Identify the conditions at which the van der | vedclass

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

Question

(C) The van der Waals equation is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$. For $1 \ mol$ of gas,this becomes $(P + \frac{a}{V^2})(V - b) = RT$

Vedclass · Accepted Answer

High temperature and low pressure

Vedclass · Answer

(C) The van der Waals equation is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$. For $1 \ mol$ of gas,this becomes $(P + \frac{a}{V^2})(V - b) = RT$. At high temperature,the kinetic energy of gas molecules is very high,making the intermolecular forces of attraction (represented by the constant $a$) negligible. At low pressure,the volume $V$ is very large,making the volume occupied by gas molecules (represented by the constant $b$) negligible compared to the total volume. Under these conditions,$(P + 0)(V - 0) = RT$,which simplifies to $PV = RT$,the ideal gas equation.

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

Similar Questions

Calculate the compressibility factor $(Z)$ of $1 \text{ mole}$ of a certain real gas if it occupies $0.4 \text{ dm}^3$ at $300 \text{ K}$ and $40 \text{ atm}$. [Given: $R = 0.082 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$]

Which statement is correct regarding a real gas?

Derive the van der Waals equation.

Why do real gases deviate from ideal behavior in the $pV$ vs $p$ and $p$ vs $V$ graphs at specific temperatures?

The compressibility factor for a real gas at high pressure is

Vedclass Test Series

Exam Paper Generator

Online Exam Module