When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litres to 20 litre, the magnitude of the maximum work obtained is ________ J. (nearest integer) (Given : R = 8.3 J `K^(-1) "mol"^(-1) and log 2 = 0.3010`)

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles of He gas, \( n = 5 \) moles
- Temperature, \( T = 300 \) K
- Initial volume, \( V_i = 10 \) L
- Final volume, \( V_f = 20 \) L
- Gas constant, \( R = 8.3 \) J K\(^{-1}\) mol\(^{-1}\)
- Logarithm value, \( \log 2 = 0.3010 \)

2. **Use the Formula for Work Done in Isothermal Reversible Process:**
The work done \( W \) in an isothermal reversible process is given by the formula:
\[ W = -2.303 \cdot n \cdot R \cdot T \cdot \log\left(\frac{V_f}{V_i}\right) \]

3. **Calculate the Volume Ratio:**
\[ \frac{V_f}{V_i} = \frac{20}{10} = 2 \]

4. **Substitute Values into the Formula:**
\[ W = -2.303 \cdot 5 \cdot 8.3 \cdot 300 \cdot \log(2) \]

5. **Calculate Each Component:**
- First, calculate \( 2.303 \cdot 5 \cdot 8.3 \cdot 300 \):
\[ 2.303 \cdot 5 = 11.515 \]
\[ 11.515 \cdot 8.3 = 95.7645 \]
\[ 95.7645 \cdot 300 = 28729.35 \]

- Now, multiply by \( \log(2) \):
\[ 28729.35 \cdot 0.3010 = 8657.32 \]

6. **Calculate the Work Done:**
\[ W = -8657.32 \text{ J} \]

7. **Round to the Nearest Integer:**
The nearest integer value of the work done is:
\[ W \approx -8657 \text{ J} \]

### Final Answer:
The magnitude of the maximum work obtained is **8657 J**.

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