3 mole of a gas at temperature 400 K expands isothermally from initial volume of 4 litre to final volume of 8 litre. Find the work done by the gas. `(R = 8.31 J mol^(-1) K^(-1))`

Answer

To find the work done by the gas during an isothermal expansion, we can use the formula for work done in an isothermal process for an ideal gas:

\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]

Where:
- \( W \) = work done by the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (8.31 J/mol·K)
- \( T \) = absolute temperature in Kelvin
- \( V_1 \) = initial volume
- \( V_2 \) = final volume

### Step-by-Step Solution:

1. **Identify the given values**:
- Number of moles, \( n = 3 \, \text{moles} \)
- Temperature, \( T = 400 \, \text{K} \)
- Initial volume, \( V_1 = 4 \, \text{liters} \)
- Final volume, \( V_2 = 8 \, \text{liters} \)
- Universal gas constant, \( R = 8.31 \, \text{J/mol·K} \)

2. **Calculate the ratio of volumes**:
\[ \frac{V_2}{V_1} = \frac{8 \, \text{liters}}{4 \, \text{liters}} = 2 \]

3. **Calculate the natural logarithm of the volume ratio**:
\[ \ln\left(\frac{V_2}{V_1}\right) = \ln(2) \]

We can convert this to a logarithm base 10:
\[ \ln(2) = 2.303 \times \log_{10}(2) \]
Using the value \( \log_{10}(2) \approx 0.3010 \):
\[ \ln(2) \approx 2.303 \times 0.3010 \approx 0.693 \]

4. **Substitute the values into the work done formula**:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) = 3 \times 8.31 \, \text{J/mol·K} \times 400 \, \text{K} \times \ln(2) \]
\[ W = 3 \times 8.31 \times 400 \times 0.693 \]

5. **Calculate the work done**:
\[ W = 3 \times 8.31 \times 400 \times 0.693 \approx 3 \times 8.31 \times 400 \times 0.693 \approx 6912.06 \, \text{J} \]

6. **Convert to kilojoules**:
\[ W \approx 6.912 \, \text{kJ} \]

### Final Answer:
The work done by the gas during the isothermal expansion is approximately \( 6.912 \, \text{kJ} \).

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