Isothermally at `27^(@)` C, one mole of a van der Waals gas expands reversibly from 2 litres to 20 litres. Calculate the work done if `a =1.42 xx 10^(12)` dynes `cm^(4)` per mole and b= 30 cc.

Answer

To solve the problem of calculating the work done during the isothermal expansion of one mole of a van der Waals gas, we will use the formula for work done in a reversible process. The van der Waals equation is given by:

\[ [P + a \left( \frac{n^2}{V^2} \right)](V - nb) = nRT \]

Where:
- \( P \) = pressure
- \( a \) = van der Waals constant for attraction between particles
- \( b \) = van der Waals constant for volume occupied by particles
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin
- \( V \) = volume

### Step 1: Convert the temperature to Kelvin
Given that the temperature is \( 27^\circ C \):

\[ T = 27 + 273.15 = 300.15 \, K \]

### Step 2: Identify the values
From the problem, we have:
- \( n = 1 \, \text{mole} \)
- \( a = 1.42 \times 10^{12} \, \text{dynes} \cdot \text{cm}^4/\text{mole} \)
- \( b = 30 \, \text{cc} = 30 \, \text{cm}^3 \)
- Initial volume \( V_1 = 2 \, \text{litres} = 2000 \, \text{cm}^3 \)
- Final volume \( V_2 = 20 \, \text{litres} = 20000 \, \text{cm}^3 \)

### Step 3: Calculate the initial pressure using the van der Waals equation
Using the van der Waals equation at the initial state:

\[ P_1 = \frac{nRT}{V - nb} - a \left( \frac{n^2}{V^2} \right) \]

Substituting the values:

\[ P_1 = \frac{1 \times 0.0821 \times 300.15}{2000 - 1 \times 30} - 1.42 \times 10^{12} \left( \frac{1^2}{2000^2} \right) \]

Calculating \( V - nb \):

\[ V - nb = 2000 - 30 = 1970 \, \text{cm}^3 \]

Now substituting into the equation:

\[ P_1 = \frac{0.0821 \times 300.15}{1970} - 1.42 \times 10^{12} \left( \frac{1}{4000000} \right) \]

Calculating \( P_1 \):

\[ P_1 \approx \frac{24.63}{1970} - 3.55 \times 10^6 \]

Calculating the first term:

\[ P_1 \approx 0.0125 - 3.55 \times 10^6 \approx -3.55 \times 10^6 \, \text{dynes/cm}^2 \]

### Step 4: Calculate the work done during the expansion
The work done \( W \) in a reversible process for an ideal gas is given by:

\[ W = -\int_{V_1}^{V_2} P \, dV \]

For a van der Waals gas, we can approximate this as:

\[ W = -nRT \ln \left( \frac{V_2}{V_1} \right) + a \left( \frac{n^2}{V_1} - \frac{n^2}{V_2} \right) \]

Substituting the values:

\[ W = -1 \times 0.0821 \times 300.15 \ln \left( \frac{20000}{2000} \right) + 1.42 \times 10^{12} \left( \frac{1^2}{2000} - \frac{1^2}{20000} \right) \]

Calculating \( \ln \left( \frac{20000}{2000} \right) = \ln(10) \approx 2.302 \):

\[ W = -24.63 \times 2.302 + 1.42 \times 10^{12} \left( \frac{1}{2000} - \frac{1}{20000} \right) \]

Calculating the second term:

\[ \frac{1}{2000} - \frac{1}{20000} = \frac{10 - 1}{20000} = \frac{9}{20000} = 0.00045 \]

Thus,

\[ W = -56.75 + 1.42 \times 10^{12} \times 0.00045 \]

Calculating the second part:

\[ W \approx -56.75 + 639000000 \]

### Final Calculation
The work done during the expansion is approximately:

\[ W \approx 638999943.25 \, \text{dynes/cm}^2 \]

### Conclusion
The work done during the isothermal expansion of the van der Waals gas is approximately \( 639 \, \text{million dynes/cm}^2 \).