Question

The work done by a gas molecule in an isolated system is given by, $\alpha .{\beta }^2.e^{-\frac{x^2}{aKT}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and$T$ is the temperature.$\alpha$ and $\beta$ are constants. Then the dimensions 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

$\left[ML{T}^{–2}\right]$

Step 1: Given information

Work done$=\alpha .{\beta }^2.e^{-\frac{x^2}{aKT}}$,

Where $x$ is the displacement, $k$ is the Boltzmann constant, and$T$ is the temperature.$\alpha$ and $\beta$ are constants

Step 2: Explanation

Let, $\frac{x^2}{aKT}$ = dimensionless

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

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

$\Rightarrow \left[\kappa \right]=\left[M^1{L}^2{T}^{-2}{K}^{-1}\right]\left[T\right]$

$$\begin{gathered} \Rightarrow \left[\alpha \right]=\left[\frac{L^2}{M^1{L}^2{T}^{-2}{K}^{-1}\times K}\right]\left[\because \left[\kappa \right]=M^1{L}^2{T}^{-2}{K}^{-1}\right]\mathrm{and}\left[\because \left[\mathrm{K}\right]=\left[\mathrm{T}\right]\right] \\ \Rightarrow \left[\alpha \right]=\left[M^{-1}{T}^2\right] \end{gathered}$$

$\Rightarrow \omega =\alpha .{\beta }^2$

So,

$$\begin{gathered} \Rightarrow \frac{\left[M^1{L}^1{T}^{-2}\right]\left[L^{-1}\right]}{\left[M^{-1}{T}^2\right]}=\left[{\beta }^2\right] \\ \Rightarrow \left[{\beta }^2\right]=\left[M^2{L}^2{T}^{-4}\right]\left[\beta \right] \\ \Rightarrow \left[{\beta }^2\right]=\left[ML{T}^{-2}\right] \end{gathered}$$

Hence, the correct option is (C).