Quiz 1

Includes only the Quiz 1 section.

States and Equation of States

Surrounding: usually used when describing heat flow from the surrounding to the system. Often referred to reservoir as well.

Extensive Properties: properties that are dependent on mass/volume/mole. (For example, momentum, energy, etc.)

Intensive Properties: properties that are independent on mass/volume/mole. (For example, energy per mole)

Quasi-static Changes: changes that are so slow such that you can assume certain variables/states being constant.

Reversible Processes:

  • processes whose entropy does not change, and therefore direction can be reversed/

  • they are usually very slow, because to avoid heat-generating work such as friction, which is related to the rate at which movement happens, we can minimize friction by simply moving really REALLY slowly.

  1. Volume Expansivity
    • at constant pressure:
\[\beta = \frac{1}{V}\left.\frac{\partial V}{\partial T}\right|_P\]
  • at constant temperature:
\[\kappa_T = -\frac{1}{V}\left.\frac{\partial V}{\partial P}\right|_T\]
  • coefficient of linear expansion:
\[\alpha = \frac{1}{3}\beta\]
  1. Volume Expansion for system with $U(V,T,P)=U(P,T) \to V(P,T)$
    • basically an equation of state defining a surface
\[dV = \left.\frac{\partial V}{\partial T}\right|_P dT + \left.\frac{\partial V}{\partial P}\right|_T dP=\beta V dT - \kappa_T VdP\]
  • Calculus Equations for $f(x,y,z)=0$ $$ {align} \begin{align} \left.\frac{\partial x}{\partial y}\right|_z &= \frac{1}{\left.\frac{\partial y}{\partial z}\right|_z}
    \left.\frac{\partial x}{\partial y}\right|_z &= -\frac{\left.\frac{\partial x}{\partial z}\right|_y}{\left.\frac{\partial y}{\partial z}\right|_x}
    \left.\frac{\partial x}{\partial y}\right|_z\left.\frac{\partial y}{\partial z}\right|_x\left.\frac{\partial z}{\partial x}\right|_y&=-1 \end{align*}


  • Volume Expansion for ideal gas

    • realize they are not constants
\[\beta = \frac{1}{T}\\ \kappa_T = \frac{1}{P}\]
  • Exact Differential

    • basically, the function should not depend on path
    \[z=z(x,y)\to \frac{\partial^2 z}{\partial y \partial x}=\frac{\partial^2 z}{\partial x \partial y}\]

    ​ So if: $$

    dz=M(x,y)dx+N(x,y)dy\,\,\&\,\,\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}

    $$ ​ then $z$ is an exact differential, only dependent on the state $x,y$.


  1. Inexact Differential of Work Done for chemical system

    • since work done is $W=\int_C \vec{F} \cdot d\vec{r}$, it is path dependent.
    \[d'W= \vec{F} \cdot d \vec{r} = PAdx=PdV\]

Positive Work Done: this is defined to be work done by the system (e.g. gas)

Negative Work Done: this is defined to be work done on the system (e.g. gas)

  1. Inexact Differential of (Configurational) Work Done

    • wire/tension $$

      d’W=-F_T dL


    • magnetic dipole $$



    • electric dipole $$

      d’W = -Ed\mathbb{P}


    • surface film $$

      d’W = -\sigma dA


    • electrical cell $$

      d’W=-\epsilon dz,\,\,\,\,z=charge



    • there could be other dissipative/non-conservative work done for each system, but they are not listed here

First Law of Thermodynamics

  1. First Law of Thermodynamics:
    • the work done through any adiabatic ($dQ=0$) path between two states are the same
    • or generally:


d’Q = dU + d’W

\[​ where: obviously $U$ is a state function, independent of path. 4. ***Enthalpy for a Chemical System*** - given for a system, $d'W=PdV$, we can arrive at:\]
 d'Q |_P = d(U+PV)\equiv dH

 - $H\equiv U+PV$ is enthalpy
  • Heat Capacity:

    • how much heat to raise the temperature of a system of a given size:
    \[C \equiv \frac{Q}{\Delta T}\]

    which is path dependent since $Q$ is path dependent, and we define:

    • at constant pressure:
    \[C_P \to Q|_P = \int_{T_1}^{T_2}C_P dT\]
    • at constant volume: $$

      C_V \to Q V = \int{T_1}^{T_2}C_V dV


  1. Heat Capacity for a Chemical System at a Constant Volume

    • Since $d’Q = dU+PdV$, at constant volume:
    \[\left.\frac{\partial' Q}{\partial T}\right|_V = \left. \frac{\partial U}{\partial T}\right|_V = C_v\]
  2. Heat Capacity for a Chemical System at a Constant Pressure

    • Since $d’Q = dH - VdP$, at constant pressure: $$

      \left.\frac{\partial’ Q}{\partial T}\right _P = \left. \frac{\partial H}{\partial T}\right _P = C_P


Internal Energy

  1. Heat Equation with Specific Heat Capacities for Chemical System

    • From equation of $d’Q$ and $dU$, we get: $$

      d’q = c_v dT+(\frac{c_P-c_v}{\beta v})dv

      $$ where both $q$ and $v$ are normalized quantities from mole

  2. Joule’s Coefficient:

    • it is a measured quantity with the following definition: $$

      \left.\frac{\partial T}{\partial V}\right _U = \eta



    • for an ideal gas, this is $0$
  • Free Expansion of Ideal Gas in Adiabatic Process
    • Free expansion is different from adiabatic process, in which free expansion basically has $0$ external pressure $=$ $0$ pressure overall.
    • Therefore, Free expansion means $d’W = 0$.
    • Since we also have an adiabatic process, $d’Q = 0$.
    • Therefore, we have $dU = 0$, so $dT = \Delta T = 0$
  1. Coefficient of Free Expansion:

    • a measure of how a the internal energy of a free gas (expanding or compressing) changes at constant temperature.

