Question

An equation is given here $\left(P+\frac{a}{V^{2}}\right)=b \frac{\theta}{V}$ where $P=$ Pressure, $V=$ Volume and $\theta=$ Absolute temperature. If $a$ and $b$ are constants, the dimensions of $a$ will be

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

Answer 1

Answer: (A)

Equation $\left(P+\frac{a}{V^{2}}\right)=b \frac{\theta}{V}$.
Since $\frac{a}{V^{2}}$ is added to the pressure,
therefore dimensions of $\frac{a}{V^{2}}$ and pressure $(P)$ will be the same.
And dimensions of $\frac{a}{V^{2}}=\frac{a}{\left[ L ^{3}\right]^{2}}=\left[ ML ^{-1} T ^{-2}\right]$
or $a=\left[ ML ^{5} T ^{-2}\right]$.