Question

An equation is given here $(P+\frac{a}{V^2})=b\frac{9}{V}$ where $P=$pressure, $V=$Volume and $\theta =$Absolute temperature. if $a$ and $b$ are constants, the dimension of a will be

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

Answer 1

Answer: (C)

$\left[M{L}^5{T}^{-2}\right]$

Given that,

Pressure =$P$

Volume = $V$

Absolute temperature = $\theta$

$P+\frac{a}{V^2}=\frac{b\theta }{V}.....\left(I\right)$

Now,

Dimension of $P=\left[M{L}^{-1}{T}^{-2}\right]$

Dimension of $V=\left[L^3\right]$

Dimension of $\theta =K$

Now, from equation (I)

$P+\frac{a}{V^2}\Rightarrow a=P{V}^2$

$a=\left[M{L}^{-1}{T}^{-2}\right]\left[L^6\right]$

$a=\left[M{L}^5{T}^{-2}\right]$

Hence, the dimension of $a$ is $\left[M{L}^5{T}^{-2}\right]$