Medium MCQ +4 / -1 PYQ · JEE Mains 2025

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 :

  1. A Compressibility
  2. B Planck's constant
  3. C Energy density Correct answer
  4. D Strain

Solution

<p>The equation for a real gas is given by:</p> <p>$ \left(\mathrm{P} + \frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V} - \mathrm{b}) = \mathrm{RT} $</p> <p>Here, $ \mathrm{P} $, $ \mathrm{V} $, $ \mathrm{T} $, and $ R $ are the pressure, volume, temperature, and gas constant, respectively.</p> <p>To find the dimension of $ ab^{-2} $, we proceed with the following steps:</p> <p>Rearrange the equation for dimensions:</p> <p><p>$ [\mathrm{a}] = [\mathrm{P}][\mathrm{V}^2] $</p></p> <p><p>Since $[\mathrm{P}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}$ and $[\mathrm{V}] = \mathrm{L}^3$, we have:</p> <p>$ [\mathrm{a}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}(\mathrm{L}^6) = \mathrm{ML}^5\mathrm{T}^{-2} $</p></p> <p>As derived from the equation:</p> <p>$[\mathrm{b}] = [\mathrm{V}] = \mathrm{L}^3$</p> <p>Now, to find $[\mathrm{ab}^{-2}]$:</p> <p>$[\mathrm{ab}^{-2}] = \frac{[\mathrm{a}]}{[\mathrm{b}]^2} = \mathrm{ML}^5\mathrm{T}^{-2} \cdot \mathrm{L}^{-6} = \mathrm{ML}^{-1}\mathrm{T}^{-2}$</p> <p>This dimension, $\mathrm{ML}^{-1}\mathrm{T}^{-2}$, is equivalent to the dimension of energy density.</p>

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Thermodynamic Processes

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