The entropy of any system is given by $\mathrm{S}=\alpha^{2} \beta \ell \mathrm{n}\left[\frac{\mu \mathrm{kR}}{\mathrm{J} \beta^{2}}+3\right]$ Where $\alpha$ and $\beta$ are the constants. $\mu, \mathrm{J}, \mathrm{k}$ and R no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively. $\left[\right.$ Take $\left.\mathrm{S}=\frac{\mathrm{dQ}}{\mathrm{T}}\right]$ Choose the incorrect option from the following :

(B) $\alpha$ and k have the same dimensions.

(A) $\mathrm{S}, \beta, \mathrm{k}$ and $\mu \mathrm{R}$ have the same dimensions.

(C) S and $\alpha$ have different dimensions.

(D) $\alpha$ and J have the same dimensions.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Units & Measurement

The entropy of any system is given by

S = α^{2} β ℓ n [\frac{μ kR}{J β ^{2}} + 3]

Where

α

and

β

are the constants.

μ, J, k

and R no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively.

[

Take

S = \frac{dQ}{T}]

Choose the incorrect option from the following :

Answer: (B)

The argument of a logarithm must be dimensionless, so

\frac{μ k R}{J β ^{2}}

carries no dimensions. The mechanical equivalent of heat is a ratio of work to heat and is itself dimensionless, while

μ R

and the Boltzmann constant each have the dimensions of energy per unit temperature. Hence

β^{2}

has the dimensions of (energy per temperature) squared, and

β

has those of energy per temperature, the same as entropy. Since

S = α^{2} β

and the bracket is dimensionless,

α^{2}

and therefore

α

are dimensionless. So entropy,

β

, the Boltzmann constant and

μ R

do share dimensions; entropy and

α

do differ; and

α

and the mechanical equivalent of heat are both dimensionless. The statement that fails is that

α

and the Boltzmann constant have the same dimensions.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option