QJEEM21Ph6953: JEE Main 2021 Physics - Class 11 JEEM21Ph6953 Solution: The work done by a gas molecule in an isolated system is given…

Question

Verified Solution • 2025–26 Session moderate

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

given by,

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

, 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 :

Accepted Answer

where, k is Boltzmann constant,

T is temperature and x is displacement.

We know that,

{{{x^2}} \over {\alpha kT}}

is a dimensionless quantity.

\therefore

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

\Rightarrow [\alpha ] = {{[{L^2}]} \over {[k][T]}}

..... (i)

Since, dimensions of k are

[k] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]

...... (ii)

Dimensions of temperature are

[T] = [K]

..... (iii)

Substituting Eqs. (ii) and (iii) in Eq. (i), we get

[\alpha ] = {{[{L^2}]} \over {[{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}

[\alpha ] = [{M^{ - 1}}{T^2}]

According to dimensional analysis,

[W] = [\alpha {\beta ^2}]

\Rightarrow [{\beta ^2}] = {{[W]} \over {[\alpha ]}}

\Rightarrow [{\beta ^2}] = {{{M^1}{L^2}{T^{ - 2}}]} \over {[{M^{ - 1}}{T^2}]}} = [{M^2}{L^2}{T^{ - 4}}]

\Rightarrow [\beta ] = [ML{T^{ - 2}}]

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