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JEE Main 2025 (2 Apr, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2025 (2 Apr, Shift 1)

Q12JEE Main 2025 · 2 APR · SHIFT 1Easy

Applications of Dimensional Analysis

The equation for 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,T\mathrm{P, V, T} and R are the pressure, volume, temperature and gas constant, respectively. The dimension of ab−2\mathrm{ab^{-2}} is equivalent to that of:

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Answer: (c)

  1. [a/V2]=[P]⇒[a]=[P][L6][a/V^2] = [P] \Rightarrow [a] = [P][L^6]; [b]=[V]=[L3][b] = [V] = [L^3].
  2. [ab−2]=[P][L6]/[L6]=[P]=[ML−1T−2][ab^{-2}] = [P][L^6]/[L^6] = [P] = [ML^{-1}T^{-2}].
  3. Pressure has the same dimensions as energy per volume.
  4. So the answer is (c) energy density.
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