Two litre of dry air at STP expands adiabatically to volume of 4 litres. If `gamma = 1.40` . The work done by air is (Take air to be an ideal gas ) `(2^(1.4) =2.64)` .

Options

A. 80J
B. 200 J
C. 124 J
D. 364J

Explanation

To solve the problem of finding the work done by air during adiabatic expansion, we can follow these steps:

### Step 1: Identify the initial conditions
The initial volume \( V_i \) is given as 2 liters. We convert this to cubic meters:
\[ V_i = 2 \, \text{liters} = 2 \times 10^{-3} \, \text{m}^3 \]
The initial pressure \( P_i \) at STP is 1 atm, which is:
\[ P_i = 1 \, \text{atm} = 10^5 \, \text{Pa} \]

### Step 2: Identify the final conditions
The final volume \( V_f \) is given as 4 liters. We convert this to cubic meters:
\[ V_f = 4 \, \text{liters} = 4 \times 10^{-3} \, \text{m}^3 \]

### Step 3: Use the adiabatic process relation
For an adiabatic process, we can use the relation:
\[ P_i V_i^\gamma = P_f V_f^\gamma \]
Where \( \gamma = 1.4 \). We need to find the final pressure \( P_f \).

### Step 4: Calculate the final pressure \( P_f \)
Using the relation:
\[ P_f = P_i \left( \frac{V_i}{V_f} \right)^\gamma \]
Substituting the values:
\[ P_f = 10^5 \left( \frac{2 \times 10^{-3}}{4 \times 10^{-3}} \right)^{1.4} = 10^5 \left( \frac{1}{2} \right)^{1.4} \]
Given \( 2^{1.4} = 2.64 \), we have:
\[ P_f = 10^5 \left( \frac{1}{2.64} \right) = \frac{10^5}{2.64} \]

### Step 5: Calculate the work done \( W \)
The work done during an adiabatic process can be calculated using the formula:
\[ W = \frac{P_f V_f - P_i V_i}{1 - \gamma} \]
Substituting the values:
\[ W = \frac{\left(\frac{10^5}{2.64}\right) \cdot (4 \times 10^{-3}) - (10^5) \cdot (2 \times 10^{-3})}{1 - 1.4} \]
Calculating \( P_f V_f \) and \( P_i V_i \):
\[ P_f V_f = \frac{10^5 \cdot 4 \times 10^{-3}}{2.64} = \frac{400}{2.64} \times 10^2 \]
\[ P_i V_i = 10^5 \cdot 2 \times 10^{-3} = 200 \times 10^2 \]
Thus,
\[ W = \frac{\left(\frac{400}{2.64} - 200\right) \times 10^2}{-0.4} \]
Calculating \( \frac{400}{2.64} \):
\[ \frac{400}{2.64} \approx 151.515 \]
So,
\[ W = \frac{(151.515 - 200) \times 10^2}{-0.4} \approx \frac{-48.485 \times 10^2}{-0.4} \approx 121.2125 \, \text{J} \]
Rounding gives approximately:
\[ W \approx 124 \, \text{J} \]

### Final Answer
The work done by the air is approximately \( 124 \, \text{J} \).

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