Question

In the low-temperature limit $(kT<<\in )$, each term in the rotational partition function is much smaller than the one before. Since the first term is independent of $T$, cut off the sum after the second term and compute the average energy and the heat capacity in this approximation. Keep only the largest T-dependent term at each stage of the calculation. Is your result consistent with the third law of thermodynamics? Sketch the behavior of the heat capacity at all temperature, interpolating between the high-temperature and low- temperature expressions.

Short answer

The graph is as follows,

Step-by-step solution

Step 1. Given Information

We are given that in the low-temperature limit $(kT<<\in )$, each term in the rotational partition function is much smaller than the one before.

Step 2. Expanding the partition function 

Expanding the partition function by neglecting all higher order terms in the low temperature limit $(kT<<\in )$.

$$\begin{gathered} Z_{rot}=1+\left(2\right(1)+1)e^{-1(1+1)\in /kT} \\ =1+3e^{-2e/kT} \end{gathered}$$

The average energy of the system is given as follows:

$$\begin{gathered} {\overline{E}}_{rot}=-\frac{1}{Z_{rot}}\frac{\partial Z_{rot}}{\partial \beta } \\ =-\frac{1}{1+3e^{-2\beta \in }}\frac{\partial }{\partial \beta }(1+3e^{-2\beta \in }) \\ =-\frac{1}{1+3e^{-2\beta \in }}\left(3e^{-2\beta \in }\right(-2\in \left)\right) \\ =\frac{6\in e^{-2\beta \in }}{1+3e^{-2\beta \in }} \end{gathered}$$

Neglecting the term $3e^{-2\beta \in }$in the denominator of $\frac{6\in e^{-2\beta \in }}{1+3e^{-2\beta \in }}$,

Since the function $3e^{-2\beta \in }$approaches to zero as $\beta \to \infty$because $e^{-\infty }=0$.

Therefore, the average energy of the system is $6\in e^{-2\beta \in }$.

Step 3. Specific heat capacity of the system

The specific heat capacity of the system is given by,

$$\begin{gathered} C=\frac{\partial {\overline{E}}_{rot}}{\partial T} \\ =\frac{\partial }{\partial T}(6\in e^{-2\beta \in \in }) \\ =6\in \frac{\partial }{\partial T}\left(e^{-2\in /kT}\right) \\ =6\in \left(e^{-2\in /kT}\right)\left(-\frac{2\in }{k}\right)\left(\frac{-1}{T^2}\right) \\ =\frac{12{\in }^2}{k{T}^2}{e}^{-2\in /kT} \\ =3k{\left(\frac{2\in }{kT}\right)}^2{e}^{-2\in /kT} \end{gathered}$$

Therefore, the heat capacity of the system in the low temperature limit is $C=3k{\left(\frac{2\in }{kT}\right)}^2{e}^{-2\in /kT}$.

In the low temperature limit, the heat capacity of the system decreases exponentially to zero.

Step 4. Behavior ho heat capacity

The low and high temperature limit of $\frac{C}{k}$are plotted against the dimensionless parameter $\frac{kT}{\in }$.

The graph is as follows,