10 mole of ideal gas expand isothermally and reversibly from a pressure of `10atm` to `1atm` at `300K`. What is the largest mass which can lifted through a height of `100` meter?

Options

A. 31842 kg
B. 58.55 kg
C. 342.58 kg
D. None of these

Explanation

To solve the problem, we need to calculate the work done by the gas during the isothermal and reversible expansion, and then use that work to determine the maximum mass that can be lifted through a height of 100 meters.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of moles of gas, \( n = 10 \) moles
- Initial pressure, \( P_1 = 10 \) atm
- Final pressure, \( P_2 = 1 \) atm
- Temperature, \( T = 300 \) K
- Height, \( h = 100 \) m

2. **Convert Pressures to SI Units:**
- \( 1 \) atm = \( 101.325 \) kPa
- Thus,
\[ P_1 = 10 \times 101.325 \, \text{kPa} = 1013.25 \, \text{kPa} \]
\[ P_2 = 1 \times 101.325 \, \text{kPa} = 101.325 \, \text{kPa} \]

3. **Calculate Work Done (W):**
- The formula for work done during isothermal reversible expansion is:
\[ W = -2.303 \, nRT \log \left(\frac{P_1}{P_2}\right) \]
- Using \( R = 8.314 \, \text{J/mol K} \):
\[ W = -2.303 \times 10 \times 8.314 \times 300 \times \log \left(\frac{10}{1}\right) \]
- Calculate \( \log(10) = 1 \):
\[ W = -2.303 \times 10 \times 8.314 \times 300 \times 1 \]
\[ W = -57441.426 \, \text{J} \]

4. **Calculate the Maximum Mass (m):**
- The work done is equal to the change in potential energy:
\[ W = mgh \]
- Rearranging for mass \( m \):
\[ m = \frac{|W|}{gh} \]
- Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h = 100 \, \text{m} \):
\[ m = \frac{57441.426}{9.8 \times 100} \]
\[ m = \frac{57441.426}{980} \approx 58.56 \, \text{kg} \]

5. **Final Answer:**
- The largest mass that can be lifted through a height of 100 meters is approximately \( 58.56 \, \text{kg} \).