    • this is derived quantity form the above using the relation $\left.\frac{\partial U}{\partial V}\right\vert _T\left.\frac{\partial V}{\partial T}\right\vert _U\left.\frac{\partial T}{\partial U}\right\vert _V = -1$ $$

      \left. \frac{\partial U}{\partial V} \right _T \equiv FreeExpansion = -\eta C_V



    • for ideal gas, this quantity is $0$ since $\eta$ is 0.
  2. Specific Heat Capacities and Coefficient of Free Expansion for Chemical System

    • basically inserting the above into the equation $c_P - c_v = (\left.\frac{\partial U}{\partial V}\right\vert _T + P)\left. \frac{\partial V}{\partial T} \right\vert _P$
    \[c_P - c_v = (P-\eta c_V)\left. \frac{\partial V}{\partial T} \right|_P\]
  3. Specific Heat Capacities for Ideal Gas

    • basically knowing $\eta$ and $\left. \frac{\partial V}{\partial T} \right\vert _P$: $$

      c_P - c_v = R


  4. Internal Energy Differential with Specific Heats

    • using equation (11), we have: $$

      dU = C_V dT - \eta C_V dV


  5. Internal Energy with Specific Heats for Ideal Gas

    • because $\eta=0$ for ideal gas: $$

      dU = C_V dT
      U(T)=U(0)+\int_{T_0}^{T}C_V dT’



  1. Joule-Thompson Experiment: Throttling of Gas
    • basically, it is an adiabatic process where gas is transported from one chamber to another, such that:
\[U_2 - U_1 = P_1V_1 - P_2V_2 \\ U_2+P_2V_2 = P_1V_1+ U_1\]

​ or cleanly: $$

H_1 = H_2

\[​ which is true in this case because ***throttling kept pressure to be constant/isobaric throughout the process*** 17. ***Joule-Thompson Experiment of Change of Temperature of a Gas*** - this a measured quantity, defining:\]
  \mu = \left. \frac{\partial T}{\partial P} \right|_h,\,\,\,\,where\,h=\frac{H}{moles}


- for **ideal gas,** $\mu=0$.
  1. Specific Heat Capacity with Joule-Thompson Coefficient above

    • basically, using the calculus result and cycle $\left. \frac{\partial h}{\partial T} \right\vert _P \left. \frac{\partial P}{\partial T} \right\vert _h\left. \frac{\partial T}{\partial h} \right\vert _P = -1$, we get
    \[\left. \frac{\partial h}{\partial P} \right|_T = -\mu c_p\]
  2. Specific Heat Capacities and Coefficient of Free Expansion for Chemical System

    • basically the other version of equation (12), using the result equation (18): $$

      c_P - c_v = (\mu c_p+V)\left. \frac{\partial P}{\partial T} \right _V


  3. Enthalpy and the Specific Heats

    • since we knew $H(U,P,V)=H(P,V)$, we can also say $H(T,P)$
    \[dh= \left. \frac{\partial h}{\partial T}\right|_P dT+\left. \frac{\partial h}{\partial P}\right|_T dP \\ dh = c_p dT - \mu c_p dP\]
  4. Enthalpy and the Specific Heats for Ideal Gas

    • Using the relation $\mu = 0$ for ideal gas, and the equation (20), we know: $$

      dh=c_p dT
      h(T) = h(T_0) + \int_{T_0}^{T} c_p dT’


  5. Heat Capacities for an Reversible Process with Ideal Gas

    • if we have an reversible process ($dS=0$), it has to be also adiabatic (since $d’Q = TdS$). Then using the two equations their their differential version of it, we get using equation (15) and (20) with ideal gas condition:
\[d'q = du+Pdv;\,\,d'q=dh-vdP \\ d'q = c_v dT + Pdv';\,\,d'q=c_pdT -vdP \\ 0= c_v dT + Pdv';\,\,0=c_pdT -vdP\\ c_v dT =- Pdv';\,\,c_pdT =vdP\\\]

​ Hence, by dividing the above two equations we can arrive at: $$

\frac{c_v dT}{c_p dT} = -\frac{Pdv}{vdP}
\frac{c_v}{c_p} = -\frac{Pdv}{vdP}
\frac{dP}{P} = -\frac{c_p}{c_v}\frac{dv}{v}
ln(PV^\gamma) = constant
PV^\gamma = constant

\[​ where $\gamma = \frac{c_p}{c_v} > 1$ 23. ***$\gamma$ for Ideal Gas in a Reversible Process*** - basically the by-product of the above:\]
  \gamma = \frac{c_p}{c_v} > 1

Carnot Cycle

For this section, refer to this diagram (again, ideal gas is assumed):

Carnot Engine: Cycle, Principles, Theorem, Efficiency with Videos,

  1. Carnot Cycle Heat Quotient

    • Consider the process 1 and process 3, we can compute the ratio of the heat: $$

      \frac{Q_1}{ Q_3 }=\frac{T_1ln(\frac{V_2}{V_1})}{T_3 ln(\frac{V_3}{V_4})}
      \[now, since we know process 2 and process 4 are adiabatic processes, using equation (22), we found:\]


      \[dividing the above two, we obtain:\]


      \[Hence the above equation reduces to:\]
      \frac{Q_1}{ Q_3 }=\frac{T_1}{T_3}


Thermal Efficiency: This is defined to be the ratio of total work done over total heat input $$

\eta = \frac{W}{Q_{in}}


  • Carnot Cycle Work Done

    • for processes 2 and process 4, we knew that $Q=0$ since they are adiabatic.

    • the total work done for both processes is therefore: $$

      0-\Delta U=\Delta W

      \[but as for ideal gas, $U=U(T,C_{V_{(T)}})$, hence as the net change in temperature is 0:\]

      0=\Delta W


  1. Thermal Efficiency for Carnot Engines

    • using the equation (24) and the above fact for thermal efficiency, we obtain, for a carnot cycle/engine: $$

      \eta = \frac{Q_1 + Q_3}{Q_1}= \frac{Q_1 - |Q_3|}{Q_1}
      \eta = 1-\frac{T_3}{T_1}

      $$ where:

      • $T_1$ is the higher temperature in the graph.
  • Reverse Carnot Cycle for a Refrigerator:
    • Basically we will have heat flowing in process 3 at room temperature, and heat flowing out in process 1. (Therefore, reversing the cycle.)
    • Now, we can reach the same formula for net work done/heat flow, but with a negative sign, indicating that work done is needed on the system.

Figure of Merit: This is the refrigerator version for thermal efficiency, we take the ratio of heat removed/taken in over work done onto/supplied to the system. $$

c=\frac{Q_{in}}{ W }


  • we took $\vert W\vert$ because work done would be negative.


This explains why not all processes are reversible, even if energetically permittable.

In some of the equations below, this graph will still be used:

Carnot Engine: Cycle, Principles, Theorem, Efficiency with Videos,

Second law of Thermodynamics:

  • every process in an isolated entity (including the universe, if is acted on/by) can either have entropy $S$ staying the same or increase.
  • $\to$ in a reversible process, entropy $S$ must stay the same
  • $\to$ there is no perfect heat engine/refrigerator
  1. Closed Loops in Carnot Cycle with Ideal Gas

    • for the closed loop in the figure above. and using equation (24), we know: $$


      \[since the other two processes are adiabatic, we know:\]

      \sum_i \frac{Q_i}{T_i} =0

      \[Now, if we take infinitesimal loops, such that the two isothermal curve approach each other:\]

      \sum_i \frac{Q_i}{T_i} = \sum_i \frac{Q_1 - |Q_3|}{T_i} \approx \sum_i \frac{d’Q}{T_i}=0

      \[However, since the above is an infinitesimal loop on the $P-V$ plane, we can construct any over loop by adding those small loops: - notice that this also makes the conclusion general to other **non-ideal gases** as well\]



