PH503 Homework
1.
(a)
Liouville's equation $\frac{\partial \rho}{\partial t} = -\{\rho, \mathcal{H}\}$, or
$$\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t} + \sum_i \left( \frac{\partial \rho}{\partial q_i}\dot{q}_i + \frac{\partial \rho}{\partial p_i}\dot{p}_i \right) = \frac{\partial \rho}{\partial t} + \{\rho, \mathcal{H}\}$$
Which is $\frac{d\rho}{dt} = 0$, volume element $d\Gamma = dpdq$ is time invariant.
Entropy is $S = - \int \rho \ln \rho \, d\Gamma$.
Then $\rho$ invariant, $\rho \ln \rho$ invariant, $(\rho \ln \rho) \times d\Gamma$ invariant.
$\frac{dS}{dt} = 0$
(b)
Normalization: $C_1 = \int dpdq \, \rho(p, q) = 1$
Fixed average energy: $C_2 = \int dpdq \, \rho(p, q) \mathcal{H}(p, q) = E$
Lagrange multipliers:
$$L[\rho] = S - \alpha C_1 - \beta C_2$$
$$L[\rho] = - \int \rho \ln \rho \, dpdq - \alpha \left(\int \rho \, dpdq - 1\right) - \beta \left(\int \rho \mathcal{H} \, dpdq - E\right)$$
$$L[\rho] = \int (-\rho \ln \rho - \alpha \rho - \beta \rho \mathcal{H}) \, dpdq + \alpha + \beta E$$
To find the function $\rho$ that maximizes $L$,
$$\frac{\partial}{\partial \rho} (-\rho \ln \rho - \alpha \rho - \beta \rho \mathcal{H}) = 0$$
$$\ln \rho = -1 - \alpha - \beta \mathcal{H}$$
Then
$$\rho_{\text{max}} = e^{-1-\alpha} e^{-\beta \mathcal{H}}$$
$$\int \rho_{\text{max}} \, dpdq = \int e^{-1-\alpha} e^{-\beta \mathcal{H}} \, dpdq = e^{-1-\alpha} \int e^{-\beta \mathcal{H}} \, dpdq = 1$$
Partition function $Z = \int e^{-\beta \mathcal{H}} \, dpdq$.
$$e^{-1-\alpha} Z = 1 \quad \implies \quad e^{-1-\alpha} = \frac{1}{Z}$$
Finally,
$$\rho_{\text{max}}(p, q) = \frac{1}{Z} e^{-\beta \mathcal{H}(p, q)}$$
This is the probability distribution of the canonical ensemble.
(c)
A density function is stationary if its partial time derivative is zero $\frac{\partial \rho}{\partial t} = 0$.
Use Liouville's equation($\frac{\partial \rho}{\partial t} = -\{\rho, \mathcal{H}\}$).
Goal is $\{\rho_{\text{max}}, \mathcal{H}\}=0$.
$$\rho_{\text{max}} = f(\mathcal{H}) = \frac{1}{Z} e^{-\beta \mathcal{H}}$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \sum_i \left( \frac{\partial f(\mathcal{H})}{\partial q_i} \frac{\partial \mathcal{H}}{\partial p_i} - \frac{\partial f(\mathcal{H})}{\partial p_i} \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
Using the chain rule:
$$\frac{\partial f(\mathcal{H})}{\partial q_i} = \frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial q_i} \quad \text{and} \quad \frac{\partial f(\mathcal{H})}{\partial p_i} = \frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial p_i}$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \sum_i \left( \left(\frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial q_i}\right) \frac{\partial \mathcal{H}}{\partial p_i} - \left(\frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial p_i}\right) \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \frac{df}{d\mathcal{H}} \sum_i \left( \frac{\partial \mathcal{H}}{\partial q_i} \frac{\partial \mathcal{H}}{\partial p_i} - \frac{\partial \mathcal{H}}{\partial p_i} \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
$$\{\rho_{\text{max}}, \mathcal{H}\} = 0$$
Since $\frac{\partial \rho_{\text{max}}}{\partial t} = -\{\rho_{\text{max}}, \mathcal{H}\}$:
$$\frac{\partial \rho_{\text{max}}}{\partial t} = 0$$
2.
(a)
Define the Systems:
- Subsystem 'a' has energy $E_a$, particle number $N_a$, and its number of accessible microstates is given by the phase space volume $\Gamma_a(E_a)$.
- Subsystem 'b' has energy $E_b$, particle number $N_b$, and its number of accessible microstates is $\Gamma_b(E_b)$.
- The total system has energy $E = E_a + E_b$ and particle number $N = N_a + N_b$.
Calculate the Total Entropy:
$$S_{total} = k_B \log \Gamma(E)$$
If system is statistically independent,
$$\Gamma(E) = \Gamma_a(E_a) \cdot \Gamma_b(E_b)$$
$$S_{total} = k_B \log(\Gamma_a(E_a) \cdot \Gamma_b(E_b))$$
Using $\log(xy) = \log(x) + \log(y)$:
$$S_{total} = k_B (\log \Gamma_a(E_a) + \log \Gamma_b(E_b))$$
$$S_{total} = k_B \log \Gamma_a(E_a) + k_B \log \Gamma_b(E_b)$$
$$S_{total} = S_a + S_b$$
(b)
$$\Gamma(E) = \Sigma(E+\Delta) - \Sigma(E)$$
Since $\Delta \ll E$, first-order Taylor expansion:
$$\Sigma(E+\Delta) \approx \Sigma(E) + \frac{\partial \Sigma(E)}{\partial E} \Delta$$
$$\Gamma(E) \approx \left(\Sigma(E) + \frac{\partial \Sigma(E)}{\partial E} \Delta\right) - \Sigma(E) = \frac{\partial \Sigma(E)}{\partial E} \Delta$$
Also,
$$\Gamma(E) \approx g(E) \Delta$$
Compare $S_1$ and $S_3$:
$$S_1 = k_B \log \Gamma(E) \approx k_B \log(g(E) \Delta)$$
$$S_1 \approx k_B (\log g(E) + \log \Delta) = S_3 + k_B \log \Delta$$
Since the entropy $S$ is extensive ($S \propto N$), this constant difference is negligible for large $N$. Then, $S_1 \approx S_3$.
