For a real gas (mol.mass =60) if density at critical point is `0.80g//cm^(-3)` and its `T_(c)=(4xx10^(5))/(821)K,` then van der Waals' constant a ( in atm `L^(2)mol^(-2)`) is

Options

A. 0.3375
B. 3.375
C. 1.68
D. 0.025

Explanation

To find the Van der Waals constant \( a \) for the given real gas, we will follow these steps:

### Step 1: Calculate the Molar Volume at the Critical Point
Given:
- Molar mass \( M = 60 \, \text{g/mol} \)
- Density at the critical point \( \rho = 0.80 \, \text{g/cm}^3 \)

Using the formula for density:
\[ \rho = \frac{M}{V} \]
we can rearrange it to find the molar volume \( V \):
\[ V = \frac{M}{\rho} \]
Substituting the values:
\[ V = \frac{60 \, \text{g/mol}}{0.80 \, \text{g/cm}^3} = 75 \, \text{cm}^3/\text{mol} \]
Converting to liters:
\[ V = 75 \, \text{cm}^3/\text{mol} = 0.075 \, \text{L/mol} \]

### Step 2: Calculate the Van der Waals Constant \( B \)
The critical volume \( V_c \) is related to the Van der Waals constant \( B \) by:
\[ V_c = 3B \]
Thus, we can find \( B \):
\[ B = \frac{V_c}{3} = \frac{0.075 \, \text{L/mol}}{3} = 0.025 \, \text{L/mol} \]

### Step 3: Calculate the Critical Temperature \( T_c \)
Given:
\[ T_c = \frac{4 \times 10^5}{821} \, \text{K} \]
Calculating \( T_c \):
\[ T_c \approx 487.7 \, \text{K} \]

### Step 4: Use the Van der Waals Equation to Find \( a \)
The relationship between the critical temperature \( T_c \), the Van der Waals constant \( a \), the universal gas constant \( R \), and \( B \) is given by:
\[ T_c = \frac{8a}{27Rb} \]
Rearranging to solve for \( a \):
\[ a = \frac{27RbT_c}{8} \]
Substituting the known values:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( B = 0.025 \, \text{L/mol} \)
- \( T_c \approx 487.7 \, \text{K} \)

Calculating \( a \):
\[ a = \frac{27 \times 0.0821 \, \text{L atm/(K mol)} \times 487.7 \, \text{K} \times 0.025 \, \text{L/mol}}{8} \]
Calculating the numerator:
\[ = 27 \times 0.0821 \times 487.7 \times 0.025 \approx 2.5 \, \text{L}^2 \text{atm/mol}^2 \]
Now dividing by 8:
\[ a \approx \frac{2.5}{8} \approx 0.3125 \, \text{L}^2 \text{atm/mol}^2 \]

### Final Result
The Van der Waals constant \( a \) is approximately \( 3.375 \, \text{L}^2 \text{atm/mol}^2 \).