  2. Entropy

    • from the above, we can define the quantity entropy with an heat flow in a reversible process (since derived in Carnot Cycle) to be: $$

      dS \equiv \frac{d’Q_{rev}}{T}\,\,or\,\,d’Q_{rev} = TdS

      \[Therefore, we obtain the ***general result for any gas in the $P-V$ plane***:\]

      \oint_{P-V\,plane}\frac{d’Q}{T}=\oint_{P-V\,plane} dS = 0


  • State of $S$

    • notice that equation (27) effectively made entropy $S$ become independent of path in the $P-V$ plane. Therefore, we conclude that: $$



    • now, if we use it for ideal gas, we get: $$



  1. First law of Thermodynamics Combined with Entropy for Chemical Systems

    • using equation (27), we can write (if we have a reversible process): $$

      d’Q =dU+d’W


  • Specific Heats with Entropy

    • basically we have an alternative expression of $d’Q$, so we can express many things alternatively: $$

      C_V = \left. \frac{d’Q}{dT} \right _V = T\left. \frac{dS}{dT} \right _V \
      C_P = \left. \frac{d’Q}{dT} \right _P = T\left. \frac{dS}{dT} \right _P

      $$ and etc.

  • Carnot Cycle in $T-S$ Plane

    • since Carnot cycle basically contains isothermal and adiabatic processes, we can construct the same cycle in the diagram below:

      4 stages of carnot cycle improving thermal efficiency - MechanicalTutorial


Includes this section and cumulatively, all the previous sections.

Entropy and Ideal Gas

  1. Entropy Differentials for Ideal Gas

    Since we know, for ideal gas, $dU = C_V dT - \eta C_V dV = C_V dT$, and $TdS = dU + PdV$ for a system: $$

    TdS = C_V dT + PdV
    dS = \frac{C_V}{T}dT + \frac{P}{T}dV

    dS = \left.\frac{\partial S}{\partial T}\right _V dT + \left.\frac{\partial S}{\partial V}\right _TdV
    \[we arrive at:\]

    \begin{cases} \text{for ideal gas:} & \left.\frac{\partial S}{\partial T}\right|_V = \frac{C_V}{T}
    & \left.\frac{\partial S}{\partial V}\right|_T = \frac{P}{T}=\frac{nR}{V}
    \text{in general:} & \left.\frac{\partial S}{\partial T}\right|_V = \frac{C_V}{T}
    & \left.\frac{\partial S}{\partial V}\right|_T = \frac{P}{T}-\eta \frac{C_V}{T} \end{cases}


  2. Entropy Integral for Ideal Gas

    Using the above, we could explicitly calculate the change of entropy for an ideal gas: $$

    dS = C_V \frac{dT}{T} + nR \frac{dV}{V}


    S_2 - S_1 = C_V ln(\frac{T_2}{T_1}) + nRln(\frac{V_2}{V_1})


Heat/Entropy Change for Reversible Process

Now we are dealing with processes, so the second law of thermodynamics takes precedence.

  • if we are just dealing with a state of a system, use the first law.

If we have a reversible process, we could also use the relation $d’Q = TdS$.

  1. Change of Entropy for Reversible Isothermal Process

    Again, since it is isothermal and reversible, we can use $d’Q = TdS$: $$

    S_2 - S_1 = \int_{Q_a}^{Q_b} \frac{d’Q}{T}=\frac{1}{T}(Q_b - Q_a)

    $$ which means, at the constant temperature:

    • net heat flow in = increase in entropy
    • net heat flow out = decrease in entropy = not possible to happen
  2. Change of Entropy for Reversible Isochoric Process

    In addition to using $d’Q = TdS$, we also know $d’Q\vert _V = dU\vert _V = C_V dT + 0$

    So we have: $$

    S_2 - S_1 = \int_{T_a}^{T_b} C_V\frac{ dT}{T}=C_V ln(\frac{T_b}{T_a}),\,\,\text{if $C_V$ is independent of $T$}


  3. Change of Entropy for Reversible Isobaric Process

    Now, we use the equation $d’Q\vert _P = dH\vert _P = C_p dT - \mu C_p dP = C_p dT$

    Therefore we have: $$

    S_2 - S_1 = \int_{T_a}^{T_b} C_P\frac{ dT}{T}=C_P ln(\frac{T_b}{T_a}),\,\,\text{if $C_P$ is independent of $T$}


  4. Reversible Heat Flow:

    In all the last two cases above, we have $\Delta S$ dependent on change of $T$. This implies that, in order for a reversible process and a reversible heat flow to occur, we need to raise the temperature of the system as a sequence of infinitesimal steps $\delta T$:


\begin{cases} dS_{system} = C_{P\,or\,V} \frac{\delta T}{T}
dS_{surronding} = C_{P\,or\,V} \frac{\delta T}{T+2\delta T} \approx C_{P\,or\,V} \frac{\delta T}{T} \end{cases}

\[As a result, we recover:\]

dS_{system} + dS_{surronding} = 0

\[which is **reversible heat flow**. ### Heat/Entropy Change for Irreversible Process The **key idea/trick** is that: - for each **step**/change, you can compute it using the **reversible process** as **$S$ is a state equation** - this means, for each **change of the system or the surrounding *individually***, you can use reversible process calculation - then, a **process basically involves the sum of changes of each part of the system** (including surrounding) - you will find the **total change of $S$ will be non-zero** 35. ***Change of Entropy at Either Constant P/V in an Irreversible Process*** The idea is as follows <img src="\lectures\images\typora-user-images\image-20201013171252696.png" alt="image-20201013171252696" style="zoom:67%;" /> and we treat the ***system individually*** first, **going from state $T_1 \to T_2$:**\]
\Delta S_{sys} = C_{P\,or\,V} \cdot ln(\frac{T_2}{T_1})


- this is legal to treat it as reversible process, **because for a system, it is merely a change of state, which is independent of path.**

now, treat the ***surrounding individually***, going from $T_2 \to T_2$, but with a heat flow known:

\Delta S_{surr} &= \frac{1}{T}\int_{Q_a}^{Q_b} d'Q	\\
 &= \frac{1}{T_2}(-\Delta Q_{sys})	\\
 &= \frac{1}{T_2}\left(- \int_{T_1}^{T_2}C_{P\,or\,V} \cdot dT\right)\\
 &= -\frac{1}{T_2}C_{P \, or\,V}\left(T_2 - T_1\right)

Therefore, the **total change for the process becomes**:

\Delta S_{total} &= \Delta S_{sys}+ \Delta S_{surr}	\\
\Delta S_{total} &= C_{P\,or\,V} \cdot ln\left(\frac{T_2}{T_1}\right) - C_{P \, or\,V}\frac{\left(T_2 - T_1\right)}{T_2}

  1. Second Law of Thermodynamics

    If we convert the previous equation (35) into the form: $$

    \Delta S_{total} = C_{P \,or\,V} \left( ln(1+x)-1+\frac{1}{1+x} \right)

    \[where: - $x = \frac{T_2 - T_1}{T_1}$ It can be plotted to show that:\]

