If 1 mole of an ideal gas expands isothermally at `37^(@)C` from 15 litres to 25 litres, the maximum work obtained is `:`

Options

A. `1316.8J`
B. `6.43J`
C. `8.57J`
D. `2.92J`

Explanation

To find the maximum work obtained when 1 mole of an ideal gas expands isothermally from 15 liters to 25 liters at a temperature of 37°C, we can use the formula for work done during an isothermal expansion:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
\[ T = 37°C + 273 = 310 \, K \]

2. **Identify Given Values**:
- Number of moles, \( n = 1 \, \text{mole} \)
- Initial volume, \( V_1 = 15 \, \text{liters} \)
- Final volume, \( V_2 = 25 \, \text{liters} \)
- Ideal gas constant, \( R = 8.314 \, \text{J/(mol·K)} \)

3. **Use the Work Done Formula**:
The work done \( W \) in an isothermal process can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]

4. **Substitute the Values into the Formula**:
\[ W = -1 \times 8.314 \, \text{J/(mol·K)} \times 310 \, K \times \ln\left(\frac{25}{15}\right) \]

5. **Calculate the Natural Logarithm**:
\[ \ln\left(\frac{25}{15}\right) = \ln\left(\frac{5}{3}\right) \approx 0.5108 \]

6. **Calculate the Work Done**:
\[ W = -1 \times 8.314 \times 310 \times 0.5108 \]
\[ W \approx -1316.8 \, \text{J} \]

7. **Interpret the Result**:
The negative sign indicates that work is done by the system (expansion).

### Final Answer:
The maximum work obtained is approximately \(-1316.8 \, \text{J}\).

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