At low pressure, the Vander Waal’s equation is reduced to

Options

A. `Z=(PV_(m))/(RT)=1-(aP)/(RT)`
B. `Z=(PV_(m))/(RT)=1+(bP)/(RT)`
C. `PV_(m)=RT`
D. `Z=(PV_(m))/(RT)=1-(a)/(RT)`

Answer

To derive the reduced form of the Van der Waals equation at low pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Start with the Van der Waals Equation**:
The Van der Waals equation for one mole of a gas is given by:
\[ \left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT \]
where \( P \) is the pressure, \( V_m \) is the molar volume, \( a \) and \( b \) are Van der Waals constants, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

2. **Consider Low Pressure**:
At low pressure, the volume \( V_m \) becomes much larger than the constant \( b \) (the excluded volume). Therefore, we can approximate:
\[ V_m - b \approx V_m \]
This simplification allows us to ignore the volume correction due to the size of the molecules.

3. **Substitute the Approximation into the Equation**:
Substituting this approximation into the Van der Waals equation gives:
\[ \left( P + \frac{a}{V_m^2} \right) V_m = RT \]

4. **Rearranging the Equation**:
Expanding this equation, we have:
\[ P V_m + \frac{a}{V_m} = RT \]
Rearranging gives:
\[ P V_m = RT - \frac{a}{V_m} \]

5. **Divide by RT**:
Dividing the entire equation by \( RT \) yields:
\[ \frac{P V_m}{RT} = 1 - \frac{a}{RT V_m} \]

6. **Define the Compressibility Factor (Z)**:
The compressibility factor \( Z \) is defined as:
\[ Z = \frac{PV_m}{RT} \]
Thus, we can rewrite the equation as:
\[ Z = 1 - \frac{a}{RT V_m} \]

7. **Conclusion**:
Therefore, at low pressure, the Van der Waals equation reduces to:
\[ Z = 1 - \frac{a}{RT V_m} \]