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 finding the work done on a diatomic gas during an adiabatic compression, we can follow these steps:

### Step 1: Understand the Process
Since the gas is compressed adiabatically, there is no heat exchange with the surroundings. Therefore, the heat added to the system \( Q = 0 \).

### Step 2: Use the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
For an adiabatic process, \( Q = 0 \), so:
\[ \Delta U = -W \]
This means the work done on the gas \( W \) is equal to the negative change in internal energy \( \Delta U \).

### Step 3: Calculate the Change in Internal Energy
The change in internal energy for an ideal gas can be calculated using the formula:
\[ \Delta U = n C_V \Delta T \]
Where:
- \( n \) = number of moles (1 gm mol = 1 mol)
- \( C_V \) = molar specific heat at constant volume
- \( \Delta T \) = change in temperature in Kelvin

For a diatomic gas, the degrees of freedom \( F \) is given by:
\[ \gamma = \frac{C_P}{C_V} \quad \text{and} \quad C_P = C_V + R \]
Given \( \gamma = 1.4 \), we can find \( C_V \):
\[ \gamma = \frac{5}{3} \Rightarrow C_V = \frac{R}{\gamma - 1} = \frac{R}{0.4} = 2.5R \]
Thus, \( C_V = \frac{5}{2} R \).

### Step 4: Calculate the Change in Temperature
Convert the temperatures from Celsius to Kelvin:
\[ T_1 = 27^\circ C = 300 K \]
\[ T_2 = 127^\circ C = 400 K \]
So, the change in temperature \( \Delta T \) is:
\[ \Delta T = T_2 - T_1 = 400 K - 300 K = 100 K \]

### Step 5: Substitute Values into the Change in Internal Energy Formula
Now we can calculate \( \Delta U \):
\[ \Delta U = n C_V \Delta T = 1 \cdot \left(\frac{5}{2} R\right) \cdot 100 \]
Using \( R = 8.31 \, J/(mol \cdot K) \):
\[ \Delta U = \frac{5}{2} \cdot 8.31 \cdot 100 = 2077.5 \, J \]

### Step 6: Calculate the Work Done
Since \( W = -\Delta U \):
\[ W = -2077.5 \, J \]
This indicates that the work is done on the gas.

### Final Answer
The work done on the gas is:
\[ W = -2077.5 \, J \]