Compare $S_2$ and $S_3$:
For large DoF $f$ system, let
$$\Sigma(E) \propto E^f$$
$$g(E) = \frac{\partial \Sigma(E)}{\partial E} \propto \frac{\partial}{\partial E}(E^f) = f E^{f-1} = \frac{f}{E} E^f \propto \frac{f}{E} \Sigma(E)$$
$$S_3 = k_B \log g(E) \approx k_B \log \left(C \frac{f}{E} \Sigma(E)\right) $$
$$S_3 \approx k_B \log \Sigma(E) + k_B \log\left(C\frac{f}{E}\right) = S_2 + k_B \log\left(C\frac{f}{E}\right)$$
Both the number of degrees of freedom ($f \propto N$) and the energy ($E \propto N$) are extensive. Then second term $O(\log N)$ at most. Since $S_2$ is of order $O(N)$, the difference is negligible for large $N$. Then, $S_2 \approx S_3$.
3.
The entropy of a monatomic ideal gas is:
$$S = N k_B \ln \left[ V \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{3}{2} N k_B$$
The entropy with the Gibbs correction factor ($S = S_{dist} - k_B \ln(N!)$) is given by the Sackur-Tetrode equation:
$$S(E, V, N) = N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} N k_B$$
(a)
Let's scale the variables $E \to \lambda E$, $V \to \lambda V$, $N \to \lambda N$:
$S(\lambda E, \lambda V, \lambda N) = (\lambda N) k_B \ln \left[ \frac{(\lambda V)}{(\lambda N)} \left( \frac{4\pi m (\lambda E)}{3(\lambda N) h^2} \right)^{3/2} \right] + \frac{5}{2} (\lambda N) k_B$
$S(\lambda E, \lambda V, \lambda N) = \lambda N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} \lambda N k_B$
$S(\lambda E, \lambda V, \lambda N) = \lambda \left( N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} N k_B \right)$
$$S(\lambda E, \lambda V, \lambda N) = \lambda S(E, V, N)$$
(b)
The Gibbs correction is a term that depends only on $N$ ($S_{corr} = -k_B \ln(N!)$).
Pressure is defined as $p = T \left( \frac{\partial S}{\partial V} \right)_{E,N}$.
Since the correction term, $-k_B \ln(N!)$, does not depend on volume $V$, its derivative with respect to $V$ is zero, pressure is unchanged.
Heat capacity is defined as $C_V = \left( \frac{\partial E}{\partial T} \right)_{V,N}$. Monatomic ideal gas, $E = \frac{3}{2} N k_B T$.
Since the relationship between $E$ and $T$ is unchanged, the heat capacity is also unchanged.
The exponent $\gamma$ is the ratio of specific heats, $\gamma = C_p/C_V$. Since $C_V$ is unchanged and Mayer's relation for an ideal gas ($C_p = C_V + N k_B$) still holds, $C_p$ is also unchanged. Consequently, $\gamma$ is unchanged, and the adiabatic relation remains the same.
(c)
Let's use the final Sackur-Tetrode equation:
$S = N k_B \left( \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right] + \frac{5}{2} \right) = N k_B \left( \ln(V E^{3/2} C) - \frac{5}{2}\ln N + \frac{5}{2} \right)$ where C is a constant.
$\frac{\partial S}{\partial N} = k_B \left( \ln(V E^{3/2} C) - \frac{5}{2}\ln N + \frac{5}{2} \right) + N k_B \left(-\frac{5}{2N}\right)$
$\frac{\partial S}{\partial N} = k_B \ln\left[\frac{V}{N^{5/2}} (\dots)\right] + \frac{5}{2}k_B - \frac{5}{2}k_B = k_B \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right]$
$\mu = -T \frac{\partial S}{\partial N} = -k_B T \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right]$
Since the entropy $S$ changes, the Helmholtz free energy $F$ must also change.
$F_{new} = E - T S_{new} = E - T(S_{old} - k_B \ln N!) = (E - T S_{old}) + k_B T \ln N!$
$$F_{new} = F_{old} + k_B T \ln(N!)$$
4.
(a)
Use the entropy formula for an ideal gas, $S(N,V) = N k_B \ln(V/N) + \text{constant}$.
Initial Entropy: $S_i = S(N_1, V_1) + S(N_2, V_2) = N_1 k_B \ln(V_1/N_1) + N_2 k_B \ln(V_2/N_2)$.
Final Entropy: $S_f = S(N_1+N_2, V_1+V_2) = (N_1+N_2) k_B \ln\left(\frac{V_1+V_2}{N_1+N_2}\right)$.
$$(\Delta S)_{1=2} = S_f - S_i = k_B \left[ (N_1+N_2) \ln\left(\frac{V_1+V_2}{N_1+N_2}\right) - N_1 \ln\left(\frac{V_1}{N_1}\right) - N_2 \ln\left(\frac{V_2}{N_2}\right) \right]$$
Set volume $v_1 = V_1/N_1$ and $v_2 = V_2/N_2$:
$$v_f = \frac{V_1+V_2}{N_1+N_2} = \frac{N_1 v_1 + N_2 v_2}{N_1+N_2}$$
Then it changes with $N = N_1+N_2$, $x_1 = N_1/N$, and $x_2 = N_2/N$:
$$\ln(v_f) - x_1 \ln(v_1) - x_2 \ln(v_2) \ge 0$$
$$\ln(x_1 v_1 + x_2 v_2) \ge x_1 \ln(v_1) + x_2 \ln(v_2)$$
Use Jensen's inequality ($f(x) = \ln(x)$, is strictly concave), it's true.
