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. 200J
C. 324J
D. 364J

Explanation

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

### Step 1: Understand the Given Information
We have:
- Initial volume, \( V_1 = 2 \, \text{liters} = 2 \times 10^{-3} \, \text{m}^3 \)
- Final volume, \( V_2 = 4 \, \text{liters} = 4 \times 10^{-3} \, \text{m}^3 \)
- \( \gamma = 1.4 \)
- Initial pressure at STP, \( P_1 = 1 \, \text{atm} = 1.01325 \times 10^5 \, \text{Pa} \)

### Step 2: Use the Adiabatic Condition
For an adiabatic process, the relationship between pressure and volume is given by:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

### Step 3: Rearrange to Find \( P_2 \)
We can rearrange the equation to find \( P_2 \):
\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma \]
Substituting the values:
\[ P_2 = 1.01325 \times 10^5 \left(\frac{2 \times 10^{-3}}{4 \times 10^{-3}}\right)^{1.4} \]
\[ P_2 = 1.01325 \times 10^5 \left(\frac{1}{2}\right)^{1.4} \]
Using the given value \( 2^{1.4} = 2.64 \):
\[ P_2 = 1.01325 \times 10^5 \left(\frac{1}{2.64}\right) \]
Calculating \( P_2 \):
\[ P_2 \approx 1.01325 \times 10^5 \times 0.3788 \approx 3.83 \times 10^4 \, \text{Pa} \]

### Step 4: Calculate the Work Done
The work done \( W \) during an adiabatic expansion can be calculated using the formula:
\[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \]
Substituting the values:
\[ W = \frac{(1.01325 \times 10^5 \times 2 \times 10^{-3}) - (3.83 \times 10^4 \times 4 \times 10^{-3})}{1.4 - 1} \]
Calculating each term:
\[ W = \frac{(202.65) - (153.2)}{0.4} \]
\[ W = \frac{49.45}{0.4} \approx 123.625 \, \text{J} \]

### Final Answer
The work done by the air during the adiabatic expansion is approximately:
\[ W \approx 124 \, \text{J} \]

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