Work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1atm at 300 k is

Options

A. -4938.8J
B. 4138.8J
C. -5744.1J
D. 6257.2J

Answer

To find the work done during the isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 300 K, we can use the formula for work done in an isothermal process:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of moles (n) = 1 mole
- Initial pressure (P1) = 10 atm
- Final pressure (P2) = 1 atm
- Temperature (T) = 300 K
- Universal gas constant (R) = 8.314 J/(mol·K)

2. **Use the Formula for Work Done:**
The work done (W) during isothermal expansion is given by:
\[ W = -2.303 \cdot n \cdot R \cdot T \cdot \log\left(\frac{P1}{P2}\right) \]

3. **Substitute the Values into the Formula:**
\[ W = -2.303 \cdot 1 \cdot 8.314 \cdot 300 \cdot \log\left(\frac{10}{1}\right) \]

4. **Calculate the Logarithm:**
\[ \log\left(\frac{10}{1}\right) = \log(10) = 1 \]

5. **Plug in the Logarithm Value:**
\[ W = -2.303 \cdot 1 \cdot 8.314 \cdot 300 \cdot 1 \]

6. **Calculate the Work Done:**
\[ W = -2.303 \cdot 8.314 \cdot 300 \]
\[ W = -5744.1 \, \text{J} \]

7. **Final Result:**
The work done during the isothermal expansion is:
\[ W \approx -5744.1 \, \text{J} \]