Question

The relation $P=\frac{\alpha }{\beta }{e}^{\left(\frac{-\alpha z}{k\theta }\right)}$, where $p$ is pressure, $z$ is distance, $k$ is Bolzmann constant and $\theta$ is temperature. The dimensional formula of $\beta$ will be

• A. $\left[M{L}^{2}T\right]$
• B. $\left[M^0{L}^2{T}^0\right]$
• C. $\left[M^0{L}^2{T}^{-1}\right]$
• D. $\left[M{L}^0{T}^{-1}\right]$

Answer 1

The correct option is B $\left[M^0{L}^2{T}^0\right]$

Given,
$P=\frac{\alpha }{\beta }{e}^{\left(\frac{-\alpha z}{k\theta }\right)}$

Since the quantity in exponentials i.e., $e^{\left(\frac{-\alpha z}{k\theta }\right)}$ is dimensionless.

$\therefore$ $\left(\frac{-\alpha z}{k\theta }\right)$ should be a dimensionless quantity

So, $P=\frac{\alpha }{\beta }$

$\Rightarrow \alpha =\beta P\dots \dots .\left(1\right)$

and also from $\frac{\alpha z}{k\theta }=\text{constant}$

$\alpha$ can be written as; $\alpha =\frac{k\theta }{z}\times \text{constant}\dots \dots ..\left(2\right)$

$\therefore$ Dimension of $\alpha$

$\left[\alpha \right]=\frac{\left[M{L}^2{T}^{-2}{K}^{-1}\right]\left[K^1\right]}{\left[L\right]}=\left[ML{T}^{-2}\right]$

Now, dimension formula of $\beta ,$

$\left[\beta \right]=\frac{\left[\alpha \right]}{\left[P\right]}=\frac{\left[ML{T}^{-2}\right]}{\left[M{L}^{-1}{T}^{-2}\right]}=\left[M^0{L}^2{T}^0\right]$

Hence , option $\left(A\right)$ is correct.