_{AB} \) is equal to the work done \( W_{AB} \).
- The work done during isothermal expansion is given by:
\[ W_{AB} = nRT \ln\left(\frac{V_2}{V_1}\right) = nRT \ln(7) \]
- Substituting the values:
\[ W_{AB} = 4 \times 8.314 \, \text{J/mol K} \times 300 \, K \times 1.95 \]
- Calculating \( W_{AB} \):
\[ W_{AB} = 4 \times 8.314 \times 300 \times 1.95 = 19,450.14 \, J = 19.45 \, kJ \]
- Therefore, \( Q_{AB} = 19.45 \, kJ \).

3. **Calculate Heat Absorbed During Constant Volume Heating (Q_BC):**
- For a constant volume process, the heat absorbed is given by:
\[ Q_{BC} = nC_V \Delta T \]
- We need to find \( \Delta T \). The pressure at point B (after expansion) can be found using the ideal gas law:
\[ P_B = \frac{P_A}{7} \]
- Since \( P_B / T_B = P_C / T_C \) and \( P_C = P_A \):
\[ \frac{P_A/7}{300} = \frac{P_A}{T_C} \]
- Rearranging gives:
\[ T_C = 7 \times 300 = 2100 \, K \]
- Now, \( \Delta T = T_C - T_B = 2100 - 300 = 1800 \, K \).
- Substituting into the equation for \( Q_{BC} \):
\[ Q_{BC} = 4C_V \times 1800 \]
\[ Q_{BC} = 7200C_V \, J = 7.2C_V \, kJ \]

4. **Total Heat Absorbed:**
- The total heat absorbed is the sum of the heat absorbed in both processes:
\[ Q_{total} = Q_{AB} + Q_{BC} \]
\[ 110.85 = 19.45 + 7.2C_V \]
- Rearranging to find \( C_V \):
\[ 7.2C_V = 110.85 - 19.45 \]
\[ 7.2C_V = 91.4 \]
\[ C_V = \frac{91.4}{7.2} = 12.7 \, J/(mol \cdot K) \]

5. **Convert \( C_V \) to Calorie:**
- To convert from Joules to calories, use the conversion factor \( 1 \, cal = 4.184 \, J \):
\[ C_V = \frac{12.7}{4.184} \approx 3.02 \, cal/(mol \cdot K) \]

### Final Answer:
The value of \( C_V \) for the gas is approximately \( 3.02 \, cal/(mol \cdot K) \).