Solution:
From dimensional homogeneity, \([a/V^2] = [P]\). \([a] = [P][V^2]\). Pressure \(P = [ML^{-1}T^{-2}]\), Volume \(V = [L^3]\). So, \([a] = [ML^{-1}T^{-2}][L^3]^2 = [ML^{-1}T^{-2}L^6] = [ML^5T^{-2}]\).
An equation is given here \[\left(P + \frac{a}{V^2}\right) = b\frac{\theta}{V}\] where P = Pressure, V = Volume and \(\theta =\) Absolute temperature. If (a) and (b) are constants, the dimensions of (a) will be:
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