To solve this question, we need to understand how to calculate the rotational partition function for a diatomic molecule. The rotational partition function is crucial for understanding molecular thermodynamics and is given by:
\(q_{\text{rot}} = \sum_{J} (2J+1) e^{- \frac{\epsilon_J}{kT}}\)
Here, \(J\) is the rotational quantum number, \(\epsilon_J\) is the energy of the level with quantum number \(J\), \(k\) is the Boltzmann constant, and \(T\) is the temperature.
In this problem, we consider the energy levels corresponding to \(J = 0\) and \(J = 1\). Let's calculate each term individually:
- For \(J = 0\):
The energy is \(\epsilon_0 = 0\).
The degeneracy, \((2J+1) = 1\).
Contribution to the partition function: \(1 \cdot e^{0} = 1\). - For \(J = 1\):
The energy is \(\epsilon_1 = \epsilon\).
The degeneracy, \((2J+1) = 3\).
Contribution to the partition function: \(3 \cdot e^{-\frac{\epsilon}{kT}}\).
Therefore, the total rotational partition function can be calculated as:
\(q_{\text{rot}} = 1 + 3e^{-\frac{\epsilon}{kT}}\)
In the options, \(\epsilon\) is considered as a constant energy difference, and \(kT = \epsilon\) for simplification. Substituting, we simplify:
\(q_{\text{rot}} = 1 + 3e^{-2\epsilon}\)
This matches the provided correct answer:
$1+3e^{-2\epsilon}$
Conclusion: The correct answer is \(1 + 3e^{-2\epsilon}\) which accounts for the degeneracy of the rotational states of a diatomic molecule with energy levels corresponding to \(J = 0\) and \(J = 1\).
Important Questions from Partition Functions and Their Relation
Six distinguishable particles are distributed over 3 non‐degenerate levels, of energies 0, ε and 2ε. The most probable value for the total energy is
The partition function for a gas is given by
Q(N, V, T) = \(\frac{1}{N!}\left(\frac{2\pi m}{h^2\beta}\right)^{3N/2}\) (v - Nb)Ne \(\frac{\beta aN^2}{V}\)
The internal energy of the gas is
A three-state system with energies E = −ε0, 0, +ε0 is in a thermal equilibrium at a temperature T. If β ε0 = x, the probability of finding the system with energy E = 0 is [recall, cosh x = \(\frac{1}{2}\)(ex + e−x)]
The translational, vibrational, and rotational molecular partition functions for a system containing ideal diatomic gas molecules in the canonical ensemble (N, V, T) are written as, $q_{trans}$, $q_{vib}$, and $q_{rot}$, respectively. The option that correctly defines their thermodynamic variable(s) dependency is
If $q_t$ and $Q_{t,m}$ are the molecular and molar translational partition functions of $X_2$, respectively, then $ln(Q_{t,m})$ =
(N is the Avogadro number)
