Question

In the equation $ \left( P+\frac{a}{{{V}^{2}}} \right)(V-b)=RT, $ where P = pressure, V= volume, a and b are constants, the dimensions of a are:

• A. $ \text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{5}}}{{\text{T}}^{\text{-1}}}\text{ }\!\!]\!\!\text{ } $
• B. $ \text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{-5}}}{{\text{T}}^{\text{-1}}}\text{ }\!\!]\!\!\text{ } $
• C. $ \text{ }\!\![\!\!\text{ M}{{\text{L}}^{5}}{{\text{T}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ } $
• D. $ \text{ }\!\![\!\!\text{ }{{\text{M}}^{\text{-1}}}{{\text{L}}^{5}}{{\text{T}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ } $

Answer 1

Answer: (C)

According to principle of homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same.
We have given
$\left(p+\frac{a}{V^{2}}\right)(V-b)=R T$
According to principle of homogeneity,
$[P]=\left[\frac{a}{V^{2}}\right]$
or $[a]=[P]\left[V^{2}\right]$
or $[a]=\left[M L^{-1} T^{-2}\right]\left[L^{6}\right]$
$\therefore [a]=\left[M L^{5} T^{-2}\right]$
Note: The physical quantities separated by the symbols $+,-,=,>,<$ etc., have the same dimensions.