Equal at
$$v_1 = v_2 \implies \frac{V_1}{N_1} = \frac{V_2}{N_2}$$
(b)
$$(\Delta S)^* = -k_B (N_1 \ln x_1 + N_2 \ln x_2)$$
where $N = N_1 + N_2$ and $x_1 = N_1/N$ and $x_2 = N_2/N$.
Let $N = N_1+N_2$. then $x = x_1 = N_1/N$, so that $x_2 = 1-x$,
$$(\Delta S)^*(x) = -N k_B [x \ln x + (1-x) \ln(1-x)]$$
To find the maximum,
$$\frac{d(\Delta S)^*}{dx} = -N k_B \frac{d}{dx}[x \ln x + (1-x) \ln(1-x)]$$ $$\frac{d(\Delta S)^*}{dx} = -N k_B \left[ (\ln x + 1) + (-\ln(1-x) - 1) \right] = -N k_B [\ln x - \ln(1-x)]$$
$$\ln x - \ln(1-x) = 0 \implies \ln x = \ln(1-x) \implies x = 1-x \implies 2x=1$$
at $x = 1/2$.
The second derivative is $\frac{d^2(\Delta S)^*}{dx^2} = -N k_B (\frac{1}{x} + \frac{1}{1-x})$, which is negative for $x \in (0,1)$, then $x=1/2$ is a maximum.
Maximum when $x_1 = x_2 = 1/2$, which means $N_1 = N_2$.
$$(\Delta S)^*_{max} = -N k_B \left[ \frac{1}{2} \ln\left(\frac{1}{2}\right) + \frac{1}{2} \ln\left(\frac{1}{2}\right) \right]$$ $$(\Delta S)^*_{max} = -N k_B \left[ \ln\left(\frac{1}{2}\right) \right] = -N k_B (-\ln 2) = N k_B \ln 2$$
$$(\Delta S)^* \le (N_1+N_2)k_B \ln 2$$
5.
For ideal gas, the adiabatic exponent $\gamma$ and the molar heat capacities $C_{P,m}$ and $C_{V,m}$ are related by:
$$\gamma = \frac{C_{P,m}}{C_{V,m}}$$
$$C_{P,m} - C_{V,m} = R$$
$$\gamma = \frac{C_{V,m} + R}{C_{V,m}} = 1 + \frac{R}{C_{V,m}}$$
$$\gamma - 1 = \frac{R}{C_{V,m}} \quad \implies \quad C_{V,m} = \frac{R}{\gamma - 1}$$
For a mixture of ideal gases,
$$C_{V,mix} = f_1 C_{V,m1} + f_2 C_{V,m2}$$
where $f_1$ and $f_2$ are the mole fractions, and $C_{V,m1}$ and $C_{V,m2}$ are the molar heat capacities of gas 1 and gas 2.
Use $C_{V,mix} = \frac{R}{\gamma - 1}$, $C_{V,m1} = \frac{R}{\gamma_1 - 1}$, $C_{V,m2} = \frac{R}{\gamma_2 - 1}$
$$\frac{R}{\gamma - 1} = f_1 \left(\frac{R}{\gamma_1 - 1}\right) + f_2 \left(\frac{R}{\gamma_2 - 1}\right)$$
$$\frac{1}{\gamma - 1} = \frac{f_1}{\gamma_1 - 1} + \frac{f_2}{\gamma_2 - 1}$$
1
(a)
The classical partition function ($Z_{cl}$) is defined as the integral over all of phase space. Let $\beta = \frac{1}{k_BT}$.
$$Z_{cl} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\beta\mathcal{H}} \,dp\,dx$$
$$Z_{cl} = \int_{-\infty}^{\infty} e^{-\frac{\beta p^2}{2m}} \,dp \int_{-\infty}^{\infty} e^{-\frac{\beta m\omega^2x^2}{2}} \,dx$$
Both integrals are Gaussian integrals of the form $\int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}$.
Integral over momentum ($p$): Here, $a = \frac{\beta}{2m}$.
$$\int_{-\infty}^{\infty} e^{-\frac{\beta p^2}{2m}} \,dp = \sqrt{\frac{\pi}{\beta/2m}} = \sqrt{\frac{2\pi m}{\beta}}$$
Integral over position ($x$): Here, $a = \frac{\beta m\omega^2}{2}$.
$$\int_{-\infty}^{\infty} e^{-\frac{\beta m\omega^2x^2}{2}} \,dx = \sqrt{\frac{\pi}{\beta m\omega^2/2}} = \sqrt{\frac{2\pi}{\beta m\omega^2}}$$
Multiplying the two results gives the partition function:
$$Z_{cl} = \left(\sqrt{\frac{2\pi m}{\beta}}\right) \left(\sqrt{\frac{2\pi}{\beta m\omega^2}}\right) = \frac{2\pi}{\beta\omega}= \frac{2\pi k_BT}{\omega}$$
(b)
$$\langle E \rangle = -\frac{\partial}{\partial\beta} \ln Z_{cl} = -\frac{\partial}{\partial\beta} \ln\left(\frac{2\pi}{\beta\omega}\right) = -\left(-\frac{1}{\beta}\right) = \frac{1}{\beta} = k_BT$$
The Hamiltonian is separable: $\mathcal{H} = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$. The average energy is the sum of the averages of its parts:
$$\langle E \rangle = \left\langle \frac{p^2}{2m} \right\rangle + \left\langle \frac{1}{2}m\omega^2x^2 \right\rangle$$
The partition function is also separable into a product $Z_{cl} = Z_p \cdot Z_x$, where $Z_p = \sqrt{\frac{2\pi m}{\beta}}$ and $Z_x = \sqrt{\frac{2\pi}{\beta m\omega^2}}$. We can calculate the average of each energy term from its corresponding part of the partition function.
