The unit of the van der Waals gas equation pa | vedclass

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

Question

(D) In the van der Waals equation $(P + \frac{an^{2}}{V^{2}})(V - nb) = nRT$,the term $\frac{an^{2}}{V^{2}}$ is added to pressure $P$.According to the

Vedclass · Accepted Answer

$atm \, dm^{6} \, mol^{-2}$

Vedclass · Answer

(D) In the van der Waals equation $(P + \frac{an^{2}}{V^{2}})(V - nb) = nRT$,the term $\frac{an^{2}}{V^{2}}$ is added to pressure $P$.According to the principle of dimensional homogeneity,only quantities with the same units can be added.Therefore,the unit of $\frac{an^{2}}{V^{2}}$ must be the same as the unit of pressure ($atm$ or $bar$ or $Pa$).$\frac{an^{2}}{V^{2}} = 	ext{Pressure} \Rightarrow a = 	ext{Pressure} 	imes \frac{V^{2}}{n^{2}}$.Substituting the units: $a = atm 	imes \frac{(dm^{3})^{2}}{(mol)^{2}} = atm \, dm^{6} \, mol^{-2}$.

The unit of the van der Waals gas equation parameter $a$ in $(P + \frac{an^{2}}{V^{2}})(V - nb) = nRT$ is :

Similar Questions

Under which conditions does the $van \ der \ Waals$ equation reduce to the ideal gas equation?

For which of the following gases will the van der Waals constant $'a'$ have the highest value?

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

$A$ gas deviates from ideal behaviour at a high pressure because its molecules

Vedclass Test Series

Exam Paper Generator

Online Exam Module