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
Partition Function and Internal Energy
The internal energy \(U\) of a thermodynamic system is related to its partition function \(Q\) by the formula:
\[ U = -\left(\frac{\partial \ln Q}{\partial \beta}\right)_V \]
where \(\beta = \frac{1}{k_B T}\), \(k_B\) is the Boltzmann constant, and \(T\) is the temperature. The partial derivative is taken with respect to \(\beta\) while keeping the volume \(V\) constant.
The given partition function for the gas is:
\[ Q(N, V, T) = \frac{1}{N!}\left(\frac{2\pi m}{h^2\beta}\right)^{3N/2}(V - Nb)^N e^{\frac{\beta aN^2}{V}} \]
First, let's find the natural logarithm of \(Q\):
\[ \ln Q = \ln \left[ \frac{1}{N!}\left(\frac{2\pi m}{h^2\beta}\right)^{3N/2}(V - Nb)^N e^{\frac{\beta aN^2}{V}} \right] \]
Using the properties of logarithms \(\ln(AB) = \ln A + \ln B\) and \(\ln(A/B) = \ln A - \ln B\), we can expand this:
\[ \ln Q = \ln\left(\frac{1}{N!}\right) + \ln\left[\left(\frac{2\pi m}{h^2\beta}\right)^{3N/2}\right] + \ln\left[(V - Nb)^N\right] + \ln\left[e^{\frac{\beta aN^2}{V}}\right] \]
\[ \ln Q = -\ln(N!) + \frac{3N}{2}\ln\left(\frac{2\pi m}{h^2\beta}\right) + N\ln(V - Nb) + \frac{\beta aN^2}{V} \]
Now we need to differentiate \(\ln Q\) with respect to \(\beta\), treating \(N\) and \(V\) as constants:
\[ \frac{\partial \ln Q}{\partial \beta} = \frac{\partial}{\partial \beta}\left[-\ln(N!) + \frac{3N}{2}\ln\left(\frac{2\pi m}{h^2\beta}\right) + N\ln(V - Nb) + \frac{\beta aN^2}{V}\right] \]
Let's differentiate each term with respect to \(\beta\):
- The term \(-\ln(N!)\) is a constant with respect to \(\beta\), so its derivative is 0.
- For the term \(\frac{3N}{2}\ln\left(\frac{2\pi m}{h^2\beta}\right)\), we use the chain rule. Let \(c = \frac{2\pi m}{h^2}\). The term is \(\frac{3N}{2}\ln(c\beta^{-1})\). The derivative of \(\ln(x)\) is \(1/x\), and the derivative of \(c\beta^{-1}\) with respect to \(\beta\) is \(-c\beta^{-2}\). \[ \frac{\partial}{\partial \beta}\left[\frac{3N}{2}\ln\left(\frac{c}{\beta}\right)\right] = \frac{3N}{2} \times \frac{1}{c/\beta} \times \left(-\frac{c}{\beta^2}\right) = \frac{3N}{2} \times \frac{\beta}{c} \times \left(-\frac{c}{\beta^2}\right) = -\frac{3N}{2\beta} \]
- The term \(N\ln(V - Nb)\) is a constant with respect to \(\beta\), so its derivative is 0.
- For the term \(\frac{\beta aN^2}{V}\), the derivative with respect to \(\beta\) is simply \(\frac{aN^2}{V}\).
Combining these derivatives, we get:
\[ \frac{\partial \ln Q}{\partial \beta} = 0 - \frac{3N}{2\beta} + 0 + \frac{aN^2}{V} \]
\[ \frac{\partial \ln Q}{\partial \beta} = -\frac{3N}{2\beta} + \frac{aN^2}{V} \]
Now, substitute this into the formula for internal energy \(U\):
\[ U = -\left(-\frac{3N}{2\beta} + \frac{aN^2}{V}\right) \]
\[ U = \frac{3N}{2\beta} - \frac{aN^2}{V} \]
Finally, substitute \(\beta = \frac{1}{k_B T}\) back into the expression:
\[ U = \frac{3N}{2(1/k_B T)} - \frac{aN^2}{V} \]
\[ U = \frac{3}{2}Nk_B T - \frac{aN^2}{V} \]
Comparing this result with the given options, we find that it matches option 3.
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
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)- At temperature T, the canonical partition function of a harmonic oscillator with fundamental frequency ($\nu$) is given by
$q_{vib} (T) = \frac{e^{-h\nu/2k_BT}}{1-e^{-h\nu/k_BT}}$
For $\frac{h\nu}{k_BT} = 3$, the probability of finding the harmonic oscillator in its ground vibrational state is ____________ (Up to two decimal places)
