# In a typical combustion engine the workdone by a gas molecule is given by

Question:

In a typical combustion engine the workdone by a gas molecule is given by

$\mathrm{W}=\alpha^{2} \beta \mathrm{e}^{\frac{-\mathrm{Bx}^{2}}{\mathrm{kT}}}$, where $x$ is the displacement, $k$ is the Boltzmann constant

and $T$ is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be :

1. $\left[M^{0} L T^{0}\right]$

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

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

4. $\left[M L T^{-1}\right]$

Correct Option: 1

Solution:

(1)

$\frac{\beta_{\mathrm{X}}^{2}}{\mathrm{KT}}$ is dimension less

So

$\mathrm{KT}=\beta \mathrm{x}^{2} \quad \Rightarrow \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}$

$\beta=\frac{\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}}{\mathrm{~L}^{2}} \Rightarrow \mathrm{M}^{1} \mathrm{~T}^{-2}$

$\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}=\alpha^{2} \mathrm{M}^{1} \mathrm{~T}^{-2}$

$\alpha^{2}=\mathrm{L}^{2}$

$\alpha=\mathrm{L}$

$\alpha=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$