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 step by step, we will analyze the two processes involved: the isothermal expansion and the heating at constant volume.

### Step 1: Understanding the Isothermal Expansion
Given:
- Number of moles (n) = 4 moles
- Initial temperature (T) = 27°C = 300 K
- Initial volume (V₀)
- Final volume (V_f) = 7V₀

For an isothermal process, the heat absorbed (Q) can be calculated using the formula:
\[ Q = nRT \ln\left(\frac{V_f}{V_0}\right) \]

Substituting the values:
- R = 8.314 J/(mol·K)
- \(\ln(7) = 1.95\)

Calculating \(Q\):
\[ Q_{A \to B} = 4 \times 8.314 \times 300 \times 1.95 \]
\[ Q_{A \to B} = 4 \times 8.314 \times 300 \times 1.95 \approx 19.455 \text{ kJ} \]

### Step 2: Analyzing the Heating at Constant Volume
In the second process (from B to C), the gas is heated at constant volume until the pressure returns to its original value (P₀).

Using the ideal gas law:
- At point B: \(P_B = \frac{P_0}{7}\) and \(V_B = 7V_0\)
- At point C: \(P_C = P_0\) and \(V_C = 7V_0\)

Since the volume is constant, the heat absorbed during this process is given by:
\[ Q_{B \to C} = nC_V \Delta T \]

### Step 3: Finding the Temperature Change
The temperature at point B (T_B) can be calculated using the ideal gas law:
\[ T_B = \frac{P_B V_B}{nR} = \frac{\left(\frac{P_0}{7}\right)(7V_0)}{4R} = \frac{P_0 V_0}{4R} \]
Since T_A = T_B = 300 K, we can find T_C:
\[ T_C = \frac{P_C V_C}{nR} = \frac{P_0 (7V_0)}{4R} = 7 \times T_B = 7 \times 300 = 2100 \text{ K} \]

### Step 4: Calculate the Heat for the Process from B to C
Now substituting the temperatures into the heat equation:
\[ Q_{B \to C} = nC_V (T_C - T_B) = 4C_V (2100 - 300) = 4C_V \times 1800 \]
\[ Q_{B \to C} = 7200C_V \text{ J} = 7.2C_V \text{ kJ} \]

### Step 5: Total Heat Absorbed
The total heat absorbed in both processes is given as:
\[ Q_{total} = Q_{A \to B} + Q_{B \to C} \]
\[ 110.85 = 19.455 + 7.2C_V \]
\[ 7.2C_V = 110.85 - 19.455 = 91.395 \]
\[ C_V = \frac{91.395}{7.2} \approx 12.7 \text{ J/(mol·K)} \]

### Step 6: Convert to Calorific Units
To convert \(C_V\) from J/(mol·K) to cal/(mol·K):
\[ C_V = \frac{12.7}{4.2} \approx 3.02 \text{ cal/(mol·K)} \]

### Final Answer
Thus, the value of \(C_V\) for the gas is approximately:
\[ \boxed{3.02 \text{ cal/(mol·K)}} \]