One gm mol of a diatomic gas `(gamma=1.4)` is compressed adiabatically so that its temperature rises from `27^(@)C` to `127^(@)C` . The work done will be

Options

A. `2077.5` joules
B. `207.5` joules
C. `207.5` ergs
D. `205.5` joules

Explanation

To solve the problem of calculating the work done during the adiabatic compression of a diatomic gas, we can follow these steps:

### Step 1: Identify the parameters
We have:
- Number of moles, \( n = 1 \) gm mol = 1 mol
- Initial temperature, \( T_1 = 27^\circ C = 300 \, K \) (convert to Kelvin by adding 273)
- Final temperature, \( T_2 = 127^\circ C = 400 \, K \) (convert to Kelvin by adding 273)
- Heat capacity at constant volume for a diatomic gas, \( C_v = \frac{R}{\gamma - 1} \)
- Given \( \gamma = 1.4 \)

### Step 2: Calculate \( C_v \)
Using the ideal gas constant \( R = 8.314 \, J/(mol \cdot K) \):
\[ C_v = \frac{R}{\gamma - 1} = \frac{8.314}{1.4 - 1} = \frac{8.314}{0.4} = 20.785 \, J/(mol \cdot K) \]

### Step 3: Calculate the change in temperature \( \Delta T \)
\[ \Delta T = T_2 - T_1 = 400 \, K - 300 \, K = 100 \, K \]

### Step 4: Calculate the change in internal energy \( \Delta U \)
Using the formula for change in internal energy:
\[ \Delta U = n C_v \Delta T = 1 \, mol \times 20.785 \, J/(mol \cdot K) \times 100 \, K = 2078.5 \, J \]

### Step 5: Calculate the work done \( W \)
Since the process is adiabatic, the work done on the system is equal to the negative change in internal energy:
\[ W = -\Delta U = -2078.5 \, J \]

### Step 6: State the magnitude of work done
The magnitude of work done is:
\[ |W| = 2078.5 \, J \]

### Final Answer
The work done during the adiabatic compression is \( -2078.5 \, J \) (indicating work is done on the gas).

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