`5mol` of an ideal gas at `27^(@)C` expands isothermally and reversibly from a volume of `6L` to `60L`. The work done in `kJ` is

Options

A. `-"14.7 KJ"`
B. `-"28.72 KJ"`
C. `"27.72 KJ"`
D. `-"56.72 KJ"`

Answer

To solve the problem of calculating the work done during the isothermal and reversible expansion of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (n) = 5 mol
- Initial temperature (T) = 27°C = 27 + 273 = 300 K
- Initial volume (V1) = 6 L
- Final volume (V2) = 60 L

2. **Formula for Work Done:**
The work done (W) during an isothermal and reversible expansion of an ideal gas is given by the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
Alternatively, it can be expressed using logarithm base 10:
\[ W = -2.303 \cdot nRT \log_{10}\left(\frac{V_2}{V_1}\right) \]

3. **Substituting the Values:**
- The gas constant (R) = 8.314 J/(mol·K)
- Substitute the values into the formula:
\[ W = -2.303 \cdot 5 \cdot 8.314 \cdot 300 \cdot \log_{10}\left(\frac{60}{6}\right) \]

4. **Calculate the Logarithm:**
- First, calculate \(\frac{V_2}{V_1} = \frac{60}{6} = 10\)
- Then, calculate \(\log_{10}(10) = 1\)

5. **Final Calculation:**
\[ W = -2.303 \cdot 5 \cdot 8.314 \cdot 300 \cdot 1 \]
\[ W = -2.303 \cdot 5 \cdot 8.314 \cdot 300 \]
\[ W = -28720.5 \text{ J} \]

6. **Convert Joules to Kilojoules:**
\[ W = \frac{-28720.5}{1000} \text{ kJ} = -28.72 \text{ kJ} \]

### Final Answer:
The work done during the isothermal expansion is \(-28.72 \text{ kJ}\).