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

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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

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

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(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:

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