One mole of an ideal gas expands at a constant temperature of `300 K` from an initial volume of `10` litres to a final volume of `20` liters. The work done in expanding the gas is `(R=8.31 J// "mole"-K)` (in joules)

Options

A. `750`
B. `1728`
C. `1500`
D. `3456`

Explanation

To solve the problem of calculating the work done by one mole of an ideal gas during isothermal expansion from an initial volume of 10 liters to a final volume of 20 liters at a constant temperature of 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (n) = 1 mole
- Temperature (T) = 300 K
- Initial volume (V1) = 10 liters
- Final volume (V2) = 20 liters
- Ideal gas constant (R) = 8.31 J/(mol·K)

2. **Convert Volumes to Cubic Meters (if necessary):**
- Since we are working in liters, we can keep them as is for this calculation, but it's good practice to convert to cubic meters if needed (1 liter = 0.001 m³). However, for the logarithmic calculation, we can use liters directly.

3. **Use the Formula for Work Done in Isothermal Expansion:**
The work done (W) during isothermal expansion of an ideal gas can be calculated using the formula:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]

4. **Calculate the Ratio of Volumes:**
\[ \frac{V_2}{V_1} = \frac{20 \text{ L}}{10 \text{ L}} = 2 \]

5. **Calculate the Natural Logarithm:**
\[ \ln(2) \approx 0.693 \]

6. **Substitute the Values into the Work Done Formula:**
\[ W = 1 \text{ mol} \times 8.31 \text{ J/(mol·K)} \times 300 \text{ K} \times 0.693 \]

7. **Calculate the Work Done:**
\[ W = 8.31 \times 300 \times 0.693 \]
\[ W \approx 8.31 \times 300 \times 0.693 \approx 1728.3 \text{ J} \]

### Final Answer:
The work done in expanding the gas is approximately **1728.3 J**.

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