Calculate the work done (in joules) when `0.2` mole of an ideal gas at `300 K` expands isothermally and reversible from an initial volume of `2.5` litres to the final volume of `25` litres.

Options

A. Isothermally
B. Isobarically
C. Adiabatically
D. Same for all three process

Answer

To calculate the work done when `0.2` moles of an ideal gas expands isothermally and reversibly from an initial volume of `2.5` liters to a final volume of `25` liters, we can follow these steps:

### Step 1: Identify the given values
- Number of moles (n) = `0.2` moles
- Temperature (T) = `300 K`
- Initial volume (V1) = `2.5 L`
- Final volume (V2) = `25 L`
- Universal gas constant (R) = `8.314 J/(mol·K)`

### Step 2: Use the formula for work done in an isothermal reversible process
The formula for work done (W) during isothermal reversible expansion is given by:
\[ W = -2.303 \times n \times R \times T \times \log\left(\frac{V_2}{V_1}\right) \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ W = -2.303 \times 0.2 \times 8.314 \times 300 \times \log\left(\frac{25}{2.5}\right) \]

### Step 4: Calculate the logarithm
Calculate the ratio of volumes:
\[ \frac{V_2}{V_1} = \frac{25}{2.5} = 10 \]
Now, calculate the logarithm:
\[ \log(10) = 1 \]

### Step 5: Substitute the logarithm back into the equation
Now substitute the value of the logarithm back into the equation:
\[ W = -2.303 \times 0.2 \times 8.314 \times 300 \times 1 \]

### Step 6: Calculate the work done
Now perform the multiplication:
1. Calculate \( 0.2 \times 8.314 = 1.6628 \)
2. Then, calculate \( 1.6628 \times 300 = 498.84 \)
3. Finally, calculate \( -2.303 \times 498.84 \)

\[ W = -1148.83 \, \text{J} \]

### Final Answer
The work done during the isothermal expansion is approximately:
\[ W \approx -1148.83 \, \text{J} \]

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