A gas `(C_(v.m) = (5)/(2)R)` behaving ideally is allowed to expand reversibly and adiabatically from `1` litre to `32` litre. Its initial temperature is `327^(@)C`. The molar enthalpy change (in `J//mol`) for the process is :

Options

A. `-1125R`
B. `-675`
C. `-1575R`
D. None of these

Explanation

To solve the problem of calculating the molar enthalpy change for the given ideal gas undergoing a reversible adiabatic expansion, we will follow these steps:

### Step 1: Identify the given data
- Initial volume \( V_1 = 1 \, \text{L} \)
- Final volume \( V_2 = 32 \, \text{L} \)
- Initial temperature \( T_1 = 327^\circ C = 600 \, \text{K} \) (after converting to Kelvin)
- Molar heat capacity at constant volume \( C_{V,m} = \frac{5}{2} R \)

### Step 2: Calculate \( C_{P,m} \)
Using the relationship between \( C_P \) and \( C_V \):
\[ C_{P,m} = C_{V,m} + R = \frac{5}{2} R + R = \frac{7}{2} R \]

### Step 3: Determine the value of \( \gamma \)
\[ \gamma = \frac{C_{P,m}}{C_{V,m}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} = \frac{7}{5} \]

### Step 4: Use the adiabatic relation to find \( T_2 \)
For an adiabatic process, the relation between temperatures and volumes is given by:
\[ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]
Substituting the known values:
\[ \frac{T_2}{600} = \left( \frac{1}{32} \right)^{\frac{7}{5} - 1} = \left( \frac{1}{32} \right)^{\frac{2}{5}} \]
Calculating \( \left( \frac{1}{32} \right)^{\frac{2}{5}} \):
\[ \left( \frac{1}{32} \right)^{\frac{2}{5}} = \frac{1}{2^6}^{\frac{2}{5}} = \frac{1}{2^{\frac{12}{5}}} = \frac{1}{2^{2.4}} \approx \frac{1}{5.278} \approx 0.189 \]
Now, substituting back:
\[ T_2 = 600 \times 0.189 \approx 113.4 \, \text{K} \]

### Step 5: Calculate the change in temperature \( \Delta T \)
\[ \Delta T = T_2 - T_1 = 113.4 - 600 = -486.6 \, \text{K} \]

### Step 6: Calculate the molar enthalpy change \( \Delta H \)
Using the formula for molar enthalpy change:
\[ \Delta H = C_{P,m} \Delta T = \frac{7}{2} R \times (-486.6) \]
Calculating:
\[ \Delta H = -1701.1 R \, \text{J/mol} \]

### Final Answer
The molar enthalpy change for the process is approximately \( -1701.1 R \, \text{J/mol} \).

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