Question

The van der Waals' equation of state for real gases is given as $\left(p+\frac{a}{V^{2}}\right)(V-b)=n R T$. Which of the following terms has dimensions different from that of energy?

• A. $ pV $
• B. $ \frac{a}{{{V}^{2}}} $
• C. $\frac{a b}{V^{2}}$
• D. $ Vp $

Answer 1

Answer: (B)

(i) Dimensions of $pV =[$ dimensions of $p ][$ dimension of $V ]$
$=\left[M L^{-1} T^{-2}\right]\left[L^{3}\right]$
$=\left[M L^{2} T^{-2}\right]$
Dimensions of energy $=[$ dimensions of force $][$ dimension of distance $]$
$=\left[M L T^{-2}\right][L]$
$=\left[M L^{2} T^{-2}\right]$
Hence, dimensions of energy $=$ dimensions of $pV$
(ii) From the given equation
$\left[p+\frac{a}{V^{2}}\right](v-b)=n R T$
Dimensions of $\frac{a}{V^{2}}$ should be dimensionally equal to pressure $p_1$.
(iii) Dimensions of $a = [$ dimensions of $p] [$dimensions of $\left.V^{2}\right]$
$=\left[M L^{-1} T^{-2}\right]\left[L^{6}\right]$
Dimensions of $b =[$ dimensions of volume $]$
$=\left[L^{3}\right]$
$\therefore $ Dimensions of $\frac{a b}{V^{2}}=\frac{\left[M L^{5} T^{-2}\right]\left[L^{3}\right]}{\left[L^{6}\right]}=\left[M L^{2} T^{-2}\right]$
(iv) Dimensions of $V p=\left[L^{3}\right]\left[M L^{-1} T^{-2}\right]$
$=\left[M L^{2} T^{-2}\right]$