Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

The workdone by a gas molecule in an isolated

Question:

The workdone by a gas molecule in an isolated

system is given by, $\mathrm{W}=\alpha \beta^{2} \mathrm{e}^{-\frac{x^{2}}{\alpha k T}}$, 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 dimension of $\beta$ will be :

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

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

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

  4. $\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$


Correct Option: , 2

Solution:

$\frac{x^{2}}{\alpha \mathrm{kT}} \rightarrow$ dimensionless

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

Now $[\mathrm{W}]=[\alpha][\beta]^{2}$

$[\beta]=\sqrt{\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{M}^{-1} \mathrm{~T}^{2}}}=\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}$

Leave a comment

None
Free Study Material