热学

Thermal Physics¶

Basic Concepts¶

The Laws of Thermodynamics¶

The Zeroth Law of Thermodynamics: If two systems are in thermal equilibrium with a third system, then they must be in thermal equilibrium with each other.

The Third Law of Thermodynamics: It is impossible for any procedure to lead to the isotherm $T= 0$ in a finite number of steps.

Temperature Scales¶

Absolute temperature (Kelvin) scale: $\displaystyle T = T_{\text{triple}} \frac{p}{p_{\text{triple}}} \approx \frac{273.16}{610} p $

$273.16$ is the difference between absolute zero and the temperature of the triple point of water .

Celsius scale: $T_C = T - 273.15$

Fahrenheit scale: $\displaystyle T_F = \frac{9}{5}T_C + 32^{\circ}F$

Thermal Expansion¶

Linear expansion: $\Delta L = \alpha L_0 \Delta T$ Volume expansion: $\Delta V = \beta V_0 \Delta T$

\[ \begin{gathered} L^3 = [L_0(1 + \alpha \Delta T)]^3 = L_0^3(1 + 3\alpha \Delta T + O(\Delta T)^2) \\\\ V = V_0(1 + \beta \Delta T) \\\\ \Rightarrow \beta = 3\alpha \end{gathered} \]

Ideal Gases¶

Equation of state for an ideal gas:

\[ \begin{gathered} PV = nRT \\\\ PV = \frac{N}{N_A}RT = Nk_BT \end{gathered} \]

where $R = 8.315 \text{ J/}(\text{mol}\cdot\text{K})$, $\displaystyle k_B = \frac{R}{N_A} = 1.38\times 10^{-23} \text{ J/K}$.

Therefore $\(\ln V = \ln T + \ln(nR / P)$\)

Then we can get $\(\beta = \left( \frac{1}{V} \frac{dV}{dT} \right)_p = \left( \frac{d(\ln V)}{dT} \right)_p = \frac{d(\ln T)}{dT} = \frac{1}{T}$\)

Real Gases¶

The van der Waals equation of state is $\((P + \frac{an^2}{V^2})(V - nb) = nRT$\)

$a$ is about the attractive forces between the gas molecules. $b$ is about the volume of gas molecule.

Microscopic Model of Ideal Gases¶

Pressure on the Wall¶

The total momentum generated by molecules colliding with wall during $dt$ is $\(dP_x = \frac{1}{2} \cdot \frac{N}{V} \cdot Av_xdt \cdot 2mv_x = \frac{NAmv_x^2}{V} dt$\)

So the pressure on the wall is $\(p = \frac{1}{A}\frac{dP_x}{dt} = \frac{Nmx_v^2}{V} = \rho v_x^2$\)

Therefore, we can get

\[ \begin{gathered} \overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2} \\\\ p = \frac{1}{3} \rho \overline{v^2} \end{gathered} \]

Furthermore, we can get

\[ \begin{gathered} pV = Nm \overline{v_x^2} = Nk_BT \\\\ \overline{\frac{1}{2} mv_x^2} = \overline{\frac{1}{2} mv_y^2} = \overline{\frac{1}{2} mv_z^2} = \frac{1}{2} k_BT \end{gathered} \]

So for ideal gas, temperature is a measure of internal energy. It measures the average energy per degree of freedom per molecule/atom.

Heat Capacity¶

Equipartition Theorem¶

The energy of each molecule is $\(E = \sum_i \frac{p_i^2}{2m} + \sum_i \frac{1}{2}I_i\omega_i^2 + \sum_i \frac{1}{2}k_iq_i^2 + \cdots$\)

The mean value of each independent quadratic term in the energy is equal to $k_BT/2$. So the average energy of each molecule is $\(f\cdot \frac{k_BT}{2}$\)

where $f$ is the degrees of freedom of the molecule. Vibration has both kinetic energy and potential energy, so it should be counted twice.

Internal energy: $\displaystyle U = \frac{f}{2} Nk_BT = \frac{f}{2} nRT = nC_VT$

Heat capacity: $\displaystyle c_V = \left( \frac{\Delta U}{\Delta T} \right)_V = \frac{f}{2}Nk_B$

Molar specific heat: $\displaystyle C_V = \frac{c_V}{n} = \frac{f}{2}R$

Mode Counting¶

$N$-atom linear molecule
- Translation: $3$
- Rotation: $2$
- Vibration: $3N - 5$
$N$-atom nonlinear molecule
- Translation: $3$
- Rotation: $3$
- Vibration: $3N - 6$

For single atom, it only has three degrees of freedom. For solid, Each atom vibrates around its equilibrium position and has three degrees of freedom, so $f=6$.

