Question

The results of either of the two preceding problems can also be applied to the vibrational motions of gas molecules. Looking only at the vibrational contribution to the heat capacity graph for ${\mathrm{H}}_2$shown in Figure 1.13, estimate the value of $\epsilon$for the vibrational motion of an${\mathrm{H}}_2$ molecule.

Short answer

The value of $\epsilon$for the vibrational motion of an ${\mathrm{H}}_2$molecule is$0.44eV$.

Step-by-step solution

Given Information

The given molecule is $H_2$.

Heat capacity:

role="math" localid="1647281996407" $C=\frac{{\epsilon }^{2}N{e}^{\frac{\epsilon }{kT}}}{k{T}^2{\left(e^{\frac{\epsilon }{kT}}-1\right)}^2}$

Calculation

The heat capacity is given as:

$C=\frac{{\epsilon }^{2}N{e}^{\frac{\epsilon }{kT}}}{k{T}^2{\left(e^{\frac{\epsilon }{kT}}-1\right)}^2}$

Let $t=\frac{kT}{\epsilon }$,

Hence, the above equation becomes,

$$\begin{gathered} C=\frac{{\epsilon }^{2}N{e}^{\frac{\epsilon }{kT}}}{k{T}^2{\left(e^{\frac{\epsilon }{kT}}-1\right)}^2} \\ C=\frac{Nk{e}^{\frac{1}{t}}}{t^2{\left(e^{\frac{1}{t}}-1\right)}^2} \\ \frac{C}{Nk}=\frac{e^{\frac{1}{t}}}{t^2{\left(e^{\frac{1}{t}}-1\right)}^2} \end{gathered}$$

The value of $t$at which the heat capacity is half of its maximum value is:

$C=\frac{1}{2}Nk$

Hence, the above equation becomes,

$$\begin{gathered} \frac{C}{Nk}=\frac{e^{\frac{1}{t}}}{t^2{\left(e^{\frac{1}{t}}-1\right)}^2} \\ \frac{1}{2}{t}^2{\left(e^{\frac{1}{t}}-1\right)}^2=e^{\frac{1}{t}} \\ t=0.335 \end{gathered}$$

Now, by resubstituting the value of $t$, we get,

role="math" localid="1647282058861" $\frac{kT}{\epsilon }=0.335$

For vibrational motion of ${\mathrm{H}}_2$gas molecules at its heat capacity, half of its maximum value, the temperature is, $T=1700\mathrm{K}$.

Hence, $\epsilon$can be calculated as:

$$\begin{gathered} ϵ=\frac{kT}{0.335} \\ \epsilon =\frac{8.62\times {10}^{-5}\times 1700}{0.335} \\ \epsilon =0.44eV \end{gathered}$$

Final answer

Hence, the value of$\epsilon$can be calculated as$0.44eV$.