Question

The equation of state of a gas is given by $(P+\frac{a}{V^3})(V-b^2)$ = cT, where P, V, Tare pressure, volume and temperature respectively, and a, b, c are constants. The dimensions of a and b are respectively:

• A. $\left[M{L}^8{T}^{-2}\right]$ and $\left[L^{3/2}\right]$
• B. $\left[M{L}^5{T}^{-2}\right]$ and $\left[L^3\right]$
• C. $\left[M{L}^5{T}^{-2}\right]$ and $\left[L^6\right]$
• D. $\left[M{L}^6{T}^{-2}\right]$ and $\left[L^{3/2}\right]$

Answer 1

Answer: (A)

$\left[M{L}^8{T}^{-2}\right]$ and $\left[L^{3/2}\right]$

Given, $[P+\frac{a}{V^3}](V-b^2)=cT$
Dimension of $\frac{a}{V^3}$ = Dimensions of $P$
$\therefore$ Dimensions of a = dimensions of $P{V}^3$
$\left[a\right]=\left[\frac{F}{A}{V}^3\right]$ $(\therefore P=\frac{F}{A})$
$=\frac{\left[ML{T}^{-2}\right]}{L^2}\times [L^3{]}^3=[M{L}^8{T}^{-2}]$
Dimensions of $b^2$ = dimensions of V
$\therefore \left[b\right]=[V{]}^{1/2}=[L^3{]}^{1/2}$ or $\left[b\right]=\left[L^{3/2}\right]$