Maxwell Distribution¶

Define the distribution function $f(v)$ as $\(f(v) = \frac{1}{N} \frac{dN}{dv}$\)

Then we have

\[ \begin{gathered} dN = Nf(v)dv \\\\ \int f(v)dv = 1 \end{gathered} \]

The Maxwell distribution function is $\(f(v) = 4\pi \left( \frac{m}{2\pi k_BT} \right)^{3/2} v^2 e^{-mv^2/2k_BT}$\)

Characteristic Speed¶

Gauss’s Probability Integral

\[ \begin{gathered} I_{2n} = (-1)^n \frac{d^n}{da^n} I_0 \\\\ I_{2n+1} = (-1)^n \frac{d^n}{da^n} I_1 \end{gathered} \]

So we can get

\[ \begin{gathered} I_0 = \int_0^{\infty} e^{-ax^2} = \frac{1}{2} \sqrt{\frac{\pi}{a}}, \quad I_1 = \int_0^{\infty} xe^{-ax^2} = \frac{1}{2a} \\\\ I_2 = \int_0^{\infty} x^2e^{-ax^2} = \frac{1}{4} \sqrt{\frac{\pi}{a^3}}, \quad I_3 = \int_0^{\infty} x^3e^{-ax^2} = \frac{1}{2a^2} \\\\ I_4 = \int_0^{\infty} x^4e^{-ax^2} = \frac{3}{8} \sqrt{\frac{\pi}{a^5}}, \quad I_5 = \int_0^{\infty} x^5e^{-ax^2} = \frac{1}{a^3} \\\\ \end{gathered} \]

Most Probable Speed

\[ \begin{gathered} \frac{df(v)}{dv} = 0 \\\\ \Rightarrow (2v - \frac{mv^3}{k_BT}) e^{-mv^2/2k_BT} = 0 \\\\ \Rightarrow v_p = \sqrt{\frac{2k_BT}{m}} \end{gathered} \]

Average Speed

\[ \begin{aligned} \bar{v} &= \frac{1}{N} \int vdN = \int vf(v)dv \\ &= \int_0^{\infty} 4\pi \left( \frac{m}{2\pi k_BT} \right)^{3/2} v^3 e^{-mv^2/2k_BT} \\ &= \sqrt{\frac{8k_BT}{\pi m}} \end{aligned} \]

Root Mean Sqaure Speed

\[ \begin{aligned} v_{\text{rms}} &= \sqrt{\overline{v^2}} = \sqrt{\frac{1}{N} \int v^2dN} = \sqrt{\int v^2f(v)dv} \\ &= \int_0^{\infty} 4\pi \left( \frac{m}{2\pi k_BT} \right)^{3/2} v^4 e^{-mv^2/2k_BT} \\ &= \sqrt{\frac{3k_BT}{m}} \end{aligned} \]

Heat and The 1st Law of Thermodynamics¶

Heat ,Latent Heat, and Specific Heat¶

Definition of Heat Heat is defined as the transfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings.

The ways of heat transfer are conduction, convection, and radiation.

Unit of heat includes $\mathrm{J}$ and $\mathrm{cal}$, where $1 \mathrm{cal} = 4.186\mathrm{J}$.

Latent Heat Latent heat is energy released or absorbed during a constant-temperature process.

The latent heat of vaporization for a given substance is usually somewhat higher than the latent heat of fusion.

The specific latent heat is defined as $\(L = \frac{Q}{m}$\)

Specific Heat Specific heat capacity is defined as $\(c = \frac{1}{M} \cdot \frac{dQ}{dT}$\)

Molar heat capacity is defined as $\(C = \frac{1}{n} \cdot \frac{dQ}{dT}$\)

Molar heat capacity at constant volume is denoted by $C_V$. Molar heat capacity at constant pressure is denoted by $C_p$.

Heat Conduction¶

Fourier heat conduction law: $$\frac{Q}{\Delta t} = -\kappa_t A \frac{dT}{dx} $$

where $\lambda$ or $\kappa$ is coefficient of thermal conductivity.

Mean Free Path¶

Assume there is only one moving molecule. It will collide with the stationary molecules whose centers of mass are located within a cylinder of diameter $2d$.

