At low pressures,van der Waals' equation is w | vedclass

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

Question

(A) The van der Waals' equation for one mole of gas is $(P + \frac{a}{V^2})(V - b) = RT$. At low pressures,the volume $V$ is large,so $b$ is negligibl

Vedclass · Accepted Answer

$(1 - \frac{a}{RTV})$

Vedclass · Answer

(A) The van der Waals' equation for one mole of gas is $(P + \frac{a}{V^2})(V - b) = RT$. At low pressures,the volume $V$ is large,so $b$ is negligible compared to $V$. The equation simplifies to $(P + \frac{a}{V^2})V = RT$. Expanding this,we get $PV + \frac{a}{V} = RT$. Dividing both sides by $RT$,we get $\frac{PV}{RT} + \frac{a}{VRT} = 1$. Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z + \frac{a}{VRT} = 1$. Therefore,$Z = 1 - \frac{a}{VRT}$.

At low pressures,van der Waals' equation is written as $(P + \frac{a}{V^2})V = RT$. The compressibility factor $Z$ will be

Similar Questions

The deviation will be maximum in the behaviour of a gas from the ideal gas equation $PV = nRT$?

At relatively high pressure,van der Waals' equation becomes

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

What is the correction term in the pressure for real gas in comparison to the ideal gas?

Under which conditions do intermolecular forces become effective?

Vedclass Test Series

Exam Paper Generator

Online Exam Module