Average kinetic energy:
$$\left\langle \frac{p^2}{2m} \right\rangle = -\frac{\partial}{\partial\beta} \ln Z_p = -\frac{\partial}{\partial\beta} \ln\left(\sqrt{\frac{2\pi m}{\beta}}\right) = -\frac{\partial}{\partial\beta}\left[\frac{1}{2}\ln(2\pi m) - \frac{1}{2}\ln\beta\right]$$
$$= -\left(-\frac{1}{2\beta}\right) = \frac{1}{2\beta} = \frac{1}{2}k_BT$$
Average potential energy:
$$\left\langle \frac{1}{2}m\omega^2x^2 \right\rangle = -\frac{\partial}{\partial\beta} \ln Z_x = -\frac{\partial}{\partial\beta} \ln\left(\sqrt{\frac{2\pi}{\beta m\omega^2}}\right) = -\frac{\partial}{\partial\beta}\left[\frac{1}{2}\ln\left(\frac{2\pi}{m\omega^2}\right) - \frac{1}{2}\ln\beta\right]$$
$$= -\left(-\frac{1}{2\beta}\right) = \frac{1}{2\beta} = \frac{1}{2}k_BT$$
(c)
The quantum mechanical partition function ($Z_{QM}$) is the sum of the Boltzmann factors over all quantum states:
$$Z_{QM} = \sum_{n=0}^{\infty} e^{-\beta E_n} = \sum_{n=0}^{\infty} e^{-\beta(n+\frac{1}{2})\hbar\omega}$$
$$Z_{QM} = e^{-\frac{\beta\hbar\omega}{2}} \sum_{n=0}^{\infty} e^{-n\beta\hbar\omega} = e^{-\frac{\beta\hbar\omega}{2}} \sum_{n=0}^{\infty} (e^{-\beta\hbar\omega})^n$$
The summation is a geometric series with the first term $a=1$ and the common ratio $r = e^{-\beta\hbar\omega}$. Since $|r|<1$, the sum converges to $\frac{1}{1-r}$.
$$\sum_{n=0}^{\infty} (e^{-\beta\hbar\omega})^n = \frac{1}{1-e^{-\beta\hbar\omega}}$$
Therefore, the quantum mechanical partition function is:
$$Z_{QM} = \frac{e^{-\frac{\beta\hbar\omega}{2}}}{1-e^{-\beta\hbar\omega}}$$
This can also be expressed compactly using the hyperbolic sine function:
$$\Large Z_{QM} = \frac{1}{e^{\frac{\beta\hbar\omega}{2}} - e^{-\frac{\beta\hbar\omega}{2}}} = \frac{1}{2\sinh\left(\frac{\beta\hbar\omega}{2}\right)}$$
(d)
Average Energy $\langle E \rangle$
The quantum average energy is calculated from $Z_{QM}$:
$$\langle E \rangle = -\frac{\partial}{\partial\beta} \ln Z_{QM}$$
$$\ln Z_{QM} = \ln\left(\frac{e^{-\frac{\beta\hbar\omega}{2}}}{1-e^{-\beta\hbar\omega}}\right) = -\frac{\beta\hbar\omega}{2} - \ln(1-e^{-\beta\hbar\omega})$$
Differentiating with respect to $\beta$:
$$\frac{\partial}{\partial\beta} \ln Z_{QM} = -\frac{\hbar\omega}{2} - \frac{1}{1-e^{-\beta\hbar\omega}}(-e^{-\beta\hbar\omega})(-\hbar\omega) = -\frac{\hbar\omega}{2} - \frac{\hbar\omega e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}$$
Thus, the average energy is:
$$\langle E \rangle = - \left(-\frac{\hbar\omega}{2} - \frac{\hbar\omega e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\right) = \frac{\hbar\omega}{2} + \frac{\hbar\omega e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}$$
By multiplying the numerator and denominator of the second term by $e^{\beta\hbar\omega}$, we get the standard form:
$$\langle E \rangle = \frac{\hbar\omega}{2} + \frac{\hbar\omega}{e^{\beta\hbar\omega}-1}$$
High-Temperature Limit ($T \to \infty$)
The high-temperature limit corresponds to $T \to \infty$, which means $\beta = \frac{1}{k_BT} \to 0$. In this limit, $\beta\hbar\omega \ll 1$, so we can use the Taylor expansion for the exponential,
$$e^{\beta\hbar\omega} \approx 1+\beta\hbar\omega$$
Average energy:
$$\langle E \rangle \approx \frac{\hbar\omega}{2} + \frac{\hbar\omega}{(1+\beta\hbar\omega)-1} = \frac{\hbar\omega}{2} + \frac{\hbar\omega}{\beta\hbar\omega} = \frac{\hbar\omega}{2} + \frac{1}{\beta}$$
As $T \to \infty$ (or $\beta \to 0$), the first term, the zero-point energy $\frac{\hbar\omega}{2}$, is a finite constant, while the second term $\frac{1}{\beta}$ becomes dominant.
$$\lim_{T \to \infty} \langle E \rangle = \lim_{\beta \to 0} \left( \frac{\hbar\omega}{2} + \frac{1}{\beta} \right) = \frac{1}{\beta} = k_BT$$
This result, $\langle E \rangle = k_BT$, exactly matches the average energy of the classical harmonic oscillator found in part (b).
