A) \[[M{{L}^{5}}{{T}^{-2}}]\]
B) \[[{{M}^{-1}}{{L}^{5}}{{T}^{-2}}]\]
C) \[[M{{L}^{-1}}{{T}^{-2}}]\]
D) \[[M{{L}^{-5}}{{T}^{-2}}]\]
Correct Answer: A
Solution :
In the equation \[p,\,\,\,V\] and \[T\] are pressure, volume and temperature respectively. \[\left( P+\frac{a}{{{V}^{2}}} \right)(V-b)=RT\] Dimensions of \[\frac{a}{{{V}^{2}}}\] will be same as that of\[p\]pressure. \[\therefore \]Dimensions of\[\frac{a}{{{V}^{2}}}=\]dimensions of\[p\] \[\Rightarrow \] Dimensions of\[a=\]dimensions of \[p\times \]dimensions of\[{{V}^{2}}\] \[=[M{{L}^{-1}}{{T}^{-2}}][{{L}^{6}}]\] \[=[M{{L}^{5}}{{T}^{-2}}]\]You need to login to perform this action.
You will be redirected in
3 sec