So during time interval $t$, the average number of collisions is $\(z = n_V\pi d^2vt$\)

where $\displaystyle n_V = \frac{p}{k_B T}$ is the density of the molecules.

Therefore, the mean free path is $\(l = \frac{vt}{n_V\pi d^2vt} = \frac{1}{n_V\pi d^2} = \frac{k_BT}{\pi d^2p}$\)

Because all molecules are actually moving, we must consider their relative motion when counting the number of collisions. By $\(|\mathbf{v}_1 - \mathbf{v}_2|^2 = |\mathbf{v}_1|^2 + |\mathbf{v}_2|^2 - 2\mathbf{v}_1\cdot \mathbf{v}_2$\)

We can conclude that the average relative speed is $\sqrt{2}v$. Hence, $\(l = \frac{vt}{n_V\pi d^2(\sqrt{2}v)t} = \frac{1}{\sqrt{2}n_V\pi d^2} = \frac{k_BT}{\sqrt{2}\pi d^2p}$\)

And the average molecular separation is $\(d = \frac{1}{n_V^{1/3}}$\)

The 1st Law of Thermodynamics¶

The change in the internal energy $U$ of the system can be expressed as: $\(\Delta U = Q - W$\)

The work done by the gas is

\[ \begin{gathered} dW = Fdy = PAdy = PdV \\\\ W = \int_{V_i}^{V_f}PdV \end{gathered} \]

Thermodynamic Processes¶

A quasi-static process is one that is performed sufficiently slowly so that the system is always close to equilibrium. A reversible process is one that can be run backward in time by simply reversing its inputs and outputs.

A reversible process must be quasi-static. A quasi-static process is not necessarily reversible (e.g., energy dissipation)

Isothermal vs Free Expansion¶

In isothermal expansion,

\[ \begin{gathered} W = \int_{V_i}^{V_f} \frac{Nk_BT}{V} dV = Nk_BT \ln \frac{V_f}{V_i} \\\\ Q = W \end{gathered} \]

In free expansion, $\(P= 0, W = 0$\)

Isobaric vs Isovolumetric Processes¶

In isovolumetric (isochoric) process

\[ W = 0 \\\\ Q = nC_V\Delta T \\\\ \Delta U = Q = nC_V\Delta T \]

In isobaric process,

\[ W = P\Delta V \\\\ Q = nC_P\Delta T \\\\ \Delta U = Q - W = n(C_P - R)\Delta T \]

So

\[ C_P = C_V + R \\\\ C_V = \frac{f}{2}R \\\\ C_P = \frac{f+2}{2}R \\\\ \gamma = \frac{C_P}{C_V} = \frac{f+2}{f} \]

Adiabatic Equation¶

In adiabatic process, no heat transfer into or out of the system, that is,

\[ \begin{gathered} Q = 0 \\\\ \Delta U = -W \end{gathered} \]

So

\[ \begin{gathered} dU = nC_VdT = -PdV = -\frac{nRT}{V} dV \\\\ \Rightarrow \frac{dT}{T} = -\frac{R}{C_V} \frac{dV}{V} \\\\ \Rightarrow TV^{\gamma-1} = \text{const} \\\\ \Rightarrow PV^{\gamma} = \text{const} \end{gathered} \]

Engines and The 2nd Law of Thermodynamics¶

Otto Cycle¶

Otto cycle is a four-stroke cycle consisting of two upstrokes and two downstrokes. Two of the strokes are adiabatic processes and the others are isovolumetric processes. It represents the gasoline engine.

\[ \begin{gathered} W = Q_h - Q_c \\\\ Q_h = nC_V(T_C - T_B) \\\\ Q_c = nC_V(T_D - T_A) \\\\ e = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h} = 1 - \frac{T_D - T_A}{T_C - T_B} \end{gathered} \]

Because $\(\left( \frac{V_2}{V_1} \right)^{\gamma-1} = \frac{T_A}{T_B} = \frac{T_D}{T_C}$\)

We can get $\(e = 1 - \frac{T_A}{T_B} = 1 - \frac{T_D}{T_C} = 1 - \frac{1}{(V_1/V_2)^{\gamma-1}} = 1 - \frac{1}{r^{\gamma-1}}$\)

Diesel Engine¶

\[ Q_h = nC_p(T_C - T_B) \\\\ Q_c = nC_V(T_D - T_A) \\\\ W_{\text{cycle}} = n[C_p(T_C - T_B) - C_V(T_D - T_A)] \]

Carnot Cycle¶

Carnot cycle provides an upper limit on the efficiency of any classical thermodynamic engine, which consists of four steps: isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.

