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Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 27 January, Shift 2 — Question 31

The equation of state of a real gas is given by (P+aV2)(V−b)=RT\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^{2}}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}, where P,V\mathrm{P}, \mathrm{V} and T are pressure. volume and temperature respectively and R is the universal gas constant. The dimensions of ab2\frac{a}{b^{2}} is similar to that of :

  1. Option A:

    PV

  2. Option B:

    PP

    Correct
  3. Option C:

    RT

  4. Option D:

    R

Answer: B

Step-by-step solution

[P]=[aV2]⇒[a]=[PV2][\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^{2}}\right] \Rightarrow[\mathrm{a}]=\left[\mathrm{PV}^{2}\right] And

[V]=[b][\mathrm{V}]=[\mathrm{b}]

[a][b2]=[PV2][V2]=[P]\frac{[\mathrm{a}]}{\left[\mathrm{b}^{2}\right]}=\frac{\left[\mathrm{PV}^{2}\right]}{\left[\mathrm{V}^{2}\right]}=[\mathrm{P}]

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis

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