The CO molecule has a rotational temperature \(\theta=\hbar^{2} /\left(2 I k_{\mathrm{B}}\right)=2.8 \mathrm{~K}\), where \(I\) is the moment of inertia of the CO molecule. The rotational partition function for one molecule is \(Z_{1}^{\mathrm{rot}}=\sum_{l=0}^{\infty}(2 l+1) \mathrm{e}^{-l l(1) \theta / T}\) (a) If one mole of CO molecules could freely rotate at temperature \(T=\) \(3.2 \mathrm{~K}\), what is their total rotational entropy? (b) What is the rotational entropy of one mole of \(\mathrm{CO}\) molecules at temperature \(T=320 \mathrm{~K}\) ? (Hint: At high temperature, where many angular momenta contribute \(Z_{1}^{\text {rot }} \approx \int_{0}^{\infty} \mathrm{d} l(2 l+1) \mathrm{e}^{-\| l+1) \theta / T}\).) (c) What is the translational entropy of one mole of \(\mathrm{CO}\) molecules in a box of volume \(V=1.0 \mathrm{~m}^{3}\) at temperature \(T=320 \mathrm{~K}\) ?

Step by step solution

01

Understanding the problem

We need to find the total rotational entropy for one mole of CO molecules at different temperatures using the given rotational partition function. Additionally, we need to find the translational entropy of the CO molecules.

02

Calculate the rotational partition function at T = 3.2 K

Use the given partition function: \[Z_{1}^{\text{rot}} = \sum_{l=0}^{\infty}(2l+1) \text{e}^{-ll\theta / T}\].Given \(\theta = 2.8 K\) and \(T = 3.2 K\). Since the temperature is comparable to \(\theta\), keep the partition function in its summation form.

03

Simplify the exponential term

For small values of \(l\), use \(l(l+1) \approx l^2\), so the exponential term becomes \[\text{e}^{-l(l+1)\theta / T} \approx \text{e}^{-l^2 \theta / T}\].

04

Approximate the summation

Use the integral approximation for summing terms in the partition function:\[Z_{1}^{\text{rot}} \approx \int_{0}^{\infty} \mathrm{d}l (2l+1) \text{e}^{-l^2 \theta / T}\].

05

Evaluate the integral

Integrate the simplified partition function:\[Z_{1}^{\text{rot}} \approx \int_{0}^{\infty} 2l \text{e}^{-l^2 \theta / T} \mathrm{d}l\].Change the variable with \( u = l^2\). Solve the integral.

06

Calculate rotational entropy at T = 3.2 K

Rotational entropy per molecule is given by\[S_{1}^{\text{rot}} = k_B \left(\ln Z_{1}^{\text{rot}} + T\frac{\partial \ln Z_{1}^{\text{rot}}}{\partial T}\right)\].Total rotational entropy for one mole:\[S_{\text{rot}} = N_A S_{1}^{\text{rot}}\].

07

Calculate rotational entropy at T = 320 K

At high temperature, use the approximation \(Z_{1}^{\text{rot}} \approx \frac{T}{\theta}\). Follow similar steps as in Step 6 to find \(S_{\text{rot}}\) at 320 K.

08

Calculate translational entropy at T = 320 K

Translational entropy is given by\[S_{\text{trans}} = k_B \left( \ln \left( \frac{V}{N_A} \left( \frac{2\pi m k_B T}{h^2} \right)^{3/2} \right) + \frac{5}{2} \right)\].Calculate for given \(V = 1.0 m^3\) and \(T = 320 K\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Entropy

Rotational entropy is a measure of the randomness or disorder associated with the rotational states of molecules. For a molecule like CO, which can rotate freely at a given temperature, the total rotational entropy is dependent on the rotational partition function. We calculate this using the formula:

\ S_{\text{rot}} = N_A S_{1}^{\text{rot}} = N_A k_B \left(\ln Z_{1}^{\text{rot}} + T \frac{\partial \ln Z_{1}^{\text{rot}}}{\partial T}\right)
Here, \(N_A\) is Avogadro's number, and \(S_{1}^{\text{rot}}\) is the rotational entropy per molecule. This formula uses the rotational partition function \(Z_{1}^{\text{rot}}\) to find the entropy.

Translational Entropy

Translational entropy refers to the disorder associated with the translational motions of molecules as they move in space within a given volume. For a mole of CO molecules in a box of volume \(V = 1.0 \text{ m}^3\), the translational entropy at a temperature \(T = 320 \text{ K}\) is determined using the formula:
\ S_{\text{trans}} = k_B \left( \ln \left( \frac{V}{N_A} \left( \frac{2\pi m k_B T}{h^2} \right)^{3/2} \right) + \frac{5}{2} \right)
Here, \(k_B\) is Boltzmann's constant, \(N_A\) is Avogadro's number, \(m\) is the molecular mass, and \(h\) is Planck's constant. This formula captures the translational contributions to the total entropy of the system.

Rotational Partition Function

The rotational partition function, \(Z_{1}^{\text{rot}}\), measures the number of ways a molecule can arrange its rotational states at a given temperature. For the CO molecule, it is given by the series: \br \(Z_{1}^{\text{rot}} = \sum_{l=0}^{\infty}(2l+1) \text{e}^{-ll\theta / T}\) \br Here, \(\theta\) is the rotational temperature, and \(T\) is the actual temperature. This series can be complicated to sum directly, but for high temperatures, it can be approximated using integrals.

Integral Approximation

When dealing with complex partition functions like the rotational partition function, we often approximate the discrete sum using an integral. For example, at high temperatures where many angular momenta states contribute, the approximation is: \br \(Z_{1}^{\text{rot}} \approx \int_{0}^{\infty} \text{d} l (2 l + 1) \text{e}^{-l(l+1)\theta / T}\).
By changing variables where \(u = l^2\), this integral becomes simpler to evaluate and helps in finding useful approximations for the rotational partition function.

High Temperature Approximation

At high temperatures, the rotational partition function can be greatly simplified. For CO molecules, the partition function can be approximated by:
\(Z_{1}^{\text{rot}} \approx \frac{T}{\theta}\), \br where \(\theta\) is the rotational temperature. This approximation assumes there are many rotational states available, making the entropy calculations much simpler. For example, if \(T = 320 \text{ K}\) and \(\theta = 2.8 \text{ K}\), the high-temperature approximation holds, and the partition function can be simplified, streamlining the calculation of rotational entropy.