Calculate the work done when 1 mole of a gas expands reversibly and isothermally from 5 atm to 1 atm at 300 K. [Value of log 5 = 0.6989].

Options

A. `-4015` J
B. `+4015` J
C. Zero
D. `-15` J

Answer

To calculate the work done when 1 mole of a gas expands reversibly and isothermally from 5 atm to 1 atm at 300 K, we can use the formula for work done in an isothermal reversible process:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial Pressure, \( P_1 = 5 \, \text{atm} \)
- Final Pressure, \( P_2 = 1 \, \text{atm} \)
- Temperature, \( T = 300 \, \text{K} \)
- Number of moles, \( n = 1 \, \text{mol} \)
- Universal gas constant, \( R = 8.314 \, \text{J/mol·K} \)
- Given value of \( \log 5 = 0.6989 \)

2. **Use the Formula for Work Done:**
The formula for work done during isothermal expansion is:
\[ W = -2.303 \, n \, R \, T \, \log \left( \frac{P_1}{P_2} \right) \]

3. **Substitute the Values into the Formula:**
\[ W = -2.303 \times 1 \, \text{mol} \times 8.314 \, \text{J/mol·K} \times 300 \, \text{K} \times \log \left( \frac{5}{1} \right) \]
\[ W = -2.303 \times 1 \times 8.314 \times 300 \times \log(5) \]

4. **Calculate \( \log(5) \):**
We know that \( \log(5) = 0.6989 \).

5. **Perform the Calculation:**
\[ W = -2.303 \times 1 \times 8.314 \times 300 \times 0.6989 \]
\[ W = -2.303 \times 2494.2 \times 0.6989 \]
\[ W = -2.303 \times 1742.4 \]
\[ W \approx -401.7 \, \text{J} \]

6. **Round the Result:**
The work done \( W \approx -415 \, \text{J} \) (considering significant figures and rounding).

### Final Answer:
The work done when 1 mole of gas expands isothermally and reversibly from 5 atm to 1 atm at 300 K is approximately:
\[ W \approx -415 \, \text{J} \]