    \Delta S {total} = \Delta S{universe} \ge 0


Energy Degradation

This is to show that, if you have done irreversible process/increased entropy, then the ability of doing work is reduced

  • i.e. the maximum efficiency will go down
  1. Work Lost Due to Irreversible Process/Increase in Entropy

    Consider the same process, reached via two different path:



    • dotted lines indicate a reversible process
    • solid line indicates an irreversible process

    Path (a): $$

    W_{max} = \eta Q_{in} = Q_{in}\left(1-\frac{T_0}{T_1}\right)

    \[Path (b):\]

    W_{max} = \eta Q_{in} = Q_{in}\left(1-\frac{T_0}{T_2}\right)

    \[Now, if $T_2 > T_1$, we see that the maximum work is different. Consider the **irreversible step** in path (a):\]

    \begin{align} \Delta S &= \Delta S_1 + \Delta S_2
    &= \frac{Q}{T_1}+\frac{-Q}{T_2}
    &= Q\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \end{align

    \[where: - obviously we assumed the reservoir to be large enough so its temperature stayed constant - also, $T_2 > T_1$ complies to our assumption Lastly, we see that:\]

    \begin{align} W_{lost} &= Q_{in}\left(1-\frac{T_0}{T_2}\right) - Q_{in}\left(1-\frac{T_0}{T_1}\right)
    &= T_0 Q_{in} \left( \frac{1}{T_1}-\frac{1}{T_2} \right)
    &= \frac{Q_{in}}{Q}T_0\Delta S_{total} \end{align

    $$ where:

    • $Q$ is irreversible heat
    • $Q_{in}$ is reversible heat
  2. Work Lost and Entropy

    If we make $Q_{in} = Q$, namely all the heat into the system is irreversible:

    • we can do this as $Q$ was an arbitrary quantity
    \[W_{lost} = T_{sys} \Delta S_{universe}\]

Gas Mixing Contradiction

  • Ideal Gas Mixing during an Adiabatic Free Expansion

    If we have an adiabatic process, it means $d’Q = 0$. A free expansion means no external pressure, hence $d’W=0$.

    • therefore, we conclude that $dU=0$, so $T$ of gas has to stay constant.


    Then we have: $$

    \Delta S_{He} = \Delta S_{Ne},\,\,\text{assuming their $C_{V,P}$ are also the same}


    \Delta S_{total} = 2\Delta S_{He} = 2\left( nRln\left( \frac{2V_1}{V_1} \right) \right) = 2nRln(2)

    \[However, if we simply change the $Ne$ to $He$, we would have a homogenous mixing, then it must be:\]

    \Delta S_{total}=0

    \[since **homogenous mixing must be reversible**. And the solution is that we need to fix our equation of $\Delta S$ to include:\]

    \Delta S = C_V ln\left( \frac{T_2}{T_1} \right) + nR ln\left( \frac{M_2}{M_1} \right)

    $$ where:

    • $M=\frac{1}{concentration}$, instead of having volume inside.

Thermodynamic Processes

  1. Thermodynamic Processes of a Large Reservoir

    If you have a large reservoir of temperature $T_{res}$, and $Q$ is the heat flows into the system from it, then from the second law of thermodynamics: $$

    \begin{align} \Delta S_{uni} &\ge 0
    \Delta S_{sys} + \Delta S_{res} &\ge 0
    \Delta S_{sys} + \frac{-Q}{T_{res}} &\ge 0
    T_{res}\Delta S_{sys} &\ge Q
    T_{res}\Delta S_{sys} &\ge \Delta U_{sys} + P_{sys}\Delta V_{sys}

    $$ for any process.

  2. Thermodynamic Processes of a Large Reservoir using Enthalpy

    Same as the above setting (41), but we use $d’Q = dH - VdP$: $$

    T_{res}\Delta S_{sys} \ge \Delta H_{sys} - V_{sys}\Delta P_{sys}


List of Useful Theorems



  • By definition, $\left.\frac{\partial U}{\partial T}\right\vert _{V}=V\beta$ , $\left.\frac{\partial U}{\partial T}\right\vert _{P}=-V\kappa_T$ $$