2
(a)
The Helmholtz free energy $F$ is defined as $F = U - TS$, where $U$ is the internal energy, $T$ is the temperature, and $S$ is the entropy.
The internal energy $U$ is the average energy of the system: $U = \langle E \rangle = \sum_s P_s E_s$.
The Gibbs entropy is given by $S = -k_B \sum_s P_s \ln P_s$.
The probability $P_s$ of the system being in a microstate $s$ with energy $E_s$ is $P_s = \frac{e^{-\beta E_s}}{Z}$, where $\beta = \frac{1}{k_B T}$ and $Z = \sum_s e^{-\beta E_s}$ is the partition function.
$$S = -k_B \sum_s \left( \frac{e^{-\beta E_s}}{Z} \right) \ln \left( \frac{e^{-\beta E_s}}{Z} \right)$$
$$S = -k_B \sum_s \frac{e^{-\beta E_s}}{Z} (-\beta E_s - \ln Z)$$
$$S = k_B \beta \sum_s \frac{E_s e^{-\beta E_s}}{Z} + k_B \ln Z \sum_s \frac{e^{-\beta E_s}}{Z}$$
Note $\sum_s \frac{E_s e^{-\beta E_s}}{Z} = U$, $\sum_s \frac{e^{-\beta E_s}}{Z} = \frac{1}{Z} \sum_s e^{-\beta E_s} = \frac{Z}{Z} = 1$.
$$S = k_B \beta U + k_B \ln Z$$
$$S = \frac{U}{T} + k_B \ln Z$$
$$TS = U + k_B T \ln Z$$
$$U - TS = -k_B T \ln Z$$
$$\boldsymbol{F = -k_B T \ln Z}$$
(b)
$$Z_1 = \frac{1}{h^2} \int d^2q \int d^2p \ e^{-\beta H_1(\vec{p})}$$
Here, $H_1(\vec{p}) = c|\vec{p}|$ is the Hamiltonian for a single particle, and the integral over the position coordinates $\int d^2q$ is simply the area $A$.
Use polar coordinates, where $d^2p = p \, dp \, d\theta$ and $|\vec{p}| = p$.
$$Z_1 = \frac{A}{h^2} \int_0^{2\pi} d\theta \int_0^\infty dp \ p e^{-\beta c p}$$
The integral over $\theta$ gives $2\pi$. The integral over $p$ is a standard form $\int_0^\infty x e^{-ax} dx = 1/a^2$. Here, $x=p$ and $a = \beta c$.
$$\int_0^\infty p e^{-\beta c p} dp = \frac{1}{(\beta c)^2}$$
Combining the parts, the single-particle partition function is:
$$Z_1 = \frac{A}{h^2} \cdot 2\pi \cdot \frac{1}{(\beta c)^2} = \frac{2\pi A}{(h\beta c)^2} = \frac{2\pi A(k_B T)^2}{(hc)^2}$$
For $N$ non-interacting, indistinguishable particles, the partition function is $Z = \frac{Z_1^N}{N!}$.
$$Z = \frac{1}{N!} \left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right)^N$$
Now, we use the result from part (a), $F = -k_B T \ln Z$.
$$F = -k_B T \ln \left[ \frac{1}{N!} \left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right)^N \right]$$
$$F = -k_B T \left[ N \ln\left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right) - \ln(N!) \right]$$
Using the Stirling approximation, $\ln(N!) \approx N \ln N - N$:
$$F \approx -k_B T \left[ N \ln\left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right) - (N \ln N - N) \right]$$
$$F = -N k_B T \left[ \ln\left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right) - \ln N + 1 \right]$$
Combining the logarithmic terms, we get the final expression for the free energy:
$$\boldsymbol{F = -N k_B T \left[ \ln \left( \frac{2\pi A (k_B T)^2}{N(hc)^2} \right) + 1 \right]}$$
(c)
Entropy (S)
$S = -\left(\frac{\partial F}{\partial T}\right)_{A,N}$
Let's first write $F$ as $F = -N k_B T \left[ \ln \left( \frac{2\pi A k_B^2}{N(hc)^2} \right) + 2\ln T + 1 \right]$.
$S = - \left( -N k_B \left[ \ln \left( \dots \right) + 2\ln T + 1 \right] - N k_B T \left[ \frac{2}{T} \right] \right)$
$S = N k_B \left[ \ln \left( \frac{2\pi A (k_B T)^2}{N(hc)^2} \right) + 1 \right] + 2N k_B$
$$\boldsymbol{S = N k_B \left[ \ln \left( \frac{2\pi A (k_B T)^2}{N(hc)^2} \right) + 3 \right]}$$
Internal Energy ($U = F + TS$)
$U = -N k_B T \left[ \ln \left( \dots \right) + 1 \right] + T \left( N k_B \left[ \ln \left( \dots \right) + 3 \right] \right)$
$U = -N k_B T \ln(\dots) - N k_B T + N k_B T \ln(\dots) + 3N k_B T$
$$\boldsymbol{U = 2N k_B T}$$
Pressure (P)
$P = -\left(\frac{\partial F}{\partial A}\right)_{T,N}$
$F = -N k_B T \left[ \ln A + \ln \left( \frac{2\pi (k_B T)^2}{N(hc)^2} \right) + 1 \right]$
$P = - \left( -N k_B T \cdot \frac{1}{A} \right)$
$$\boldsymbol{P = \frac{N k_B T}{A}}$$
Chemical Potential ($\mu$)
$\mu = \left(\frac{\partial F}{\partial N}\right)_{T,A}$
$F = -N k_B T \left[ \ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) - \ln N + 1 \right]$
Let $C = k_B T \left[ \ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) + 1 \right]$. Then $F = -NC + N k_B T \ln N$.
