If Z is a compressibility factor,the van der | vedclass

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

Question

(B) The van der Waals equation for $1 \ mol$ of a gas is given by:$(P + \frac{a}{V^2})(V - b) = RT$At low pressure,the volume $V$ is very large,so $V

Vedclass · Accepted Answer

$Z=1 - \frac{a}{VRT}$

Vedclass · Answer

(B) The van der Waals equation for $1 \ mol$ of a gas is given by:$(P + \frac{a}{V^2})(V - b) = RT$At low pressure,the volume $V$ is very large,so $V >> b$. Therefore,the term $(V - b)$ can be approximated as $V$.The equation simplifies to:$(P + \frac{a}{V^2})V = RT$Expanding the equation:$PV + \frac{a}{V} = RT$Rearranging to solve for $PV$:$PV = RT - \frac{a}{V}$Dividing both sides by $RT$ to express in terms of the compressibility factor $Z = \frac{PV}{RT}$:$\frac{PV}{RT} = 1 - \frac{a}{VRT}$Thus,$Z = 1 - \frac{a}{VRT}$.

If $Z$ is a compressibility factor,the van der Waals equation at low pressure can be written as:

Similar Questions

Under which of the following conditions does a real gas behave most like an ideal gas?

The compressibility factor $(Z)$ of a gas is less than unity at $STP$. Therefore:

The graph between $P$ and $V$ below the critical temperature is:

The values of van der Waals' constant '$a$' for gases $O_2, N_2, NH_3$,and $CH_4$ are $1.360, 1.390, 4.170$,and $2.253 \ L^2 \ atm \ mol^{-2}$ respectively. The gas which can be most easily liquefied is

The van der Waals equation reduces to the ideal gas equation under which of the following conditions?

Vedclass Test Series

Exam Paper Generator

Online Exam Module