One mole of an ideal gas expands isothermally at 300 K from 1 litre to 101 litres at constant pressure of 1 atmosphere. The work done during this change is (1 lit atrn = 24.2 cals)

Options

A. `-242 kcals`
B. `-24.2 kcals`
C. `-2 kcals`
D. `-2.42 kcals `

Explanation

To solve the problem of work done during the isothermal expansion of an ideal gas, we can follow these steps:

### Step 1: Identify the Given Data
We have:
- Number of moles (n) = 1 mole
- Initial volume (V1) = 1 liter
- Final volume (V2) = 101 liters
- Temperature (T) = 300 K
- Pressure (P) = 1 atmosphere

### Step 2: Calculate the Change in Volume (ΔV)
The change in volume (ΔV) can be calculated as:
\[ \Delta V = V2 - V1 = 101 \, \text{liters} - 1 \, \text{liter} = 100 \, \text{liters} \]

### Step 3: Use the Formula for Work Done (W)
For an isothermal process, the work done (W) on the gas is given by the formula:
\[ W = -P \Delta V \]
Substituting the values:
\[ W = -1 \, \text{atm} \times 100 \, \text{liters} = -100 \, \text{liter atm} \]

### Step 4: Convert Work Done to Calories
We know that:
\[ 1 \, \text{liter atm} = 24.2 \, \text{calories} \]
Thus, we can convert the work done from liter atm to calories:
\[ W = -100 \, \text{liter atm} \times 24.2 \, \text{calories/liter atm} = -2420 \, \text{calories} \]

### Step 5: Convert Calories to Kilocalories
To express the work done in kilocalories, we convert calories to kilocalories:
\[ W = -2420 \, \text{calories} = -2.42 \, \text{kcal} \quad (\text{since } 1 \, \text{kcal} = 1000 \, \text{calories}) \]

### Final Answer
The work done during the isothermal expansion is:
\[ \text{W} = -2.42 \, \text{kcal} \]