Question

A gas is made up of diatomic molecules. At temperature ${\mathbf{T}}_{\mathbf{1}}$,the ratio of the number of molecules in vibrational energy state 2to the number of molecules in the ground state is measured, andfound to be 0.35. The difference in energy between state 2 andthe ground state is ΔE. (a) Which of the following conclusions is

correct? (1) $\mathbf{∆}\mathbf{E}\mathbf{\approx }{\mathbf{k}}_{\mathbf{B}}{\mathbf{T}}_{\mathbf{1}}$, (2) $\mathbf{∆}\mathbf{E}\mathbf{\ll }{\mathbf{k}}_{\mathbf{B}}{\mathbf{T}}_{\mathbf{1}}$, (3)$\mathbf{∆}\mathbf{E}\mathbf{\gg }{\mathbf{k}}_{\mathbf{B}}{\mathbf{T}}_{\mathbf{1}}$

(b) At a different temperature ${\mathbf{T}}_{\mathbf{2}}$, the ratio is found to be $\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$. Which of the following is true? (1) $\mathbf{∆}\mathbf{E}\mathbf{\approx }{\mathbf{k}}_{\mathbf{B}}{\mathbf{T}}_{\mathbf{2}}$, (2) $\mathbf{∆}\mathbf{E}\mathbf{\ll }{\mathbf{k}}_{\mathbf{B}}{\mathbf{T}}_{\mathbf{2}}$,(3) $\mathbf{∆}\mathbf{E}\mathbf{\gg }{\mathbf{k}}_{\mathbf{B}}{\mathbf{T}}_{\mathbf{2}}$.

Short answer

1. The correct conclusion is, (1) $∆\mathrm{E}\approx {\mathrm{k}}_{\mathrm{B}}\mathrm{T}$
2. the correct conclusion is, (3) $∆\mathrm{E}\gg {\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_2$

Step-by-step solution

Formula used to solve the problem.

The Boltzmann factor is given as

${\mathbf{e}}^{\mathbf{-}\mathbf{∆}\mathbf{E}\mathbf{/}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}$

Where, T is temperature, ${\mathbf{k}}_{\mathbf{B}}$Boltzmann constant, and E represent the energy of different molecular states.

(a) Finding the correct conclusion

Given in the question,

The ratio between the number of molecules between two vibrational energy states

Is equal to the Boltzmann factor

Hence Boltzmann factor is =3.5

Temperature is ${\mathrm{T}}_1$

We know,

The Boltzmann factor${\mathrm{e}}^{-∆\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$

Substituting the given values

$\begin{array}{rcl}3.5& =& {\mathrm{e}}^{-∆\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_1}\\ \ln \left(3.5\right)& =& -\frac{∆\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_1}\\ -1.05& =& -\frac{∆\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_1}\\ ∆\mathrm{E}& =& 1.05{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_1\\ ∆\mathrm{E}& \approx & {\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_1\\ & & \end{array}$

Hence the correct conclusion is $∆\mathrm{E}\approx {\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_1$.

(b) Finding the correct conclusion

Given in the question,

The ratio between the number of molecules between two vibrational energy states

Is equal to the Boltzmann factor

Hence Boltzmann factor is =$8\times {10}^{-5}$

Temperature is ${\mathrm{T}}_2$

We know,

Boltzmann factordata-custom-editor="chemistry" $={\mathrm{e}}^{-∆\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$

Substituting the given values

data-custom-editor="chemistry" $$\begin{gathered} 8\times {10}^{-5}={\mathrm{e}}^{-∆\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_2} \\ \ln \left(8\times {10}^{-5}\right)=-\frac{∆\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_2} \\ -9.43=-\frac{∆\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_2} \\ ∆\mathrm{E}=9.43{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_2 \\ ∆\mathrm{E}\gg {\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_2 \end{gathered}$$

Hence, the correct conclusion is data-custom-editor="chemistry" $∆\mathrm{E}\gg {\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}_2$