Physical Quantities in Statistical Mechanics

In the following posts, the expression $$\left(\frac{\partial U}{\partial T}\right)_P$$ means differentiating $U$ w.r.t $T$, holding $P$ constant.

Multiplicity

Some defintions

• Among all the possible cases, each possible case is called a microstate.
• Microstates with common features can be group to a macrostate.
• The number of microstate a macrostate has is called mutiplicity

If the probability of all the microstates are equal, then the probability for a macrostate $X$ to occur will be $$\bbox[5px,border:2px solid #666] { P(X) = \frac{\Omega(X)}{\Omega(\text{all})} }$$

Entropy

Since the multiplicity is proportional to the power of the number of particles, taking logarithm will yield a quantity proportional to the number of particles. The entropy is defined as $$\bbox[5px,border:2px solid #666] { S \equiv k \ln \Omega }$$ where $k$ is the Boltzmann constant to keep the dimension right.

Thermodynamics Laws

The thermodynamics laws govern how the physical quantities behave.

• In a closed system, the change of the internal energy of a system is equal to the heat flowed into the system and the work done on the system, i.e.
  $$\Delta U = Q + W$$
• In a closed system, entropy can never decrease with time, and can remain constant if and only if all processes underwent are reversible
• At absolute zero, entropy is a constant

Temperature

As stated in thermodynamics second law, in a closed system containing two systems $A$ and $B$, only processes that either increase or does not change entropy can be carried out. Thus, holding other macroscopic coordinates constant, when energy exchange has been finished, i.e. at equilibrium, entropy, hence multiplicity should be maximum $$\begin{align} \left(\frac{\partial \Omega_{total}}{\partial U_A}\right)_{N,V} = \frac{\partial}{\partial U_A}(\Omega_A\Omega_B) &= 0 \\ \Omega_A\frac{\partial\Omega_B}{\partial U_A} + \Omega_B\frac{\partial\Omega_A}{\partial U_A} &= 0 \end{align}$$ As energy is conserved, change in $U_A$ equals negative change in $U_B$ $$\Omega_A\frac{\partial\Omega_B}{\partial U_B} = \Omega_B\frac{\partial\Omega_A}{\partial U_A}$$ As $\frac{d}{dx}\ln x = \frac{1}{x}$, $$\begin{align} \frac{1}{\Omega_A}\frac{\partial\Omega_A}{\partial U_A} &= \frac{1}{\Omega_B}\frac{\partial\Omega_B}{\partial U_B}\\ \frac{\partial\ln\Omega_A}{\partial U_A} &= \frac{\partial\ln\Omega_B}{\partial U_B} \\ \left(\frac{\partial S_A}{\partial U_A}\right)_{N,V} &= \left(\frac{\partial S_B}{\partial U_B}\right)_{N,V} \end{align}$$ Define temperature $T$ as $$\bbox[5px,border:2px solid #666] { \frac{1}{T} \equiv \frac{\partial S}{\partial U} }$$ where $S$ is entropy and $U$ is internal energy. Therefore, at equilibrium, $$T_A = T_B$$

