Question

The van der Waals' equation of state for real gases is given as $\left(p+\frac{a}{V^2}\right)(V-b)=nRT$. Which of the following terms has dimensions different from that of energy?

• A. $pV$
• B. $\frac{a}{V^2}$
• C. $\frac{ab}{V^2}$
• 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 $]$
$=\left[ML{T}^{-2}\right][L]$
$=\left[M{L}^2{T}^{-2}\right]$
Hence, dimensions of energy $=$ dimensions of $pV$
(ii) From the given equation
$\left[p+\frac{a}{V^2}\right](v-b)=nRT$
Dimensions of $\frac{a}{V^2}$ should be dimensionally equal to pressure $p_1$.
(iii) Dimensions of $a=[$ dimensions of $p][$dimensions of $V^2]$
$=\left[M{L}^{-1}{T}^{-2}\right]\left[L^6\right]$
Dimensions of $b=[$ dimensions of volume $]$
$=\left[L^3\right]$
$\therefore$ Dimensions of $\frac{ab}{V^2}=\frac{\left[M{L}^5{T}^{-2}\right]\left[L^3\right]}{\left[L^6\right]}=\left[M{L}^2{T}^{-2}\right]$
(iv) Dimensions of $Vp=\left[L^3\right]\left[M{L}^{-1}{T}^{-2}\right]$
$=\left[M{L}^2{T}^{-2}\right]$