The bond length of the${\mathbf{N}}_{\mathbf{2}}$molecule is 0.11nm, and its effect the spring constant is $\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$ .

(a) From the size other energy jumps for rotation and vibration, determine whether either of these modes of energy storage should be active at 300K .

(b) According to the equipartition theorem, the heat capacity of a diatomic molecule storing energy in rotations but not vibrations should be$\frac{\mathbf{5}}{\mathbf{2}}\mathbf{R}$(3 translational +2rotational degrees of freedom). If it is also storing energy in vibrations. it should be$\frac{\mathbf{7}}{\mathbf{2}}\mathbf{R}$(adding 2 vibrational degrees). Nitrogen's molar heat capacity is 20.8J/mol.K at 300K. Does this agree with your findings in part (a)?

The bond length a of the $N_2$molecule is 0.11nm.

The temperature T at room level is 300K.

The value of k is $2.3\times {10}^3\mathrm{N}.\mathrm{m}$.

To solve this part of the problem, we will be applying an equation that determines energy at$\mathit{n}\mathbf{\to }\mathit{n}\mathbf{+}\mathbf{1}$as:

(a)

The expression for the rotational energy of the molecule at I= 1 and I= 0 is given as:

$$\begin{gathered} E_{0,1}-E_{0,0}=\frac{ħ\left(1\right)\left(2\right)}{2\mu a^2}-\frac{ħ\left(0\right)\left(2\right)}{2\mu a^2} \\ =\frac{ħ}{2\mu a^2} \end{gathered}$$

The expression to determine the thermal energy at room temperature is, $E=k_{B}T$.

Calculate the effective mass of the nitrogen molecule as shown below.

$$\begin{gathered} \mathrm{\mu }=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2} \\ =\frac{{\left(1.4007\mathrm{u}\right)}^2}{1.4007\mathrm{u}+1.4007\mathrm{u}} \\ =7.0035\mathrm{u} \end{gathered}$$

Calculate the difference for the rotational energy.

$$\begin{gathered} {\mathrm{E}}_{0,1}-{\mathrm{E}}_{0,0}=\frac{{\mathrm{ħ}}^2}{2{\mathrm{\mu a}}^2} \\ =\frac{{\left(1.055\times {10}^{-34}\mathrm{Js}\right)}^2}{2\left(7.0035\mathrm{u}\left({\displaystyle \frac{1.66\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}}\right)\right){\left(0.11\mathrm{nm}\left({\displaystyle \frac{{10}^{-9}\mathrm{m}}{1\mathrm{mm}}}\right)\right)}^2} \\ =7.91\times {10}^{-23}\mathrm{J} \end{gathered}$$

Calculate the difference of the vibration energy of the molecules in n = 1 and the energy of the molecules at n = 0 .

$$\begin{gathered} {\mathrm{E}}_{1,0}-{\mathrm{E}}_{0,0}=\mathrm{h}\sqrt{\frac{\mathrm{k}}{\mathrm{\mu }}} \\ =\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\sqrt{\frac{2.3\times {10}^3\mathrm{N}/\mathrm{m}}{7.0035\mathrm{u}\left({\displaystyle \frac{1.66\times {10}^{-27}\mathrm{kg}}{\mathrm{u}}}\right)}} \\ =4.69\times {10}^{-29}\mathrm{J} \end{gathered}$$

Calculate the thermal energy at room temperature.

$$\begin{gathered} \mathrm{E}={\mathrm{k}}_{\mathrm{B}}\mathrm{T} \\ =\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(300\mathrm{K}\right) \\ =4.14\times {10}^{-21}\mathrm{J} \end{gathered}$$

Therefore, the thermal energy is less than the vibration energy difference and is more than the kinetic energy difference, the rotational energy mode therefore is active and the vibration modes are inactive.

(b)

Calculate the ratio of the specific heat of the nitrogen molecule and the gas constant.

$$\begin{gathered} \mathrm{r}=\frac{20.8\mathrm{J}/\mathrm{mol}.\mathrm{K}}{8.31\mathrm{J}/\mathrm{mol}.\mathrm{K}} \\ =2.5 \end{gathered}$$

The above result agree with the energy comparison that the rotational modes are active and vibration mode are inactive.

Therefore, the ratio of the specific heat and the gas constant is 2.5 and it agrees with the energy comparison that the rotational modes are active and vibration modes are inactive.