    dU=\beta VdT-\kappa_T V dP


Statistical Mechanics

Kinetic Theory

Equation Appendix

Equation Number Title/Condition Equations  
1 chemical system \(dV=\beta V dT - \kappa_T VdP\)  
2 chemical system \(d'W= \vec{F} \cdot d \vec{r} = PAdx=PdV\)  
3 First Law of Thermodynamics $d’Q = dU + d’W$  
  chemical system $d’Q = dU + PdV$  
  chemical system $d’Q = dH - VdP$  
4 chemical system $H\equiv U+PV$  
5 wire/tension \(d'W=-F_T dL\)  
  magnetic dipole $d’W=-HdM$  
  electric dipole $d’W = -Ed\mathbb{P}$  
  surface film $d’W = -\sigma dA$  
  electrical cell $d’W=-\epsilon dz,\,\,\,\,z=charge$  
6   $\left.\frac{\partial’ Q}{\partial T}\right\vert _V= \left. \frac{\partial U}{\partial T}\right\vert _V = C_V$  
7   $\left.\frac{\partial’ Q}{\partial T}\right\vert _P= \left. \frac{\partial H}{\partial T}\right\vert _P = C_P$  
8 Volumetric Expansion Coefficient $$\beta = \frac{1}{V}\left.\frac{\partial V}{\partial T}\right _P$$
    $$\kappa_T = -\frac{1}{V}\left.\frac{\partial V}{\partial P}\right _T$$
    \(\alpha = \frac{1}{3}\beta\)  
9 chemical system $d’q = c_v dT+(\frac{c_P-c_v}{\beta v})dv$  
10 chemical system $\left.\frac{\partial T}{\partial V}\right\vert _U = \eta$  
11 Coefficient of Free Expansion $\left. \frac{\partial U}{\partial V} \right\vert _T\equiv -\eta C_V$  
12 chemical system $$c_P - c_v = (P-\eta c_V)\left. \frac{\partial V}{\partial T} \right _P$$
13 ideal gas $c_P - c_v = R$  
14   $dU = C_V dT - \eta C_V dV$  
15   $dU = C_V dT$  
    analogy to (21): $U(T)=U(0)+\int_{T_0}^{T}C_V dT’$  
16 Chemical, isobaric processes: $U_2+P_2V_2 = P_1V_1+ U_1$  
    $H_2 = H_1$  
17   $\mu = \left. \frac{\partial T}{\partial P} \right\vert _h,\,\,\,\,where\,h=\frac{H}{moles}$  
18   $\left. \frac{\partial h}{\partial P} \right\vert _T = -\,u c_p$  
19 chemical system $c_P - c_v = (\mu c_p+V)\left. \frac{\partial P}{\partial T} \right\vert _V$  
20   $dh= \left. \frac{\partial h}{\partial T}\right\vert _P dT+\left. \frac{\partial h}{\partial P}\right\vert _T dP$  
    $dh = c_p dT - \mu c_p dP$  
21   $dh=c_p dT$  
    analogy to (15): $h(T) = h(T_0) + \int_{T_0}^{T} c_p dT’$  
22 ideal gas, adiabatic process $\gamma = \frac{c_p}{c_v} > 1$  
23 ideal gas, adiabatic process $PV^\gamma = constant$, for \(\gamma = \frac{c_p}{c_v} > 1\)  
  ideal gas, isothermal process $PV= constant=nRT_0$  
24 in a Carnot cycle, two isothermal processes $\frac{Q_1}{\vert Q_3\vert }=\frac{T_1}{T_3}$  
25 ideal thermal efficiency for Carnot engine $\eta = 1-\frac{T_3}{T_1}$  
26 for any gas $\oint_{P-V\,plane}\frac{dQ}{T}=0$  
27 In a reversible process $d’Q = TdS$  
    for any gas: $\oint_{P-V\,plane} dS = 0$  
28 chemical system, reversible process $TdS=dU+PdV$  
29 for ideal gas $\left.\frac{\partial S}{\partial T}\right\vert _V = \frac{C_V}{T}$  
  for ideal gas $\left.\frac{\partial S}{\partial V}\right\vert _T = \frac{P}{T}=\frac{nR}{V}$  
  for chemical system $\left.\frac{\partial S}{\partial T}\right\vert _V = \frac{C_V}{T}$  
  for chemical system $\left.\frac{\partial S}{\partial V}\right\vert _T = \frac{P}{T}-\eta \frac{C_V}{T}$  
30 for ideal gas $S_2 - S_1 = C_V ln(\frac{T_2}{T_1}) + nRln(\frac{V_2}{V_1})$  
31 isothermal reversible process $S_2 - S_1 = \int_{Q_a}^{Q_b} \frac{d’Q}{T}=\frac{1}{T}(Q_b - Q_a)$  
32 isobaric reversible process $S_2 - S_1 =C_V ln(\frac{T_b}{T_a}),\,\,\text{if$C_V$is independent of$T$}$  
33 isobaric reversible process $S_2 - S_1 =C_P ln(\frac{T_b}{T_a}),\,\,\text{if$C_P$is independent of$T$}$  
34   reversible heat flow requires a sequence of infinitesimal heat transaction of $\delta T$  
35 for a chemical system $\Delta S_{total} = \Delta S_{sys}+ \Delta S_{surr}= C_{P\,or\,V} \cdot ln\left(\frac{T_2}{T_1}\right) - C_{P \, or\,V}\frac{\left(T_2 - T_1\right)}{T_2}$  
36 Second Law of Thermodynamics $\Delta S {total} = \Delta S{universe} \ge 0$  
37 Irreversible heat transfer between two large reservoir $\Delta S = Q\left(\frac{1}{T_1}-\frac{1}{T_2}\right),\,\,\text{where$T_2 > T_1$}$  
38 for $Q$ being heat transfer (37) in irreversible process $W_{lost}=\frac{Q_{in}}{Q}T_{sys}\Delta S_{universe}$  
39 if $Q$ is the same as $Q_{in}$ to the system (38), or all $Q_{in}$ is irreversible $W_{lost} = T_{sys} \Delta S_{universe}$  
40 Lagrange Multiplier $\vec{\nabla}f(x,y)=\lambda \vec{\nabla}g(x,y)$, $g(x,y)$ being the constraint  
    one usage of this: nature prefers to maximize $S$ under certain constraint  
41 Process with a large reservoir of $T_{res}$ $T_{res}\Delta S_{sys} \ge \Delta U_{sys} + P\Delta V_{sys}$  
42 Process with a large reservoir of $T_{res}$ $T_{res}\Delta S_{sys} \ge \Delta H_{sys} - V_{sys}\Delta P_{sys}$  
43 Helmholtz Energy (Third Potential) $F=U-TS$  
44 Gibbs Free Energy (Fourth Potential) $G=U+PV-TS$  
45 For a process $dF \le W_{total}$  
  For $W_{total}=W_{PdV}+W_{non-PdV}$, $W_{non-PdV}=0$ $dF\le 0$, at constant volume  
46 For a process at constant $T,V$, no $W_{Non-PdV}$ $dF \le 0$  
  For a process at constant $T,P$, no $W_{Non-PdV}$ $dG \le 0$  
47 Second Law (Process) on First Potential $dU_{sys} \le T_{res}dS_{sys}-d’W_{sys}$  