$\mu = \frac{\partial}{\partial N} (-NC + N k_B T \ln N)$
$\mu = -C + \left( k_B T \ln N + N k_B T \frac{1}{N} \right)$
$\mu = -C + k_B T \ln N + k_B T$
$\mu = -k_B T \left[ \ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) + 1 \right] + k_B T \ln N + k_B T$
$\mu = k_B T \left[ -\ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) -1 + \ln N + 1 \right]$
$\mu = k_B T \left[ \ln N - \ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) \right]$
$$\boldsymbol{\mu = k_B T \ln \left( \frac{N(hc)^2}{2\pi A (k_B T)^2} \right)}$$
3
(a)
The canonical partition function for $N_i$ indistinguishable particles of mass $m_i$ in a volume $V$ at temperature $T$ is given by:
$$Q_i = \frac{q_i^{N_i}}{N_i!}$$
where $q_i$ is the single-particle translational partition function:
$$q_i = V \left( \frac{2\pi m_i k_B T}{h^2} \right)^{3/2} = \frac{V}{\Lambda_i^3}$$
Here, $\Lambda_i = h/\sqrt{2\pi m_i k_B T}$ is the thermal de Broglie wavelength for species $i$, $k_B$ is the Boltzmann constant, and $h$ is Planck's constant.
Since the two types of molecules are distinguishable from each other, the total partition function for the system is the product of the individual partition functions:
$$Q(N_1, N_2, V, T) = Q_1 Q_2 = \frac{q_1^{N_1}}{N_1!} \frac{q_2^{N_2}}{N_2!}$$
Substituting the expressions for $q_1$ and $q_2$:
$$Q = \frac{1}{N_1! N_2!} \left[ V \left( \frac{2\pi m_1 k_B T}{h^2} \right)^{3/2} \right]^{N_1} \left[ V \left( \frac{2\pi m_2 k_B T}{h^2} \right)^{3/2} \right]^{N_2}$$
To find the thermodynamic properties, we first compute the natural logarithm of $Q$, using Stirling's approximation ($\ln N! \approx N \ln N - N$):
$$\ln Q = N_1 \ln q_1 + N_2 \ln q_2 - (N_1 \ln N_1 - N_1) - (N_2 \ln N_2 - N_2)$$
$$\ln Q = N_1 \ln\left(\frac{q_1}{N_1}\right) + N_2 \ln\left(\frac{q_2}{N_2}\right) + (N_1 + N_2)$$
1. Helmholtz Free Energy ($A$)
$A = -k_B T \ln Q$
$$A = -k_B T \left[ N_1 \ln\left(\frac{q_1}{N_1}\right) + N_2 \ln\left(\frac{q_2}{N_2}\right) + N_1 + N_2 \right]$$
2. Pressure ($P$)
$P = k_B T \left( \frac{\partial \ln Q}{\partial V} \right)_{T, N_1, N_2}$. Since $\ln q_i = \ln V + \text{terms independent of } V$, we have $\frac{\partial \ln q_i}{\partial V} = \frac{1}{V}$.
$$P = k_B T \left( N_1 \frac{\partial \ln q_1}{\partial V} + N_2 \frac{\partial \ln q_2}{\partial V} \right) = k_B T \left( \frac{N_1}{V} + \frac{N_2}{V} \right)$$
$$P = \frac{(N_1 + N_2)k_B T}{V}$$
3. Internal Energy ($U$)
$U = k_B T^2 \left( \frac{\partial \ln Q}{\partial T} \right)_{V, N_1, N_2}$. Since $\ln q_i = \frac{3}{2}\ln T + \text{terms independent of } T$, we have $\frac{\partial \ln q_i}{\partial T} = \frac{3}{2T}$.
$$U = k_B T^2 \left( N_1 \frac{\partial \ln q_1}{\partial T} + N_2 \frac{\partial \ln q_2}{\partial T} \right) = k_B T^2 \left( N_1 \frac{3}{2T} + N_2 \frac{3}{2T} \right)$$
$$U = \frac{3}{2}(N_1 + N_2)k_B T$$
4. Entropy ($S$)
$S = \frac{U - A}{T}$
$$S = \frac{3}{2}(N_1 + N_2)k_B + k_B \left[ N_1 \ln\left(\frac{q_1}{N_1}\right) + N_2 \ln\left(\frac{q_2}{N_2}\right) + N_1 + N_2 \right]$$
$$S = k_B \left[ N_1 \ln\left(\frac{V}{N_1\Lambda_1^3}\right) + N_2 \ln\left(\frac{V}{N_2\Lambda_2^3}\right) + \frac{5}{2}(N_1 + N_2) \right]$$
(b)
Properties of the Single-Component Gas
The partition function $Q'$ for this gas is:
$$Q' = \frac{q'^N}{N!}$$
where $q' = V \left( \frac{2\pi m k_B T}{h^2} \right)^{3/2} = \frac{V}{\Lambda'^3}$.
Using the same methods as in part (a):
Pressure ($P'$):
$$P' = \frac{N k_B T}{V} = \frac{(N_1+N_2)k_B T}{V}$$
Internal Energy ($U'$):
$$U' = \frac{3}{2}N k_B T = \frac{3}{2}(N_1+N_2)k_B T$$
Entropy ($S'$):
$$S' = k_B \left[ N \ln\left(\frac{q'}{N}\right) + \frac{5}{2}N \right] = k_B \left[ (N_1+N_2) \ln\left(\frac{V}{(N_1+N_2)\Lambda'^3}\right) + \frac{5}{2}(N_1+N_2) \right]$$
The pressure and internal energy are identical in both cases ($P=P'$ and $U=U'$).
