Q72 / 75 · JEE Main 2025 (2 Apr Shift 1) · Physics

The equation for real gas is given by $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$, where $\mathrm{P}, \mathrm{V}, \mathrm{T}$ and R are the pressure, volume, temperature and gas constant, respectively. The dimension of $\mathrm{ab}^{-2}$ is equivalent to that of :

  • A
    Planck's constant
  • B
    Compressibility
  • C
    Strain
  • Correct answer
    D
    Energy density

Solution

$\begin{aligned} & {\left[\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right](\mathrm{V}-\mathrm{b})=\mathrm{RT}} \\ & \therefore[\mathrm{a}]=[\mathrm{P}]\left[\mathrm{V}^2\right]=\mathrm{ML}^{-1} \mathrm{~T}^{-2} \mathrm{~L}^6=\mathrm{ML}^5 \mathrm{~T}^{-2} \\ & {[\mathrm{~b}]=[\mathrm{V}]=\mathrm{L}^3} \\ & {\left[\mathrm{ab}^{-2}\right]=\mathrm{ML}^5 \mathrm{~T}^{-2} \mathrm{~L}^{-6}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}}\end{aligned}$
Dimension of energy density.

Hence, the correct answer is option: 'D'.

Relevant Questions