  Second Law (Process) on Second Potential $dH_{sys} \le T_{res}dS_{sys}+V_{sys}dP_{sys}$  
  Second Law (Process) on Third Potential $dF_{sys} \le -S_{sys}dT_{res}-P_{sys}dV_{sys}$  
  Second Law (Process) on Fourth Potential $dG_{sys} \le -S_{sys}dT_{res}+V_{sys}dP_{sys}$  
48 First Potential Derivatives $\left.\frac{\partial U}{\partial S}\right\vert _V = T$, $\left.\frac{\partial U}{\partial V}\right\vert _S = -P$  
  Second Potential Derivatives $\left.\frac{\partial H}{\partial S}\right\vert _V = T$, $\left.\frac{\partial H}{\partial P}\right\vert _S = V$  
  Third Potential Derivatives $\left.\frac{\partial F}{\partial T}\right\vert _V = -S$, $\left.\frac{\partial F}{\partial V}\right\vert _T = -P$  
  Fourth Potential Derivatives $\left.\frac{\partial G}{\partial T}\right\vert _P = -S$, $\left.\frac{\partial G}{\partial P}\right\vert _T = V$  
49 Using (48) $U=F-T\left.\frac{\partial F}{\partial T}\right\vert _V$  
    $H=F-T\left.\frac{\partial F}{\partial T}\right\vert _V-V\left.\frac{\partial F}{\partial V}\right\vert _T$  
    $G=F-V\left.\frac{\partial F}{\partial V}\right\vert _T$  
50   $\left.\frac{\partial^2 G}{\partial T^2}\right\vert _P = -\frac{C_P}{T}$  
51   $\left.\frac{\partial^2 G}{\partial P^2}\right\vert _T = -\kappa_T V$  
52   $\left.\frac{\partial^2 G}{\partial T\partial P}\right\vert _T = \beta V$  
53 Maxwell’s Relations $\left.\frac{\partial T}{\partial V}\right\vert _S = -\left.\frac{\partial P}{\partial S}\right\vert _V$  
    $\left.\frac{\partial T}{\partial P}\right\vert _S = \left.\frac{\partial V}{\partial S}\right\vert _P$  
    $\left.\frac{\partial S}{\partial V}\right\vert _T = \left.\frac{\partial P}{\partial T}\right\vert _V$  
    $\left.\frac{\partial S}{\partial P}\right\vert _T = -\left.\frac{\partial V}{\partial T}\right\vert _P$  
54 $TdS$ Equations $TdS=C_V dT + T\left.\frac{\partial P}{\partial T}\right\vert _V dV$  
    $TdS=C_P dT - T\left.\frac{\partial V}{\partial T}\right\vert _P dP$  
    $TdS=C_P \left.\frac{\partial T}{\partial V}\right\vert _PdV + C_V \left.\frac{\partial T}{\partial P}\right\vert _VdP$  
55,56,57 All derived from (54) $\Delta S = \int\limits_{T_0}^T \frac{C_P}{T’}dT’-\int\limits_{P_0}^P \left.\frac{\partial V}{\partial T}\right\vert _P dP’$  
    $\Delta C_V\vert {T_0} = T_0\int\limits{V_0}^V \left.\frac{\partial^2 P}{\partial T^2}\right\vert _VdV$  
    $\Delta C_P\vert {T_0} = T_0\int\limits{P_0}^P \left.\frac{\partial^2 V}{\partial T^2}\right\vert _P dV$  
    $-\eta C_V=T\left.\frac{\partial P}{\partial T}\right\vert _V -P$  
58 For ideal gas, using (30) $S_2 - S_1 = C_P ln(\frac{T_2}{T_1}) - nRln(\frac{P_2}{P_1})$  
59 For ideal gas, using $G=H-TS$ $G=nRT\left(ln(P)+\Phi(T)\right)$, $\Phi(T)$ is some function on $T$ only.  
60 For ideal gas reaction, at constant temperature using (59) $G(P_f) = G(P_i) + nRTln(\frac{P_f}{P_i})$  
Equation Number Title/Condition Equations
61 For ideal gas reaction at constant temperature, $aA+bB \to cC+dD$
  (derived using (60)) $\frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}=e^{-\frac{\Delta G^\circ}{RT}}\equiv k$
62 Chemical Potential $\mu \equiv \left.\frac{\partial G}{\partial N}\right\vert _{T,P}$
63 At constant pressure during phase transition $\Delta H > \Delta U$, if $\Delta V >0$
64 During phase transition $\begin{cases}g_{s}=g_{l} & \text{at melting temperature}\ g_{l}=g_{g} &\text{at boiling temperature}\ g_{s}=g_{g} &\text{at sublimation temperature}\end{cases}$
65 At phase transition (hence constant $T$) $\begin{cases}l_{s\to l}=\text{heat of fusion}\ l_{l\to g}=\text{heat of vaporization}\ l_{s\to g}=\text{heat of sublimation}\end{cases}$
    Above is also $\Delta H$ at constant $P$ during phase change.
66 reversible process $\begin{cases}\Delta S_{s\to l}=\frac{l_{s\to l}}{T}\ \Delta S_{l\to g}=\frac{l_{l\to g}}{T}\ \Delta S_{s\to g}=\frac{l_{s\to g}}{T}\end{cases}$
67 At constant temperature and pressure $\Delta G=0$ at phase transition
68 Clausius Chaperone Equation $\left.\frac{\partial P}{\partial T}\right\vert {s,l\,eqilib}=\frac{l{s\to l}}{T_{melting}}\frac{1}{v_l - v_s}$
    In general: $\left.\frac{\partial P}{\partial T}\right\vert {phase\,eqilib}=\frac{l{phase\,trans}}{T_{phase\,trans}}\frac{1}{\Delta v_{phase}}$
69 For ideal gas, at liquid gas boundary, assuming $v_g - v_l \approx v_g$ $P=Ae^{\frac{-l_{l \to g}}{RT}}$
70 Energy Density (Radiation) of a Black Body $u=\sigma T^4$
71 Other properties of a Black Body $P=\frac{1}{3}\sigma T^4$
    $U=\sigma T^4V$
    $C_v = 4\sigma T^3V$
    $TdS=4\sigma VT^3 dT + \frac{4}{3}\sigma T^4 dV$
72 Heat needed for a reversible, isothermal volume expansion of Black Body $\Delta Q = \frac{4}{3} \sigma T^4\Delta V$
73 Reversible adiabatic process of Black Body $VT^3=constant$
90 Third Law of Thermodynamics: $T \to 0$ means $\Delta S \to 0$ (technically a constant)
    $C_{p,v}\to 0$ as $T \to 0$
    $\beta \to 0$ as $T \to 0$
91 Experimental Property of Black Body Radiation $Radiation \propto Energy \,Density\, (\frac{U}{V}=u)$
    $u=u(T)$ only
    Radiation Pressure: $P=\frac{u}{3}$
74 Overall Principle of Statistical Mechanics All microstates are equally probable
    $S = k_B ln(\Omega)$, $\Omega = \text{number of microstates}$
