Easy MCQ +4 / -1 PYQ · JEE Mains 2024

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

  1. A P Correct answer
  2. B RT
  3. C PV
  4. D R

Solution

<p>$$\begin{aligned} & {[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right] \Rightarrow[\mathrm{a}]=\left[\mathrm{PV}^2\right]} \\ & \text { And }[\mathrm{V}]=[\mathrm{b}] \\ & \frac{[\mathrm{a}]}{\left[\mathrm{b}^2\right]}=\frac{\left[\mathrm{PV}^2\right]}{\left[\mathrm{V}^2\right]}=[\mathrm{P}] \end{aligned}$$</p>

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

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