A) \[pV\]
B) \[\frac{a}{{{V}^{2}}}\]
C) \[\frac{ab}{{{V}^{2}}}\]
D) \[Vp\]
Correct Answer: B
Solution :
(i) Dimensions of pV = [dimensions of p] [dimension of V ] \[=[M{{L}^{-1}}{{T}^{-2}}][{{L}^{3}}]\] \[=[M{{L}^{2}}{{T}^{-2}}]\] Dimensions of energy = [dimensions of force] [dimension of distance] \[=[ML{{T}^{-2}}][L]\] \[=[M{{L}^{2}}{{T}^{-2}}]\] Hence, dimensions of energy = dimensions of pV (ii) From the given equation \[\left[ p+\frac{a}{{{V}^{2}}} \right](v-b)=nRT\] Dimensions of \[\frac{a}{{{V}^{2}}}\] should be dimensionally equal to pressure pi. (iii) Dimensions of a = [dimensions of p] [dimensions of \[{{V}^{2}}\]] \[=[M{{L}^{-1}}{{T}^{-2}}][{{L}^{6}}]\] Dimensions of b = [dimensions of volume] \[=[{{L}^{3}}]\] \[\therefore \] Dimensions of \[\frac{ab}{{{V}^{2}}}=\frac{[M{{L}^{5}}{{T}^{-2}}][{{L}^{3}}]}{[{{L}^{6}}]}=[M{{L}^{2}}{{T}^{-2}}]\] (iv) Dimensions of \[Vp=[{{L}^{3}}][M{{L}^{-1}}{{T}^{-2}}]\] \[=[M{{L}^{2}}{{T}^{-2}}]\]
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