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. 2075 J
B. 207.5 J
C. 207.5 erG
D. 201.1 erg

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: Understand the Given Data
We have:
- One gram mole of a diatomic gas (which means n = 1 mole).
- The specific heat ratio (gamma, γ) = 1.4.
- Initial temperature (T2) = 27°C.
- Final temperature (T1) = 127°C.

### Step 2: Convert Temperatures to Kelvin
To use the temperatures in calculations, we need to convert them from Celsius to Kelvin:
- \( T_2 = 27°C + 273 = 300 \, K \)
- \( T_1 = 127°C + 273 = 400 \, K \)

### Step 3: Use the Formula for Work Done in Adiabatic Process
The work done (W) during an adiabatic process can be calculated using the formula:
\[ W = \frac{R}{\gamma - 1} (T_1 - T_2) \]
where:
- \( R \) is the universal gas constant, approximately \( 8.31 \, J/(mol \cdot K) \).
- \( \gamma \) is the specific heat ratio.
- \( T_1 \) and \( T_2 \) are the final and initial temperatures in Kelvin.

### Step 4: Substitute the Values into the Formula
Substituting the known values into the formula:
\[ W = \frac{8.31}{1.4 - 1} (400 - 300) \]
Calculating the difference in temperature:
\[ W = \frac{8.31}{0.4} \times 100 \]

### Step 5: Calculate the Work Done
Now, calculate the values:
\[ W = \frac{8.31}{0.4} \times 100 = 20.775 \times 100 = 2077.5 \, J \]

### Step 6: Round Off the Answer
Since the problem asks for an approximate answer, we can round it off:
\[ W \approx 2075 \, J \]

### Final Answer
The work done during the adiabatic compression is approximately **2075 J**.

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