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. 100.8J
B. 90.5J
C. 45J
D. 18J

Answer

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

### Step 1: Understand the Adiabatic Process
In an adiabatic process, the relationship between pressure (P) and volume (V) is given by the equation:
\[ P V^\gamma = \text{constant} \]
where \(\gamma\) (gamma) is the heat capacity ratio.

### Step 2: Identify Initial Conditions
Given:
- Initial Volume, \( V_i = 1 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3 \)
- Final Volume, \( V_f = 3 \, \text{L} = 3 \times 10^{-3} \, \text{m}^3 \)
- Initial Pressure at STP, \( P_i = 1 \, \text{atm} = 10^5 \, \text{Pa} \)
- \(\gamma = 1.40\)

### Step 3: Calculate Final Pressure
Using the adiabatic relation:
\[ P_f = P_i \left( \frac{V_i}{V_f} \right)^\gamma \]

Substituting the values:
\[ P_f = 10^5 \left( \frac{1}{3} \right)^{1.4} \]

Using the provided value \( 3^{1.4} = 4.65 \):
\[ P_f = 10^5 \left( \frac{1}{4.65} \right) \]

Calculating \( P_f \):
\[ P_f = \frac{10^5}{4.65} \approx 21506.45 \, \text{Pa} \]

### Step 4: Calculate Work Done
The work done \( W \) during an adiabatic expansion can be calculated using the formula:
\[ W = \frac{P_f V_f - P_i V_i}{1 - \gamma} \]

Substituting the values:
\[ W = \frac{(21506.45 \times 3 \times 10^{-3}) - (10^5 \times 1 \times 10^{-3})}{1 - 1.4} \]

Calculating the numerator:
\[ W = \frac{(64.51935) - (100)}{-0.4} \]
\[ W = \frac{-35.48065}{-0.4} \]
\[ W = 88.701625 \, \text{J} \]

### Step 5: Final Result
Thus, the work done by the gas during the adiabatic expansion is approximately:
\[ W \approx 88.7 \, \text{J} \]