Isothermally at `27^(@)` C, one mole of a van der Waals gas expands reversibly from 2 litres to 20 litres. Calculate the work done.

Answer

To calculate the work done during the isothermal expansion of one mole of a van der Waals gas from 2 liters to 20 liters at a temperature of 27°C, we can follow these steps:

### Step 1: Convert the temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
\[ T = 27 + 273.15 = 300.15 \, K \]

### Step 2: Identify the values needed for the formula
We need the following values:
- Number of moles, \( n = 1 \, \text{mole} \)
- Gas constant, \( R = 8.314 \, \text{J/mol·K} \)
- Initial volume, \( V_1 = 2 \, \text{liters} \)
- Final volume, \( V_2 = 20 \, \text{liters} \)

### Step 3: Write the formula for work done
The formula for the work done during isothermal reversible expansion is given by:
\[ W = -2.303 \, nRT \log \left( \frac{V_2}{V_1} \right) \]

### Step 4: Substitute the values into the formula
Now we can substitute the known values into the formula:
\[ W = -2.303 \times 1 \times 8.314 \times 300 \log \left( \frac{20}{2} \right) \]

### Step 5: Calculate the logarithm
Calculate the ratio of the volumes:
\[ \frac{20}{2} = 10 \]
Now calculate the logarithm:
\[ \log(10) = 1 \]

### Step 6: Substitute the logarithm back into the equation
Now substitute this value back into the equation:
\[ W = -2.303 \times 1 \times 8.314 \times 300 \times 1 \]

### Step 7: Perform the multiplication
Now we can perform the multiplication:
\[ W = -2.303 \times 8.314 \times 300 \]
Calculating this gives:
\[ W = -5744.14 \, \text{J} \]

### Step 8: Convert to kilojoules
To convert joules to kilojoules, divide by 1000:
\[ W = -5.74414 \, \text{kJ} \]

### Final Answer
The work done during the isothermal expansion is:
\[ W = -5.74414 \, \text{kJ} \]

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