(µ8 kT] ū=µ coth kT Sketch this result versus & from & =0 to & = co and interpret it. %3D
(µ8 kT] ū=µ coth kT Sketch this result versus & from & =0 to & = co and interpret it. %3D
Related questions
Question
![The partition function of an ideal gas of diatomic molecules in an external electric field \(\mathcal{E}\) is
\[ Q(N, V, T, \mathcal{E}) = \frac{[q(V, T, \mathcal{E})]^N}{N!} \]
where
\[
q(V, T, \mathcal{E}) = V \left( \frac{2\pi mkT}{h^2} \right)^{3/2} \left( \frac{8\pi^2IkT}{h^2} \right) \frac{e^{-\frac{h\nu}{2kT}}}{(1-e^{-\frac{h\nu}{kT}})} \left( \frac{kT}{\mu \mathcal{E}} \right) \sinh \left( \frac{\mu \mathcal{E}}{kT} \right)
\]
Here, \( I \) is the moment of inertia of the molecule; \(\nu\) is its fundamental vibrational frequency; and \(\mu\) is its dipole moment. Using this partition function along with the thermodynamic relation,
\[ dA = -S \, dT - p \, dV - M \, d\mathcal{E} \]
where \( M = N \bar{\mu} \), where \(\bar{\mu}\) is the average dipole moment of a molecule in the direction of the external field \(\mathcal{E}\), show that
\[
\bar{\mu} = \mu \left[ \coth \left( \frac{\mu \mathcal{E}}{kT} \right) - \frac{kT}{\mu \mathcal{E}} \right]
\]
Sketch this result versus \(\mathcal{E}\) from \(\mathcal{E} = 0\) to \(\mathcal{E} = \infty\) and interpret it.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddd0f2aa-4b89-4a4a-8799-436758bb524d%2Fcc453e48-4ef8-48f8-a8cf-00415b2525f8%2F5d4a9eo_processed.png&w=3840&q=75)
Transcribed Image Text:The partition function of an ideal gas of diatomic molecules in an external electric field \(\mathcal{E}\) is
\[ Q(N, V, T, \mathcal{E}) = \frac{[q(V, T, \mathcal{E})]^N}{N!} \]
where
\[
q(V, T, \mathcal{E}) = V \left( \frac{2\pi mkT}{h^2} \right)^{3/2} \left( \frac{8\pi^2IkT}{h^2} \right) \frac{e^{-\frac{h\nu}{2kT}}}{(1-e^{-\frac{h\nu}{kT}})} \left( \frac{kT}{\mu \mathcal{E}} \right) \sinh \left( \frac{\mu \mathcal{E}}{kT} \right)
\]
Here, \( I \) is the moment of inertia of the molecule; \(\nu\) is its fundamental vibrational frequency; and \(\mu\) is its dipole moment. Using this partition function along with the thermodynamic relation,
\[ dA = -S \, dT - p \, dV - M \, d\mathcal{E} \]
where \( M = N \bar{\mu} \), where \(\bar{\mu}\) is the average dipole moment of a molecule in the direction of the external field \(\mathcal{E}\), show that
\[
\bar{\mu} = \mu \left[ \coth \left( \frac{\mu \mathcal{E}}{kT} \right) - \frac{kT}{\mu \mathcal{E}} \right]
\]
Sketch this result versus \(\mathcal{E}\) from \(\mathcal{E} = 0\) to \(\mathcal{E} = \infty\) and interpret it.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 1 images
