The compressibility factor $(Z)$ of one mole of a van der Waals gas of negligible $�a�$ value is
- $1$
- $\frac{bp}{RT}$
- $1+\frac{bp}{RT}$
- $1-\frac{bp}{RT}$
The Correct Option is C
Solution and Explanation
$\left(p+\frac{a}{V^{2}}\right)(V-b)=R T\,\,\,$(for 1 mole)
If $a$ is negligible, $p+\frac{a}{V^{2}} \approx p$
$p(V-b)=R T $
$\Rightarrow p V-p b=R T $
or $\frac{p V}{R T}-\frac{p b}{R T}=1 $
$\Rightarrow \frac{p V}{R T}=1+\frac{p b}{R T}$
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Concepts Used:
Van Der Waals Equation
Van der Waals equation is an equation relating the relationship between the pressure, volume, temperature, and amount of real gases.
Read More: Derivation of Van Der Waals Equation
Derivation of Van der Waals equation:
For a real gas containing ‘n’ moles, the equation is written as
Where, P, V, T, n are the pressure, volume, temperature and moles of the gas. ‘a’ and ‘b’ constants specific to each gas.
Where,
Vm: molar volume of the gas
R: universal gas constant
T: temperature
P: pressure
V: volume
Thus, Van der Waals equation can be reduced to ideal gas law as PVm = RT.
The equation can further be written as;
- Cube power of volume:
- Reduced equation (Law of corresponding states) in terms of critical constants:
Units of Van der Waals equation Constants
a: atm lit² mol-²
b: litre mol-¹