The compressibility factor for a real gas at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The Van der Waals equation for a real gas is $(P + \frac{a}{V^2})(V - b) = RT$. At high pressure,the term $\frac{a}{V^2}$ is very small compared t

Vedclass · Accepted Answer

$1 + \frac{Pb}{RT}$

Vedclass · Answer

(B) The Van der Waals equation for a real gas is $(P + \frac{a}{V^2})(V - b) = RT$. At high pressure,the term $\frac{a}{V^2}$ is very small compared to $P$ and can be neglected. Thus,the equation simplifies to $P(V - b) = RT$. Expanding this,we get $PV - Pb = RT$. Dividing both sides by $RT$,we get $\frac{PV}{RT} - \frac{Pb}{RT} = 1$. Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z - \frac{Pb}{RT} = 1$. Therefore,$Z = 1 + \frac{Pb}{RT}$. Hence,the correct option is $B$.

The compressibility factor for a real gas at high pressure is :-

Similar Questions

The term that accounts for intermolecular force in Van der Waals equation for non-ideal gas is

When does a real gas behave like an ideal gas?

The van der Waals equation is valid for $......$.

If $Z$ is the compressibility factor,then for $1 \, \text{mole}$ of a real gas,the van der Waals equation at low pressure can be written as:

The $Van \text{ der } Waals$ equation explains the behaviour of

Vedclass Test Series

Exam Paper Generator

Online Exam Module