In isothermal processes,

\[ \begin{gathered} Q_h = W_{AB} = nRT_h \ln \frac{V_B}{V_A} \\\\ Q_c = |W_{CD}| = nRT_c \ln \frac{V_C}{V_D} \\\\ \Rightarrow \frac{Q_c}{Q_h} = \frac{T_c}{T_h} \frac{\ln(V_C/V_D)}{\ln(V_B/V_A)} \end{gathered} \]

In adiabatic processes,

\[ \begin{gathered} T_hV_B^{\gamma-1} = T_cV_C^{\gamma-1} \\\\ T_hV_A^{\gamma-1} = T_cV_D^{\gamma-1} \\\\ \Rightarrow \frac{Q_c}{Q_h} = \frac{T_c}{T_h} \end{gathered} \]

So the efficiency of Carnot engine is $\(e_C = 1 - \frac{T_c}{T_h}$\)

Heat Pumps¶

The coefficient of performance at heating mode is $\(\frac{Q_h}{W} = \frac{Q_h}{Q_h - Q_c} = \frac{1}{1 - \dfrac{Q_c}{Q_h}} = \frac{1}{1 - \dfrac{T_c}{T_h}} = \frac{T_h}{T_h - T_c}$\)

The coefficient of performance at cooling modet is $\(\frac{Q_c}{W} = \frac{T_c}{T_h - T_c}$\)

The 2nd Law of Thermodynamics¶

Clausius Statement It is impossible to construct a cyclical machine whose sole effect is the continuous transfer of heat from one object to another object at a higher temperature without the input of energy by work.

Kelvin Statement It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of heat from a reservoir and the performance of an equal amount of work.

Entropy and Information¶

The Function of Entropy¶

In a Carnot cycle, $\displaystyle \frac{Q_c}{Q_h} = \frac{T_c}{T_h}$, now put in the proper signs, we can get $\(\frac{Q_h}{T_h} + \frac{Q_c}{T_c} = 0$\)

Any reversible process can be approximated by a sum of Carnot cycles, hence $\(\oint_{\text{reversible } C} \frac{dQ}{T} = 0$\)

Therefore, we can define entropy as a state function.

The entropy of an ideal gas is

\[ \begin{gathered} dQ = dU + Pdv \\\\ \Rightarrow TdS = dU + PdV \\\\ \Rightarrow dS = \frac{nC_V^{mol}dT}{T} + \frac{nRdV}{V} \\\\ \Rightarrow S(T, V) = S_0 + nC_V^{mol}\ln \frac{T}{T_0} + nR\ln \frac{V}{V_0} \end{gathered} \]

The Second Law in terms of Entropy¶

If the path is arbitrary,

\[ \begin{gathered} 1 - \frac{Q_c}{Q_h} \le 1 - \frac{T_c}{T_h} \\\\ \Rightarrow \frac{Q_h}{T_h} + \frac{Q_c}{T_c} \le 0 \end{gathered} \]

Then we can get

\[ \begin{gathered} \oint_C \frac{dQ}{T} = \oint_{C_1} \frac{dQ}{T} + \oint_{C_2} \frac{dQ}{T} = \oint_{C_1} \frac{dQ}{T} + S(1) - S(2) \le 0 \\\\ \Rightarrow \oint_{C_1} \frac{dQ}{T} \le S(2) - S(1) \\\\ \Rightarrow dS \ge \frac{dQ}{T} \end{gathered} \]

So the change in entropy of an isolated system must be greater than zero for an irreversible process and equal to zero for a reversible process.

Entropy: A Measure of Disorder¶

Assume that each molecule occupies some microscopic volume $V_m$, and the number of molecules is $N$, then the number of microstates is

\[ \begin{gathered} W_i = \left( \frac{V_i}{V_m} \right)^N \\\\ W_f = \left( \frac{V_f}{V_m} \right)^N \\\\ \Rightarrow W = \frac{W_f}{W_i} = \left( \frac{V_f}{V_i} \right)^N \end{gathered} \]

Hence we can suggest $\(S = k_B \ln W$\)

Shannon Entropy¶

Given a random variable $X \in \{ x_1, x_2, \cdots, x_n \}$, and its probability $P_x$, we can define Shannon entropy as $\(S = -\sum_x P_x\log_2 P_x$\)

Shannon entropy quantifies the average information/uncertainty of the random variable. Shannon entropy also quantifies the resources needed to store the information.