If the slope of 'Z' (compressibility factor) vs. 'p' curve is constant `("slope"=(pi)/(492.6)atm^(-1))` at a particular temperature (300 K) abd very high pressure, then calculate diameter of the molecules.
(Given : `N_(A)=6.0xx10^(23), R=0.0821 atm.Lmol^(-1)K^(-1)`)

Options

A. 7.5 Å
B. 5 Å
C. 2.5 Å
D. 1.25 Å

Explanation

To solve the problem, we need to calculate the diameter of the molecules based on the given slope of the compressibility factor (Z) vs. pressure (P) curve. The slope is given as \(\frac{\pi}{492.6} \, \text{atm}^{-1}\) at a temperature of 300 K and very high pressure.

### Step-by-Step Solution:

1. **Understanding the Compressibility Factor (Z)**:
The compressibility factor \(Z\) is defined as:
\[ Z = \frac{PV}{nRT} \]
where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature.

2. **Using the Van der Waals Equation**:
For real gases, the Van der Waals equation is:
\[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \]
where \(a\) accounts for intermolecular forces and \(b\) accounts for the volume occupied by the gas molecules.

3. **At High Pressure**:
At very high pressure, the term \(a\) becomes negligible, and we can simplify the equation to:
\[ PV \approx nRT \]
This leads to:
\[ Z \approx 1 + \frac{PB}{RT} \]
where \(B\) is the volume excluded due to the finite size of the molecules.

4. **Finding the Slope**:
The slope of the \(Z\) vs. \(P\) curve is given as:
\[ \text{slope} = \frac{\pi}{492.6} \, \text{atm}^{-1} \]
From the equation \(Z = 1 + \frac{PB}{RT}\), we can see that the slope is equal to \(\frac{B}{RT}\).

5. **Calculating B**:
Rearranging the slope equation gives:
\[ B = \text{slope} \times RT \]
Substituting the values:
\[ B = \left(\frac{\pi}{492.6}\right) \times (0.0821 \, \text{atm L mol}^{-1} \text{K}^{-1}) \times (300 \, \text{K}) \]

6. **Calculating B Numerically**:
\[ B = \frac{\pi \times 0.0821 \times 300}{492.6} \]
\[ B \approx \frac{77.439}{492.6} \approx 0.157 \]

7. **Relating B to Molecular Volume**:
The value of \(B\) can be expressed in terms of the volume of one molecule:
\[ B = 4 \times N_A \times V_m \]
where \(V_m\) is the volume of one molecule and \(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)).

8. **Finding the Volume of One Molecule**:
Rearranging gives:
\[ V_m = \frac{B}{4N_A} \]
Substituting the values:
\[ V_m = \frac{0.157}{4 \times 6.022 \times 10^{23}} \approx \frac{0.157}{2.4088 \times 10^{24}} \approx 6.52 \times 10^{-26} \, \text{m}^3 \]

9. **Calculating the Diameter**:
The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Setting \(V_m = \frac{4}{3} \pi r^3\) and solving for \(r\):
\[ r^3 = \frac{3V_m}{4\pi} \]
\[ r = \left(\frac{3 \times 6.52 \times 10^{-26}}{4\pi}\right)^{1/3} \]

10. **Calculating the Diameter**:
\[ d = 2r \]
After calculating \(r\) and multiplying by 2, we find:
\[ d \approx 5 \, \text{Å} \]

### Final Answer:
The diameter of the molecules is approximately \(5 \, \text{Å}\).