# The workdone by a gas molecule in an isolated system is given by,

Question:

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

$\mathrm{W}=\alpha \beta^{2} \mathrm{e}^{-\frac{x^{2}}{a k \mathrm{~T}}}$, where $\mathrm{x}$ is the displacement, $\mathrm{k}$ is the Boltzmann constant and $\mathrm{T}$ is

the temperature. $\alpha$ and $\beta$ are constants. Then the dimensions of $\beta$ will be -

1. $\left[\mathrm{M}^{0} \mathrm{LT}^{0}\right]$

2. $\mathrm{M}^{2} \mathrm{LT}^{2}$

3. $\left[\mathrm{MLT}^{-2}\right]$

4. $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$

Correct Option: 3

Solution:

(3)

given : work $=\alpha \cdot \beta^{2} \cdot e^{-\frac{x^{2}}{a \cdot k \cdot T}}$

$\mathrm{k}=$ boltzmann constant

$\mathrm{T}=$ temperature $\mathrm{x}=$ displacement we know that, $\frac{\mathrm{x}^{2}}{\alpha \cdot \mathrm{k} \cdot \mathrm{T}}=$ dimensionless

$\left[\frac{\mathrm{x}^{2}}{\alpha \cdot \mathrm{k} \cdot \boldsymbol{T}}\right]=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$

$[\alpha]=\left[\frac{\mathrm{L}^{2}}{\mathrm{~K} \cdot \mathrm{T}}\right]$

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

$[\mathrm{T}]=[\mathrm{K}]$

$\Rightarrow[\alpha]=\left[\frac{\mathrm{L}^{2}}{\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1} \times \mathrm{K}}\right] \Rightarrow[\alpha]=\left[\mathrm{M}^{-1} \mathrm{~T}^{2}\right]$

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

$\Rightarrow \frac{\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{-1}\right]}{\left[\mathrm{M}^{-1} \mathrm{~T}^{2}\right]}=\left[\beta^{2}\right]=\left[\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-4}\right]$

$[\beta]=\left[\mathrm{MLT}^{-2}\right]$