Question

The work done by a gas molecule in an isolated system is given by,

$W=\alpha {\beta }^2{e}^{-x^2/\alpha kT}$

Where $x$ is the displacement, $k$ is the Boltzmann constant, $T$ is the temperature, $\alpha$ and $\beta$ are constants, then the dimension of $\beta$ will be :

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

Answer 1

The correct option is C $\left[ML{T}^{-2}\right]$

Given:
Work done, $W=\alpha {\beta }^2{e}^{-x^2/\alpha kT}$

Where,
$x=$ Displacement
$k=$ Boltzmann constant
$T=$ Temperature
$\alpha ,\beta =$ Constants

We know that,
$-\frac{x^2}{\alpha kT}$ $=$ dimensionless

$\Rightarrow \left[\frac{x^2}{\alpha kT}\right]=\left[M^0{L}^0{T}^0\right]$

$\left[kT\right]=\left[PV\right]=\left[M{L}^2{T}^{-2}\right]$
$\left[x^2\right]=\left[L^2\right]$

$\therefore \left[\alpha \right]=\left[M^{-1}{T}^2\right]$

Also,
$\left[W\right]=\left[\alpha {\beta }^2\right]$
$\Rightarrow \left[M{L}^2{T}^{-2}\right]=\left[M^{-1}{T}^2\right]\left[{\beta }^2\right]$
$\Rightarrow \left[{\beta }^2\right]=\left[M^2{L}^2{T}^{-4}\right]$
$\Rightarrow \left[\beta \right]=\left[ML{T}^{-2}\right]$