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NEET PHYSICSMedium

The equation of state of some gases can be expressed as (P + a/V²)(V − b) = RT. Here P is the pressure, V is the volume, T is the absolute temperature and a, b, R are constants. The dimensions of ‘a’ are:

A

ML⁵T⁻²

B

ML⁻¹T⁻²

C

M⁰L³T⁰

D

M⁰L⁶T⁰

Step-by-Step Solution

According to the Principle of Homogeneity of Dimensions, only physical quantities having the same dimensions can be added to or subtracted from each other.

  1. In the term (P+aV2)(P + \frac{a}{V^2}), the quantity aV2\frac{a}{V^2} must have the same dimensions as Pressure (PP).
  2. Dimensions of Pressure (PP): Force per unit area. [P]=[Force][Area]=[MLT−2][L2]=[ML−1T−2][P] = \frac{[Force]}{[Area]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]
  3. Dimensions of Volume (VV): [L3][L^3].
  4. Calculate Dimensions of 'a': [aV2]=[P][\frac{a}{V^2}] = [P] [a]=[P]×[V]2[a] = [P] \times [V]^2 [a]=[ML−1T−2]×([L3])2[a] = [ML^{-1}T^{-2}] \times ([L^3])^2 [a]=[ML−1T−2]×[L6][a] = [ML^{-1}T^{-2}] \times [L^6] [a]=[ML5T−2][a] = [ML^5T^{-2}]
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