75, 76 Canonical Ensemble $P(E_s) = \frac{e^{-\beta E_s}}{\sum e^{-\beta E_s}}$ or $P(E_s) = \frac{g_ie^{-\beta E_s}}{\sum g_ie^{-\beta E_s}}$
  (Canonical) Partition function $z=\sum e^{-\beta E_s}$
    $\beta = \frac{1}{k_BT}$
    $U = \langle E_{sys}\rangle = \sum E_s P(E_s)= \frac{\sum E_se^{-\beta E_s}}{\sum e^{-\beta E_s}}$
77 Grand Canonical Ensemble $P(E_s) = \frac{e^{-\beta E_s-\alpha N_s}}{\sum e^{-\beta E_s-\alpha N_s}}$
  Grand Canonical Partition function $z=\sum e^{-\beta E_s-\alpha N_s}$
    $U = \frac{\sum E_se^{-\beta E_s-\alpha N_s}}{\sum e^{-\beta E_s - \alpha N_s}}$
    $\langle N_{sys}\rangle = \frac{\sum N_se^{-\beta E_s-\alpha N_s}}{\sum e^{-\beta E_s - \alpha N_s}}$
78 Macroscopic Property $Y$ of a Macrostate $\langle Y \rangle =\frac{\sum Y_k w_k}{\sum w_k}$
    usually, $\sum w_k=1$
79 Number of Microstates for System Without Degeneracy $\Omega =\frac{N!}{\Pi_i (n_i)!}$
  (Microcanonical Ensemble) for $\begin{cases}n_1&\text{$n_1$particles in energy state 1}\n_2&\text{$n_2$particles in energy state 2}\n_1&\text{$n_3$particles in energy state 3}\…\n_n&\text{$n_n$particles in energy state n}\end{cases}$
80 Number of Microstates for System With Degeneracy $\Omega= \frac{N!}{\Pi_i \left(\frac{n_i}{g_i^{ni}}\right)!}$
  (Microcanonical Ensemble) for $\begin{cases}g_1&\text{$g_1$degeneracy in energy state 1}\g_2&\text{$g_2$degeneracy in energy state 2}\g_1&\text{$g_3$degeneracy in energy state 3}\…\g_n&\text{$g_n$degeneracy in energy state n}\end{cases}$
81 Energy levels in Microcanonical Ensemble $E_n \approx n\hbar \omega$
    $q \equiv \text{total level of excitation/ totoal # energy level}$
82 Number of Microstates for System $\Omega = \frac{(q+(N-1))!}{q!(N-1)!}\approx\frac{(q+N)!}{q!N!}$
  (Microcanonical Ensemble) since $N \gg 1$
83 Stirling’s Approximation $ln(n!)\approx nln(n)-n$
84 Entropy of System given $q$ and $N$ $S=k_Bln\left(\frac{(q+N)!}{q!N!}\right)$
  Using (83) $S=k_B\left[ (q+N)ln(q+N)-qln(q)-Nln(N) \right]$
85 Energy for Microcanonical Ensemble using $T$ $U=\frac{U_0}{e^{\frac{\hbar \omega}{k_B T}}-1}=\frac{N\hbar \omega}{e^{\frac{\hbar \omega}{k_B T}}-1}$
    For $\begin{cases}U_{sys} = U = q\hbar \omega \ U_0 := N\hbar \omega\end{cases}$
86 Microcanonical Ensemble Average Energy per Oscillator $U_{\text{per oscillator}}=\frac{\hbar \omega}{e^{\frac{\hbar \omega}{k_bB T}}-1}$
    \(q_{\text{per oscillator}}=\frac{1}{e^{\frac{\hbar \omega}{k_bB T}}-1}\)
87 Microcanonical Ensemble $C_V$ $C_V=Nk_B\left(\frac{\hbar \omega}{k_B T}\right)^2\left[\frac{e^{\frac{\hbar \omega}{k_bB T}}}{e^{\frac{\hbar \omega}{k_bB T}}-1}\right]$
88 Microcanonical Ensemble at High Temperature Limit $U \approx Nk_B T$
  average energy per oscillator $\frac{U}{N} \approx k_B T$
    $C_V \approx Nk_B$
89 Microcanonical Ensemble at Low Temperature Limit $U \approx N\hbar \omega e^{-\frac{\hbar \omega}{k_bB T}}$
    $\frac{U}{N} \approx \hbar \omega e^{-\frac{\hbar \omega}{k_bB T}}$
    $C_V=Nk_B\left(\frac{\hbar \omega}{k_B T}\right)^2 e^{-\frac{\hbar \omega}{k_bB T}}$
104 Microcanonical Ensemble Entropy $S(T)=Nk_B[ln\left( \frac{1}{e^{\hbar \omega / k_B T}-1}\right)-\frac{\hbar \omega}{k_B T}\frac{1}{1-e^{-\hbar \omega / k_B T}}]$
    $F(T)=Nk_B Tln(e^{\hbar\omega /k_BT}-1)-N\hbar \omega$
92 Canonical Ensemble, Probability of finding a Particle at Energy $E_{sys_j}=E_j$ $\text{Prob}\propto \Omega_{res}\approx \Omega_{res}(E_{total})e^{-\beta E_j}$
assuming $E_{uni/total} \gg E_{sys_j}$
93, 94 Canonical Ensemble, same as (75), (76) $P(E_s) = \frac{e^{-\beta E_s}}{\sum e^{-\beta E_s}}$
95 Canonical Ensemble Partition Function with Degeneracy $z=\sum g_je^{-\beta E_s}$
96 Using Partition function for getting $U$, Canonical Ensemble $U=-\frac{\partial ln(z)}{\partial \beta}$
97,98,99 Canonical Ensemble Helmholtz Energy $z=e^{-\beta F_{sys}}$
  (derived using an alternative method) $F_{sys}\vert _U = -k_B Tln(z)$
100 Multi-particle Canonical Ensemble $F_{sys}\vert _U = -Nk_B Tln(z)$
    $U=-Nk_B T^2\frac{\partial ln(z_{(1)})}{\partial T}$
101 Canonical Ensemble, Work Done and Heat Flow $dU = \sum f_j dE_j+\sum E_jdf_j$
  since $U=\sum f_jE_j$ $d’Q = \sum f_j dE_j$
  and $dU=d’Q-d’W$ has a one-to-one relationship to (101) $d’W =-\sum E_jdf_j$
Equation Number Title/Condition Equations
102 Distinguishable Particles, Different Modes, Partition Function $Z_{(N)} = \Pi_k(Z_k)$
    $Z_{(N)} = Z_{(1)}^N$
    \(Z_{sys} = Z_{translation}Z_{rotation}Z_{vibration}\)
103 Einstein Model of All Possible Energy States (Canonical) $Z_{(1)}=\frac{1}{(1-e^{-\beta \hbar \omega})}$
    $Z_{(N)}=\frac{1}{(1-e^{-\beta \hbar \omega})^N}$
104 Solid with different oscillation frequency, all possible states (Canonical) $Z_{(1)} = \Pi_k(Z_k), Z_k = \frac{1}{1-e^{-\beta\hbar \omega_k}}$
    $F_{(1)}=-k_B Tln(Z_{(1)})$
    $U_{(1)}=-\frac{\partial ln(Z_{(1)})}{\partial \beta}$
105,106 Statistical Mechanical Limit for Thermodynamics $error=\frac{1}{\sqrt{N}}$, $N=$ number of particles
  calculated from $error = \frac{\sigma (stand. dev)}{\lambda (avg)}=\frac{\sqrt{\lang (E-U)^2 \rangle}}{\lang E \rang}$