Chemical Potential

As stated in thermodynamics second law, in a closed system containing two systems $A$ and $B$, only processes that either increase or does not change entropy can be carried out. Thus, holding other macroscopic coordinates constant, when particle exchange has been finished, i.e. at equilibrium, entropy, hence multiplicity should be maximum $$\begin{align} \left(\frac{\partial \Omega_{total}}{\partial N_A}\right)_{U,V} = \frac{\partial}{\partial N_A}(\Omega_A\Omega_B) &= 0 \\ \Omega_A\frac{\partial\Omega_B}{\partial N_A} + \Omega_B\frac{\partial\Omega_A}{\partial N_A} &= 0 \end{align}$$ As number of particles is conserved, change in $N_A$ equals negative change in $N_B$ $$\Omega_A\frac{\partial\Omega_B}{\partial N_B} = \Omega_B\frac{\partial\Omega_A}{\partial N_A}$$ As $\frac{d}{dx}\ln x = \frac{1}{x}$, $$\begin{align} \frac{1}{\Omega_A}\frac{\partial\Omega_A}{\partial N_A} &= \frac{1}{\Omega_B}\frac{\partial\Omega_B}{\partial N_B}\\ \frac{\partial\ln\Omega_A}{\partial N_A} &= \frac{\partial\ln\Omega_B}{\partial N_B} \\ \frac{\partial S_A}{\partial U_A}\frac{\partial U_A}{\partial N_A} &= \frac{\partial S_B}{\partial U_B} \frac{\partial U_B}{\partial N_B}\\ \frac{1}{T_A}\frac{\partial U_A}{\partial N_A} &= \frac{1}{T_B} \frac{\partial U_B}{\partial N_B}\\ \frac{\partial U_A}{\partial N_A} &= \frac{\partial U_B}{\partial N_B}\\ \end{align}$$ As holding other macroscopic coordinates constant, $T_A = T_B$. Define chemical potential $\mu$ as $$\bbox[5px,border:2px solid #666] { \mu \equiv \frac{\partial U}{\partial N} }$$ where $T$ is the temperature, $U$ is internal energy and $N$ is the number of particles. Therefore, at equilibrium, $$\mu_A = \mu_B$$

Pressure

As stated in thermodynamics second law, in a closed system containing two systems $A$ and $B$, only processes that either increase or does not change entropy can be carried out. Thus, holding other macroscopic coordinates constant, when exchange of the volume of the two system has been finished, i.e. at equilibrium, entropy, hence multiplicity should be maximum $$\begin{align} \left(\frac{\partial \Omega_{total}}{\partial V_A}\right)_{P,U} = \frac{\partial}{\partial V_A}(\Omega_A\Omega_B) &= 0 \\ \Omega_A\frac{\partial\Omega_B}{\partial V_A} + \Omega_B\frac{\partial\Omega_A}{\partial V_A} &= 0 \end{align}$$ As total volume is conserved, change in $V_A$ equals negative change in $V_B$ $$\Omega_A\frac{\partial\Omega_B}{\partial V_B} = \Omega_B\frac{\partial\Omega_A}{\partial V_A}$$ As $\frac{d}{dx}\ln x = \frac{1}{x}$, $$\begin{align} \frac{1}{\Omega_A}\frac{\partial\Omega_A}{\partial V_A} &= \frac{1}{\Omega_B}\frac{\partial\Omega_B}{\partial V_B}\\ \frac{\partial\ln\Omega_A}{\partial V_A} &= \frac{\partial\ln\Omega_B}{\partial V_B} \\ \frac{\partial S_A}{\partial U_A}\frac{\partial U_A}{\partial V_A} &= \frac{\partial S_B}{\partial U_B} \frac{\partial U_B}{\partial V_B}\\ \end{align}$$ From first law of thermodynamics, $\Delta U = Q + W$, as other macroscopic coordinates are held constant, $Q = 0$ $$dU = PdV$$ So, $$\frac{\partial S}{\partial U}\frac{\partial U}{\partial V} = \frac{P}{T}$$ As holding other macroscopic coordinates constant, $T_A = T_B$. Therefore, at equilibrium, $$P_A = P_B$$

Thermodynamic Identity

As multiplicity also depends on volume $V$ and number of particles $N$, entropy can be expressed as $$dS = \left(\frac{\partial S}{\partial U}\right)_{V,N}dU + \left(\frac{\partial S}{\partial V}\right)_{U,N}dV + \left(\frac{\partial S}{\partial N}\right)_{V,U}dN$$ So, $$dS = \frac{1}{T}dU + \frac{P}{T}dV -\frac{\mu}{T}dN$$ Thus, we have the thermodynamic identity $$\bbox[5px,border:2px solid #666] { dU = TdS - PdV + \mu dN }$$

Phase

A phase is a region of space in which the physical properties of the material are the same.

Heat Capacity

Heat capacity, $C$, is defined as $$C \equiv \lim_{\Delta T \to 0}\frac{\Delta Q}{\Delta T}$$ where $Q$ is the heat flowed into the system and $T$ is the temperature.

Latent Heat

The heat absorbed during a phase change is called latent heat. It does not change the temperature of the material. The latent heat per unit mass is called specific latent heat.