A gram mole of a gas at `27^@C` expands isothermally until its volume is doubled. Calculate the amount of work done. `(R=8 J mol^(-1) K^(-1))`

Answer

To solve the problem of calculating the work done when a gram mole of gas expands isothermally until its volume is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
The given temperature is \(27^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
\[ T = 27 + 273 = 300 \, K \]

2. **Identify Given Values**:
- Number of moles, \(n = 1 \, \text{mol}\)
- Gas constant, \(R = 8 \, \text{J mol}^{-1} \text{K}^{-1}\)
- Initial volume \(V_1\) and final volume \(V_2\) where \(V_2 = 2V_1\).

3. **Use the Work Done Formula for Isothermal Expansion**:
The work done \(W\) during an isothermal expansion can be calculated using the formula:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]
Since \(V_2 = 2V_1\), we have:
\[ \frac{V_2}{V_1} = 2 \]

4. **Substitute the Values into the Formula**:
Now we can substitute the known values into the work done formula:
\[ W = 1 \, \text{mol} \times 8 \, \text{J mol}^{-1} \text{K}^{-1} \times 300 \, K \times \ln(2) \]

5. **Calculate \(\ln(2)\)**:
The natural logarithm of 2 is approximately:
\[ \ln(2) \approx 0.6932 \]

6. **Calculate the Work Done**:
Now substituting \(\ln(2)\) into the equation:
\[ W = 1 \times 8 \times 300 \times 0.6932 \]
\[ W = 2400 \times 0.6932 \approx 1663.68 \, J \]

7. **Final Result**:
The work done during the isothermal expansion is:
\[ W \approx 1.66 \times 10^3 \, J \]

### Final Answer:
The amount of work done is approximately \(1.66 \times 10^3 \, J\).