The compressibility factor (Z) of one mole of | vedclass

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

Question

(C) The van der Waals' equation for $1$ mole of gas is $(p+\frac{a}{V^2})(V-b) = RT$. Since '$a$' is negligible, the equation simplifies to $p(V-b) =

Vedclass · Accepted Answer

$1+\frac{b p}{R T}$

Vedclass · Answer

(C) The van der Waals' equation for $1$ mole of gas is $(p+\frac{a}{V^2})(V-b) = RT$. Since '$a$' is negligible, 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 $Z = \frac{pV}{RT}$, the equation becomes $Z - \frac{pb}{RT} = 1$. Therefore, $Z = 1 + \frac{pb}{RT}$.

The compressibility factor $(Z)$ of one mole of a van der Waals' gas of negligible '$a$' value is

Similar Questions

The correct equation for one mole of a real gas is ($a, b$ are constants)

For a gas,the deviation from ideal behavior is maximum at:

Under which of the following conditions does a gas deviate from ideal behaviour?
$A$. Low pressure
$B$. High pressure
$C$. Low temperature
$D$. High temperature

Identify the conditions at which the van der Waals equation of state changes to the ideal gas equation.

At low pressure,the van der Waals equation becomes:

Vedclass Test Series

Exam Paper Generator

Online Exam Module