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. `58.55kg`
C. `342.58 kg`
D. none of these

Explanation

To solve the problem of determining the largest mass that can be lifted through a height of 100 meters during the isothermal expansion of an ideal gas, we will follow these steps:

### Step 1: Identify the Given Data
- Number of moles of gas (n) = 10 moles
- Initial pressure (P1) = 10 atm
- Final pressure (P2) = 1 atm
- Temperature (T) = 300 K
- Height (h) = 100 m
- Acceleration due to gravity (g) = 9.81 m/s² (standard value)

### Step 2: Calculate the Work Done by the Gas
The work done (W) during isothermal reversible expansion can be calculated using the formula:
\[ W = -2.303 \, nRT \log \left( \frac{P_1}{P_2} \right) \]

### Step 3: Convert Pressure from atm to SI Units
1 atm = 101.325 kPa
- P1 = 10 atm = 10 × 101.325 kPa = 1013.25 kPa
- P2 = 1 atm = 1 × 101.325 kPa = 101.325 kPa

### Step 4: Substitute Values into the Work Done Formula
Using R = 8.314 J/(mol·K):
\[ W = -2.303 \times 10 \times 8.314 \times 300 \log \left( \frac{10}{1} \right) \]
\[ W = -2.303 \times 10 \times 8.314 \times 300 \log(10) \]
Since \(\log(10) = 1\):
\[ W = -2.303 \times 10 \times 8.314 \times 300 \]

### Step 5: Calculate the Work Done
Calculating the above expression:
\[ W = -2.303 \times 10 \times 8.314 \times 300 \approx -5730.7 \, \text{J} \]

### Step 6: Relate Work Done to Lifting Mass
The work done in lifting a mass (m) through a height (h) is given by:
\[ W = -mgh \]
Setting the two expressions for work equal gives:
\[ -mgh = -5730.7 \]
Thus,
\[ mgh = 5730.7 \]

### Step 7: Solve for Mass (m)
Rearranging the equation to solve for m:
\[ m = \frac{5730.7}{gh} \]
Substituting g = 9.81 m/s² and h = 100 m:
\[ m = \frac{5730.7}{9.81 \times 100} \]
\[ m \approx \frac{5730.7}{981} \approx 5.83 \, \text{kg} \]

### Step 8: Final Result
The largest mass that can be lifted through a height of 100 meters is approximately:
\[ \boxed{5.83 \, \text{kg}} \]