JEE Main 2025 Apr 2 Shift 1, Physics Q2: Units And Dimensions

JEE Main2025Apr 2, Shift 1Physics

The equation for real gas is given by $(P + \frac{a}{V ^{2}}) (V - b) = R T$ , where $P, V, T$ and $R$ are the pressure, volume, temperature and gas constant, respectively. The dimension of $a b^{- 2}$ is equivalent to that of :

A
Planck's constant
B
Compressibility
C
Strain
D
Energy density

Solution

For terms inside the bracket to be dimensionally consistent, $\frac{a}{V ^{2}}$ has the dimensions of pressure, and $b$ has the dimensions of volume.

$[a] = [P] [V^{2}] = (M L^{- 1} T^{- 2}) (L^{6}) = M L^{5} T^{- 2}$

$[b] = [V] = L^{3}$

$[a b^{- 2}] = \frac{M L ^{5} T ^{- 2}}{L ^{6}} = M L^{- 1} T^{- 2}$

This matches the dimensions of pressure, which is also equivalent to energy per unit volume (energy density):

$[\frac{Energy}{Volume}] = \frac{M L ^{2} T ^{- 2}}{L ^{3}} = M L^{- 1} T^{- 2}$

$∴$ Option (4) is correct.

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JEE Main 2025 Apr 2 Shift 1, Physics Q2: Units And Dimensions

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