1 litre of a gas at STP is expanded adiabatically to 3 litre. Find work done by the gas. Given `gamma` = 1.40 and` 3^(1.4)= 4.65 `

Options

A. ` 60.7 ` J
B. ` 90.5 ` J
C. `100.8 ` J
D. `48 ` J

Answer

To solve the problem of finding the work done by a gas during an adiabatic expansion from 1 liter to 3 liters, we can follow these steps:

### Step 1: Understand the Given Data
- Initial volume (V1) = 1 liter = \(1 \times 10^{-3} \, \text{m}^3\)
- Final volume (V2) = 3 liters = \(3 \times 10^{-3} \, \text{m}^3\)
- Initial pressure (P1) = 1 atm = \(10^5 \, \text{Pa}\)
- Adiabatic exponent (γ) = 1.40
- \(3^{1.4} = 4.65\)

### Step 2: Use the Adiabatic Condition
For an adiabatic process, the relationship between pressure and volume is given by:
\[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \]
We need to find \(P_2\).

### Step 3: Rearranging the Equation to Find P2
Rearranging the equation gives:
\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma} \]
Substituting the known values:
\[ P_2 = 10^5 \left(\frac{1 \times 10^{-3}}{3 \times 10^{-3}}\right)^{1.4} \]
\[ P_2 = 10^5 \left(\frac{1}{3}\right)^{1.4} \]
Using the given value \(3^{1.4} = 4.65\):
\[ P_2 = 10^5 \div 4.65 \]

### Step 4: Calculate P2
Calculating \(P_2\):
\[ P_2 \approx \frac{10^5}{4.65} \approx 21506.45 \, \text{Pa} \]

### Step 5: Calculate Work Done (W)
The work done during an adiabatic process can be calculated using the formula:
\[ W = \frac{P_2 V_2 - P_1 V_1}{1 - \gamma} \]
Substituting the values:
\[ W = \frac{(21506.45 \times 3 \times 10^{-3}) - (10^5 \times 1 \times 10^{-3})}{1 - 1.4} \]
Calculating each term:
\[ W = \frac{(64.51935) - (100)}{-0.4} \]
\[ W = \frac{-35.48065}{-0.4} \]
\[ W = 88.701625 \approx 90 \, \text{J} \]

### Final Answer
The work done by the gas during the adiabatic expansion is approximately **90 Joules**.

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