Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are `10^5 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 (\(P_i\)) = \(10^5 \, \text{N/m}^2\)
- Initial Volume (\(V_i\)) = 6 liters = \(6 \times 10^{-3} \, \text{m}^3\)
- Final Volume (\(V_f\)) = 2 liters = \(2 \times 10^{-3} \, \text{m}^3\)
- Molar Specific Heat at Constant Volume (\(C_V\)) = \(\frac{3R}{2}\)

### Step 2: Calculate the Adiabatic Index (\(\gamma\))
The adiabatic index \(\gamma\) is defined as:
\[ \gamma = \frac{C_P}{C_V} \]
Using Mayer's relation:
\[ C_P - C_V = R \implies C_P = R + C_V \]
Substituting \(C_V\):
\[ C_P = R + \frac{3R}{2} = \frac{5R}{2} \]
Now, substituting \(C_P\) and \(C_V\) into the equation for \(\gamma\):
\[ \gamma = \frac{C_P}{C_V} = \frac{\frac{5R}{2}}{\frac{3R}{2}} = \frac{5}{3} \]

### Step 3: Use the Adiabatic Condition
For an adiabatic process, the relationship between pressure and volume is given by:
\[ P_i V_i^\gamma = P_f V_f^\gamma \]
We can rearrange this to find the final pressure \(P_f\):
\[ P_f = P_i \left(\frac{V_i}{V_f}\right)^\gamma \]
Substituting the known values:
\[ P_f = 10^5 \left(\frac{6 \times 10^{-3}}{2 \times 10^{-3}}\right)^{\frac{5}{3}} = 10^5 \left(3\right)^{\frac{5}{3}} \]

### Step 4: Calculate \(P_f\)
Calculating \(3^{\frac{5}{3}}\):
\[ 3^{\frac{5}{3}} \approx 5.196 \]
Thus,
\[ P_f \approx 10^5 \times 5.196 \approx 5.196 \times 10^5 \, \text{N/m}^2 \]

### Step 5: Calculate Work Done (\(W\))
The work done in an adiabatic process is given by:
\[ W = \frac{P_i V_i - P_f V_f}{\gamma - 1} \]
Substituting the values:
\[ W = \frac{10^5 \times 6 \times 10^{-3} - 5.196 \times 10^5 \times 2 \times 10^{-3}}{\frac{5}{3} - 1} \]
Calculating the denominator:
\[ \frac{5}{3} - 1 = \frac{2}{3} \]
Now substituting into the work done formula:
\[ W = \frac{(10^5 \times 6 \times 10^{-3}) - (5.196 \times 10^5 \times 2 \times 10^{-3})}{\frac{2}{3}} \]
Calculating the terms:
\[ 10^5 \times 6 \times 10^{-3} = 600 \]
\[ 5.196 \times 10^5 \times 2 \times 10^{-3} = 1039.2 \]
Thus:
\[ W = \frac{600 - 1039.2}{\frac{2}{3}} = \frac{-439.2}{\frac{2}{3}} = -659.8 \, \text{J} \]

### Final Answer
The work done when one mole of a perfect gas is compressed adiabatically is approximately:
\[ W \approx -660 \, \text{J} \]