The entropy is different. Let's find the difference $\Delta S = S_{mixture} - S_{single}$.
$$\frac{\Delta S}{k_B} = \left[ N_1 \ln\left(\frac{q_1}{N_1}\right) + N_2 \ln\left(\frac{q_2}{N_2}\right) \right] - (N_1+N_2) \ln\left(\frac{q'}{N_1+N_2}\right)$$
4
The virial $\mathcal{V}$ of a system of $N$ particles is defined as the ensemble average of the sum of the dot products of the force $\mathbf{F}_i$ on each particle with its position vector $\mathbf{r}_i$.
$$\mathcal{V} = \left\langle \sum_{i=1}^N \mathbf{F}_i \cdot \mathbf{r}_i \right\rangle$$
The total force $\mathbf{F}_i$ on any particle can be split into two components: the external force exerted by the walls of the container, $\mathbf{F}_i^{\text{ext}}$, and the internal force from other particles in the system, $\mathbf{F}_i^{\text{int}}$.
$$\mathcal{V} = \left\langle \sum_{i=1}^N \mathbf{F}_i^{\text{ext}} \cdot \mathbf{r}_i \right\rangle + \left\langle \sum_{i=1}^N \mathbf{F}_i^{\text{int}} \cdot \mathbf{r}_i \right\rangle$$
The external forces are due to the pressure $P$ exerted by the container walls. The force on a surface element $d\mathbf{S}$ of the container wall is $-P d\mathbf{S}$ (pointing into the volume). The sum over the particles feeling this force can be replaced by an integral over the container's surface area $S$.
$$\left\langle \sum_{i=1}^N \mathbf{F}_i^{\text{ext}} \cdot \mathbf{r}_i \right\rangle = \oint_S (-P d\mathbf{S}) \cdot \mathbf{r} = -P \oint_S \mathbf{r} \cdot d\mathbf{S}$$
Using the divergence theorem, which states $\oint_S \mathbf{A} \cdot d\mathbf{S} = \int_V (\nabla \cdot \mathbf{A}) dV$, we can convert the surface integral into a volume integral. Here, our vector field is the position vector $\mathbf{A} = \mathbf{r} = (x, y, z)$.
The divergence of $\mathbf{r}$ is:
$$\nabla \cdot \mathbf{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3$$
Substituting this back into the integral:
$$-P \oint_S \mathbf{r} \cdot d\mathbf{S} = -P \int_V (\nabla \cdot \mathbf{r}) dV = -P \int_V 3 dV = -3PV$$
So, the contribution to the virial from external forces is -3PV.
The internal force on particle $i$ is the negative gradient of the total potential energy $U_{\text{total}}$ with respect to its coordinates: $\mathbf{F}_i^{\text{int}} = -\nabla_i U_{\text{total}}$.
The contribution to the virial is:
$$\left\langle \sum_{i=1}^N \mathbf{F}_i^{\text{int}} \cdot \mathbf{r}_i \right\rangle = \left\langle \sum_{i=1}^N (-\nabla_i U_{\text{total}}) \cdot \mathbf{r}_i \right\rangle = -\left\langle \sum_{i=1}^N \mathbf{r}_i \cdot \nabla_i U_{\text{total}} \right\rangle$$
The problem states that the interparticle potential energy $u(\mathbf{r})$ is a homogeneous function of degree $n$. This implies that the total potential energy of the system, $U_{\text{total}}(\mathbf{r}_1, \dots, \mathbf{r}_N)$, is also a homogeneous function of degree $n$ with respect to all particle coordinates. That is, if we scale all coordinates by a factor $\lambda$, the potential energy scales as:
$$U_{\text{total}}(\lambda\mathbf{r}_1, \dots, \lambda\mathbf{r}_N) = \lambda^n U_{\text{total}}(\mathbf{r}_1, \dots, \mathbf{r}_N)$$
According to Euler's theorem for homogeneous functions, if a function $f(x_1, \dots, x_m)$ is homogeneous of degree $k$, then $\sum_{j=1}^m x_j \frac{\partial f}{\partial x_j} = kf$. Applying this theorem to our potential energy $U_{\text{total}}$:
$$\sum_{i=1}^N \mathbf{r}_i \cdot \nabla_i U_{\text{total}} = n U_{\text{total}}$$
Taking the ensemble average gives $\langle n U_{\text{total}} \rangle = nU$, where $U$ is the mean potential energy.
Thus, the contribution to the virial from internal forces is:
$$-\left\langle \sum_{i=1}^N \mathbf{r}_i \cdot \nabla_i U_{\text{total}} \right\rangle = -nU$$
Combining the contributions from external and internal forces, we get the total virial:
$$\mathcal{V} = -3PV - nU$$
The virial theorem provides a general relationship between the mean kinetic energy $K$ and the virial $\mathcal{V}$:
$$K = \left\langle \sum_{i=1}^N \frac{p_i^2}{2m_i} \right\rangle = -\frac{1}{2} \mathcal{V}$$
Substituting the expression for $\mathcal{V}$ we just derived:
$$K = \frac{1}{2}(3PV + nU)$$
The total mean energy of the system is the sum of the mean kinetic and mean potential energies: $E = K + U$. We can rearrange this to write the potential energy as $U = E - K$.
