In the equation (P+frac{a}{V^2})(V-b)=RT,wher | vedclass

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(B) According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be added or subtracted. In the given eq

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$ML^5 T^{-2}$

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(B) According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be added or subtracted. In the given equation $(P+\frac{a}{V^2})(V-b)=RT$,the term $\frac{a}{V^2}$ is added to pressure $P$. Therefore,the dimensions of $\frac{a}{V^2}$ must be equal to the dimensions of pressure $P$. $[P] = [\frac{a}{V^2}] \implies [a] = [P][V^2]$. The dimension of pressure $P$ is $[ML^{-1} T^{-2}]$ and the dimension of volume $V$ is $[L^3]$. Substituting these,we get $[a] = [ML^{-1} T^{-2}] 	imes [L^3]^2 = [ML^{-1} T^{-2}] 	imes [L^6] = [ML^5 T^{-2}]$.

In the equation $(P+\frac{a}{V^2})(V-b)=RT$,where $P$ is pressure,$V$ is volume,$T$ is temperature,$R$ is universal gas constant,and $a$ and $b$ are constants. The dimensions of $a$ are

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