The units of constant a in the Van der Waals | vedclass

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

Question

(A) The Van der Waals equation for $n$ moles of a gas is given by: $(P + \frac{n^2 a}{V^2})(V - nb) = nRT$. According to the principle of dimensional

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The Van der Waals equation for $n$ moles of a gas is given by: $(P + \frac{n^2 a}{V^2})(V - nb) = nRT$. According to the principle of dimensional homogeneity,the term $\frac{n^2 a}{V^2}$ must have the same units as pressure $(P)$. Therefore,the unit of $\frac{n^2 a}{V^2} = 	ext{unit of } P = atm$. Substituting the units: $\frac{(mol)^2 \cdot a}{(L)^2} = atm$. Rearranging for $a$: $a = atm \cdot L^2 \cdot mol^{-2}$. Since $1 \, L = 1 \, dm^3$,then $L^2 = (dm^3)^2 = dm^6$. Thus,the unit of $a$ is $dm^6 \, atm \, mol^{-2}$.

The units of constant $a$ in the Van der Waals equation are:

Similar Questions

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

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

For a real gas (molar mass $= 60 \, g/mol$),if the density at the critical point is $0.80 \, g/cm^3$ and its critical temperature $T_c = \frac{4 \times 10^5}{821} \, K$,then the van der Waals constant $a$ (in $atm \, L^2 \, mol^{-2}$) is:

Maximum deviation from ideal gas behavior is expected from which of the following gases?

In the van der Waals equation,which term indicates the presence of attractive forces in a gas?

Vedclass Test Series

Exam Paper Generator

Online Exam Module