Now, substitute this into our expression for $K$:
$$K = \frac{1}{2}(3PV + n(E - K))$$
$$K = \frac{1}{n+2}(3PV + nE)$$
5
(a)
The total energy of a single molecule is given as $\epsilon = \epsilon_{trans} + \epsilon_{rot} + \epsilon_{pot}$. Since these energy terms are independent, the single-particle partition function $Z_1$ is a product of the partition functions for translation, rotation, and potential energy.
$Z_1 = Z_{trans} \cdot Z_{rot,pot}$
Translational Part: The standard result for a classical gas in a volume $V$ is:
$$Z_{trans} = \frac{V}{h^3} \int e^{-\beta p^2/2m} d^3p = V \left(\frac{2\pi m k_B T}{h^2}\right)^{3/2}$$
Rotational and Potential Part: This part requires integrating over the rotational phase space $(\theta, \phi, p_\theta, p_\phi)$.
$$Z_{rot,pot} = \frac{1}{h^2} \int e^{-\beta \left( \frac{p_\theta^2}{2I} + \frac{p_\phi^2}{2I \sin^2\theta} - \mu E \cos\theta \right)} dp_\theta dp_\phi d\theta d\phi$$
Integrating over the momenta $p_\theta$ and $p_\phi$ (which are Gaussian integrals) yields:
$$Z_{rot,pot} = \frac{2\pi I k_B T}{h^2} \int_0^{2\pi} d\phi \int_0^\pi e^{\beta \mu E \cos\theta} \sin\theta \, d\theta$$
Let $x = \beta \mu E = \frac{\mu E}{k_B T}$. The integral over the angles becomes:
$$2\pi \int_{-1}^{1} e^{xu} du = 4\pi \frac{\sinh(x)}{x}$$
where $u = \cos\theta$. Combining these gives the rotational partition function:
$$Z_{rot,pot} = \frac{8\pi^2 I k_B T}{h^2} \left( \frac{\sinh(\beta \mu E)}{\beta \mu E} \right)$$
The total partition function for $N$ non-interacting molecules is $Z_N = \frac{Z_1^N}{N!}$.
The average alignment of the dipoles with the electric field gives the electric polarization $P$, defined as the average dipole moment per unit volume.
$$P = \frac{N}{V} \langle \mu \cos\theta \rangle$$
The average value $\langle \cos\theta \rangle$ is given by the Langevin function, $L(x) = \coth(x) - 1/x$, where $x = \beta \mu E$.
$$P = \frac{N\mu}{V} \left( \coth(\beta \mu E) - \frac{1}{\beta \mu E} \right)$$
Using the assumption that $|\mu E| \ll k_B T$ (i.e., $x \ll 1$), we can approximate the Langevin function by expanding it: $L(x) \approx x/3$.
$$P \approx \frac{N\mu}{V} \left( \frac{\beta \mu E}{3} \right) = \frac{N \mu^2}{3V k_B T} E$$
The dielectric constant $\epsilon_r$ relates the polarization to the electric field. In cgs units, the relation is:
$$\epsilon_r = 1 + 4\pi \frac{P}{E}$$
Substituting our approximate expression for $P$:
$$\epsilon_r = 1 + 4\pi \frac{N \mu^2}{3V k_B T}$$
Letting $n = N/V$ be the number density of the molecules, we get the final expression:
$$\boldsymbol{\epsilon_r \approx 1 + \frac{4\pi n \mu^2}{3 k_B T}}$$
This is the Debye equation for the dielectric constant of a polar gas.
(b)
We can treat steam as an ideal gas. The ideal gas law is $PV = N k_B T$, so the number density $n=N/V$ is:
$$n = \frac{P}{k_B T}$$
We will use cgs units for all quantities:
Pressure $P$: $1 \text{ atm} = 1.01325 \times 10^6 \text{ dyne/cm}^2$.
Temperature $T$: $100^\circ C = 373.15 \text{ K}$.
Boltzmann constant $k_B$: $1.3806 \times 10^{-16} \text{ erg/K}$.
First, calculate the thermal energy $k_B T$:
$k_B T = (1.3806 \times 10^{-16} \text{ erg/K}) \times (373.15 \text{ K}) = 5.152 \times 10^{-14} \text{ erg}$.
Now, calculate the number density:
$$n = \frac{1.01325 \times 10^6 \text{ dyne/cm}^2}{5.152 \times 10^{-14} \text{ erg}} = 1.9667 \times 10^{19} \text{ cm}^{-3}$$
Now we plug our values into the Debye equation:
$$\epsilon_r - 1 = \frac{4\pi n \mu^2}{3 k_B T}$$
We are given:
Dipole moment $\mu = 1.85 \times 10^{-18}$ e.s.u. (statcoulomb-cm).
Let's calculate the terms:
$\mu^2 = (1.85 \times 10^{-18})^2 = 3.4225 \times 10^{-36} \text{ (statC}\cdot\text{cm)}^2$.
$3 k_B T = 3 \times (5.152 \times 10^{-14} \text{ erg}) = 1.5456 \times 10^{-13} \text{ erg}$.
Now, combine everything:
$$\epsilon_r - 1 = \frac{4\pi (1.9667 \times 10^{19} \text{ cm}^{-3}) (3.4225 \times 10^{-36} \text{ (statC}\cdot\text{cm)}^2)}{1.5456 \times 10^{-13} \text{ erg}}$$
Recalling that in cgs units, the units combine to be dimensionless (since $\text{erg} = \text{dyne}\cdot\text{cm}$ and $\text{dyne} = \text{statC}^2/\text{cm}^2$), the calculation gives:
$$\epsilon_r - 1 \approx 0.00547$$
$$\boldsymbol{\epsilon_r \approx 1.00547}$$
Thus, the dielectric constant of steam at $100^\circ C$ and atmospheric pressure is approximately 1.0055.