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Step-by-Step Solution
Step 1: Write down the given formula for entropy
The entropy of the system is given by
$S = \alpha^2 \beta \ln \left[\frac{\mu k R}{J \beta^2} + 3\right].$
Step 2: Establish the dimensional formula for entropy
Entropy ($S$) can be expressed as
$S = \frac{Q}{T},$
where $Q$ is heat energy and $T$ is the absolute temperature.
Since the dimension of heat energy $Q$ is $[M L^2 T^{-2}]$ and the dimension of temperature is $[K]$, the dimension of entropy becomes
$[S] = \frac{[M L^2 T^{-2}]}{[K]} = [M L^2 T^{-2} K^{-1}].$
Step 3: Determine the dimension of Boltzmann’s constant $k$
Boltzmann’s constant $k$ has the units of energy per unit temperature. Hence
$[k] = \frac{[M L^2 T^{-2}]}{[K]} = [M L^2 T^{-2} K^{-1}].$
Therefore, $[k]$ matches the dimension of entropy: $[M L^2 T^{-2} K^{-1}].$
Step 4: Determine the dimension of $\alpha$ from the given formula
From the given expression,
$S = \alpha^2 \beta \ln \big(\dots\big),$
we note that $\ln(\dots)$ is dimensionless. This implies the product $\alpha^2 \beta$ must have the dimension of entropy $[S]$. Once we figure out $[\beta]$, we can find $[\alpha]$.
Step 5: Find the dimension of $\beta$
Inside the logarithm, we have
$\frac{\mu k R}{J \beta^2} + 3.$
Here,
• $\mu$ is number of moles (dimension $[mol]$),
• $k$ has dimension $[M L^2 T^{-2} K^{-1}]$,
• $R$ (gas constant) has dimension $\frac{[M L^2 T^{-2}]}{[mol\,K]}$,
• $J$ (mechanical equivalent of heat) is dimensionless $[M^0 L^0 T^0]$,
• $\beta$ is unknown.
Since the argument of the logarithm must be dimensionless,
$\frac{\mu k R}{J \beta^2}$
must be dimensionless. Substituting the known dimensions, we obtain:
$\big[mol\big] \times \big[M L^2 T^{-2} K^{-1}\big] \times \frac{\big[M L^2 T^{-2}\big]}{\big[mol\,K\big]} \times \frac{1}{\beta^2} = 1.$
Simplifying gives
$[M L^2 T^{-2} K^{-1}]^2 \times \frac{1}{[mol] \times [mol]} \times \frac{[mol]^2}{\beta^2} = 1,$
but more directly:
$\beta^2$ must have the dimension $[M L^2 T^{-2} K^{-1}],$
so
$[\beta] = [M L^2 T^{-2} K^{-1}].$
Step 6: Relate the dimensions of $\alpha$ and $\beta$ to $S$
Because
$S = \alpha^2 \beta \,(\text{dimension of } S),$
we get
$\alpha^2 \beta : [M L^2 T^{-2} K^{-1}].$
Substituting $[\beta] = [M L^2 T^{-2} K^{-1}],$
we have
$\alpha^2 \times [M L^2 T^{-2} K^{-1}] = [M L^2 T^{-2} K^{-1}].$
Thus, $\alpha^2$ is dimensionless, meaning $[\alpha]$ is also dimensionless:
$[\alpha] = [M^0 L^0 T^0].$
Step 7: Compare dimensions of $\alpha$ and $k$
We have
$[\alpha] = [M^0 L^0 T^0]$
and
$[k] = [M L^2 T^{-2} K^{-1}].$
Clearly, $\alpha$ is dimensionless, whereas $k$ has dimensions of entropy. They do not have the same dimensions.
Step 8: Identify the incorrect statement
The statement “$\alpha$ and $k$ have the same dimensions” is therefore incorrect. Hence, the correct choice (the incorrect option in the question) is:
“$\alpha$ and $k$ have the same dimensions.”
