At high pressure,the form of Van der Waals eq | vedclass

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

Question

(B) The Van der Waals equation for $1$ mole of gas is $\left( P + \frac{a}{V^2} \right) (V - b) = RT$.At high pressure,the volume $V$ is small,so the

Vedclass · Accepted Answer

$P(V - b) = RT$

Vedclass · Answer

(B) The Van der Waals equation for $1$ mole of gas is $\left( P + \frac{a}{V^2} \right) (V - b) = RT$.At high pressure,the volume $V$ is small,so the correction term $\frac{a}{V^2}$ becomes very large compared to $P$. However,the term $(V - b)$ cannot be neglected because $V$ is comparable to $b$.Thus,the pressure correction term $\frac{a}{V^2}$ is neglected relative to $P$,simplifying the equation to $P(V - b) = RT$.

At high pressure,the form of Van der Waals equation becomes

Similar Questions

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

What is compressibility factor $(Z)$? Explain $(i)$ deviation factor $(Z = 1)$,$(ii)$ $Z > 1$,$(iii)$ $Z < 1$,$(iv)$ deviation graph,and $(v)$ the relation between molar volume and $Z$.

Explain real gases showing deviations from ideal gas with correction of pressure and volume and derive van der Waals equation.

At relatively high pressure,the van der Waals equation becomes?

The variation of the compressibility factor $(Z)$ with pressure $(P$ in $bar)$ for some gases is shown in the figure below. Identify the gases $(A)$, $(B)$, and $(C)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module