One mole of `O_(2)` gas having a volume equal to 22.4 litres at `0^(@)C` and 1 atmospheric pressure is compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is

Options

A. 672.5 J
B. 1728 J
C. `-1728J`
D. `-1572.5J`

Explanation

To solve the problem of calculating the work done during the isothermal compression of one mole of \( O_2 \) gas, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of moles of gas, \( n = 1 \) mole
- Initial volume, \( V_1 = 22.4 \) liters
- Final volume, \( V_2 = 11.2 \) liters
- Temperature, \( T = 0^\circ C = 273 \) K
- Universal gas constant, \( R = 8.314 \, \text{J/(mol K)} \)

2. **Understand the Work Done Formula:**
The work done \( W \) during isothermal compression can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
Since we are using logarithm base 10, we will convert it using:
\[ \ln(x) = 2.303 \log_{10}(x) \]
Therefore, the formula becomes:
\[ W = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right) \]

3. **Calculate the Logarithmic Term:**
\[ \frac{V_2}{V_1} = \frac{11.2}{22.4} = 0.5 \]
Now calculate \( \log_{10}(0.5) \):
\[ \log_{10}(0.5) \approx -0.301 \]

4. **Substitute Values into the Work Done Formula:**
Now substitute \( n = 1 \), \( R = 8.314 \, \text{J/(mol K)} \), \( T = 273 \, \text{K} \), and \( \log_{10}(0.5) \) into the formula:
\[ W = -2.303 \times 1 \times 8.314 \times 273 \times (-0.301) \]

5. **Calculate the Work Done:**
First, calculate the product:
\[ W = -2.303 \times 8.314 \times 273 \times (-0.301) \]
Calculate:
\[ W \approx -2.303 \times 8.314 \times 273 \times (-0.301) \approx -1573.5 \, \text{J} \]

6. **Final Result:**
The work done during the isothermal compression is approximately:
\[ W \approx -1573.5 \, \text{J} \]

### Conclusion:
The work done in this process is approximately \(-1573.5\) Joules.