Question

An equation is given here P + a /v2= B theta/V where p= pressure V = volume and theta = Absolute Temperature if A and B are constant, the dimension of a will be

Answer 1

Given, $\bf{P + \frac{a}{V^2}=\frac{B\theta}{V}}$
here, P is the pressure,
V is the volume
and $\theta$ is the absolute temperature.
we know, dimension of P = $[ML^{-1}T^{-2}]$
dimension of V = $[L^3]$
dimension of $\theta$ = K

so, dimension of P = dimension of {a/V²}
dimension of P × dimension of V² = dimension of a
$[ML^{-1}T^{-2}][L^6]K$ = dimension of a
$ML^5T^{-2}K$ = dimension of a,

dimension of a = $ML^5T^{-2}K$

Answer 2

Keeping in tn equation is given here P + a /v2= B theta/V where p= pressure V = volume and theta = Absolute Temperature if A and B are constant, the dimension of a will be [M1 L5 T-2 ]

Explanation:

Here, P and a/V2 are added which means they have the same dimensions.

Therefore,

[M1 L-1 T-2 ] = a/[M0 L 6 T 0 ]

a = [M1 L-1 T-2 ] x [M0 L 6 T 0 ]

a = [M1 L5 T-2 ]