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The Van der Waals’ equation for the one mol of gas is given by $(P + \frac{a}{V^{2}}) (V - b) = R T$ where P is pressure; V is volume; T is absolute temperature, R universal gas constant and a, b are constants. The dimensional formula of $(\frac{a b}{R T})$ is

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$[M L^{2} T^{- 2}]$

$[M^{0} L^{- 3} T^{0}]$

$[M L^{- 1} T^{- 2}]$

$[M^{0} L^{6} T^{0}]$

answer is D.

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Detailed Solution

$P and \frac{a}{V^{2}}$ will have same dimensions

Similarly $V and b$ will have same dimensions

Therefore, dimensions of a are

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

Dimensions of b are

$b = V = L^{3}$

Therefore dimensions of $(\frac{a b}{R T})$ are

$(\frac{a b}{R T}) = (\frac{[M L^{5} T^{- 2}] [L^{3}]}{M L^{2} T^{- 2}}) = M^{0} L^{6} T^{0}$

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