Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are `105 N//m^2` and 6 litres respectively. The final volume of the gas are 2 litre. Molar specific heat of the gas at constant volume is `3R//2`.

Explanation

To calculate the work done when one mole of a perfect gas is compressed adiabatically, we can follow these steps:

### Step 1: Identify Given Values
- Initial Pressure, \( P_1 = 10^5 \, \text{N/m}^2 \)
- Initial Volume, \( V_1 = 6 \, \text{liters} = 6 \times 10^{-3} \, \text{m}^3 \)
- Final Volume, \( V_2 = 2 \, \text{liters} = 2 \times 10^{-3} \, \text{m}^3 \)
- Molar specific heat at constant volume, \( C_V = \frac{3R}{2} \)

### Step 2: Calculate the Value of \( \gamma \)
Using the relation \( \gamma = \frac{C_P}{C_V} \) and knowing that \( C_P - C_V = R \):
1. Calculate \( C_V = \frac{3R}{2} \)
2. Therefore, \( C_P = C_V + R = \frac{3R}{2} + R = \frac{5R}{2} \)
3. Now, calculate \( \gamma \):
\[ \gamma = \frac{C_P}{C_V} = \frac{\frac{5R}{2}}{\frac{3R}{2}} = \frac{5}{3} \]

### Step 3: Calculate the Final Pressure \( P_2 \)
Using the adiabatic condition \( P_1 V_1^\gamma = P_2 V_2^\gamma \):
1. Rearranging gives:
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]
2. Substitute the known values:
\[ P_2 = 10^5 \left( \frac{6 \times 10^{-3}}{2 \times 10^{-3}} \right)^{\frac{5}{3}} = 10^5 \left( 3 \right)^{\frac{5}{3}} \]
3. Calculate \( (3)^{\frac{5}{3}} \):
\[ (3)^{\frac{5}{3}} \approx 6.2996 \]
4. Thus,
\[ P_2 \approx 10^5 \times 6.2996 \approx 6.2996 \times 10^5 \, \text{N/m}^2 \]

### Step 4: Calculate the Work Done
Using the formula for work done in an adiabatic process:
\[ W = \frac{1}{1 - \gamma} (P_2 V_2 - P_1 V_1) \]
1. Substitute \( \gamma = \frac{5}{3} \):
\[ W = \frac{1}{1 - \frac{5}{3}} (P_2 V_2 - P_1 V_1) = \frac{1}{-\frac{2}{3}} (P_2 V_2 - P_1 V_1) \]
2. Calculate \( P_2 V_2 \) and \( P_1 V_1 \):
\[ P_2 V_2 = (6.2996 \times 10^5) \times (2 \times 10^{-3}) = 1259.92 \, \text{J} \]
\[ P_1 V_1 = (10^5) \times (6 \times 10^{-3}) = 600 \, \text{J} \]
3. Now substitute back into the work done equation:
\[ W = -\frac{3}{2} (1259.92 - 600) = -\frac{3}{2} \times 659.92 = -989.88 \, \text{J} \]

### Final Answer
The work done when one mole of a perfect gas is compressed adiabatically is approximately:
\[ W \approx -989.88 \, \text{J} \]