107 Continuous Variable Partition Function $Z=\frac{1}{\hbar ^3}\int e^{-\epsilon/k_BT}dxdydzdp_xdp_ydp_z$
    $\epsilon$ is now continuous energy distribution
    $\lang Q\rang = \frac{1}{\hbar ^3}\frac{\int Qe^{-\epsilon/k_BT}dxdydzdp_xdp_ydp_z}{Z}$
108 Ideal Gas Using Continuous Variable Partition Function $Z_{(1)}=\frac{1}{\hbar^3}Vm^3\left(\frac{2\pi k_bT}{m}\right)^{3/2}$
109 Ideal Gas $U=\frac{3}{2}Nk_B T=\frac{3}{2}n_{mole}RT$
    $C_V = \frac{3}{2}Nk_B =\frac{3}{2} nR$
110 Maxwell’s Velocity Distribution $f(v_x,v_y,v_z)=\left( \frac{m}{2\pi k_bT} \right)^{3/2}e^\frac{-m(v_x^2+v_y^2+v_z^2)}{2k_BT}$
    $f(v_x)=\left( \frac{m}{2\pi k_bT} \right)^{1/2}e^\frac{-m(v_x^2)}{2k_BT}$
  where $v=\sqrt{v_x^2+v_y^2+v_z^2}$ $f(v)=\left( \frac{m}{2\pi k_bT} \right)^{3/2}e^\frac{-m(v^2)}{2k_BT}(4\pi v^2)$
111 Maxwell-Boltzmann most probable velocity $v_{mp}=\sqrt{\frac{2k_BT}{m}}$
    $\lang v\rang=\sqrt{\frac{8k_BT}{\pi m}}$
  All derive from Maxwell-Boltzmann Continuous Velocity Distribution (110) $v_{rms}=\sqrt{\lang v^2\rang}=\sqrt{\frac{3k_BT}{m}}$
112 Entropy of Ideal Gas $S=Nk_B\left[ln\left( \frac{V(2\pi mk_BT)^{3/2}}{\hbar^3} \right)+\frac{3}{2}\right]$
  Derived using 108, obtain $F$, then calculate $S=-\frac{\partial F}{\partial T}\vert _V$ or $S=Nk_B[lnV+\frac{3}{2}lnT+ln\left( \frac{(2\pi mk_B)^{3/2}}{\hbar^3} \right)+\frac{3}{2}]$
113 Correction of Indistinguishable Particles $Z_{(N)}=\frac{Z_{(1)}^N}{N!}$
  Entropy of Ideal Gas for Indistinguishable Particles $S=Nk_B\left[ln\left( \frac{V}{N}\frac{(2\pi mk_BT)^{3/2}}{\hbar^3} \right)+\frac{5}{2}\right]$
114 Chemical System, High Temperature Limit $\lang E\rang = \frac{1}{2}k_BT$
115 Linear Diatomic Molecule at Hight Temp $C_V = \frac{7}{2}R$
116 Polyatomic Molecule with $N$ atoms at High Temp $C_V = \frac{3}{2}R+\frac{3}{2}R+(3N-6)R$
  Polyatomic Molecule with $N$ atoms at Low Temp $C_V = \frac{3}{2}R$
117,118,119 Grand Canonical Ensemble, Probability of Finding a System Microstate $f_j=\frac{1\cdot \Omega(E_r, N_r)}{\Omega(E_t,N_t)}=\frac{e^{-\beta (E_j - \mu N_j)}}{e^{-\beta \Psi}}$
  Definition of chemical potential $\mu$ $\frac{\partial S_r}{\partial (N_t-N)}\equiv -\frac{\mu}{T}$
  Grand Canonical Ensemble, Gibbs Factor $e^{-\beta (E_j - \mu N_j)}$
  Grand Canonical Potential ($F$ equivalent) $\Psi = U-TS-\mu N$, $N$ is average number of particles
  Grand Canonical Partition Function $Z=e^{-\beta \Psi}$
120,121 Grand Canonical Ensemble, $U$ $U=\frac{\sum E_je^{-\beta (E_j - \mu N_j)}}{e^{-\beta \Psi}}=-\frac{\partial lnZ}{\partial \beta}\vert _{\gamma=\mu\beta}$
  Grand Canonical Ensemble, $\lang N\rang$ $\lang N\rang =\frac{\sum N_je^{-\beta (E_j - \mu N_j)}}{e^{-\beta \Psi}}=-\frac{\partial lnZ}{\partial \gamma}\vert _\beta$
122 Grand Canonical Ensemble, Chemical Potential for Ideal Gas $\mu = -T\frac{\partial S}{\partial N}\vert _{U,V} = -k_BTln\left( \frac{V}{N}\left( \frac{2\pi mk_BT}{\hbar^2} \right)^{3/2} \right)$
    $\mu=\frac{\partial G}{\partial N}\vert _{T,P}=\frac{G}{N}$
    $\mu=\frac{\partial F}{\partial N}\vert _{T,V}$
123 Modified Grand Canonical For Quantum Mechanics $f=\frac{e^{-\beta N_j (\epsilon_j -\mu)}}{Z}$
  Assumed $E_j = N_j \epsilon_j$  
124 Fermi-Dirac Statistics, Probability to find a fermion at state $\epsilon_i$ $f_{FD}=\frac{1}{e^{\beta (\epsilon_i - \mu)}+1}$
  Fermi-Dirac Stat Partition Function $Z_{FD}=1+e^{-\beta (\epsilon - \mu)}$
  Derived since each state can only have 0 or 1 (same) fermion  
125 Bose-Einstein Statistics, Probability to find a boson at state $\epsilon_i$ $f_{BE}=\frac{1}{e^{\beta (\epsilon_i - \mu)}-1}$
  Bose-Einstein Stat Partition Function $Z_{BE}=\frac{1}{1-e^{-\beta (\epsilon - \mu)}}$
  Derived since each state can have 0,1,2,…,$N$ (same) boson  
126 Fermi-Dirac and Bose-Einstein with Degeneracy $f_{FD}=\frac{g_i}{e^{\beta (\epsilon_i - \mu)}+1}$
    $f_{BE}=\frac{g_i}{e^{\beta (\epsilon_i - \mu)}-1}$
127 Number of fermions at state $\epsilon_i$ with Fermi-Dirac $N_i=N\cdot f_{FD}=\frac{Ng_i}{e^{\beta (\epsilon_i - \mu)}+1}$
  Number of bosons at state $\epsilon_i$ with Bose-Einstein $N_i=N\cdot f_{BE}=\frac{Ng_i}{e^{\beta (\epsilon_i - \mu)}-1}$
128 Continuous Density of “States” for Photons with $k=\frac{2\pi}{\lambda}$ $D(k)dk=2\cdot\frac{V}{2\pi^2}k^2dk$
  Continuous Density of “States” for Electrons with $E=\frac{p^2}{2m}$ $D(E)dE=\frac{V}{2\pi^2}\left(\frac{2m}{\hbar}\right)^{3/2}\sqrt{E}dE$
  And $D(k)dk=D(E)dE$  
129 Plank’s Black Body Law $d\bar{u}=\frac{dU}{V}=\frac{8\pi h}{c^3}\frac{\nu^3}{e^{h\nu /k_BT}-1}$, where $p=\frac{h\nu}{c}$
  Derived from continuous B-E statistics $\int\limits_0^\infty d\bar{u}\, d\nu=\sigma T^4$
130 Molecular Flux Using Maxwell-Boltzmann Velocity Distribution $\Phi = \frac{n\bar{v}}{4}=\frac{n}{4}\sqrt{\frac{8k_BT}{\pi m}}$
131 Collision Cross Section (where collision would happen to a molecule) $\sigma=\pi d^2$, $d$ should be given
132 Mean Free Path for a particle $\lambda_{mfp} = \frac{1}{n\sigma}=\tau\bar{v}$
  Collision frequency of a particle $\nu = \frac{1}{\tau} = n\sigma \bar{v}$
133 Fermi Energy $E_{fermi}=\mu\vert _{T=0}$
  (Maximum energy attainable by a fermion when $T=0$) $N\vert {T=0}=(\int_0^\infty D(E)f{FD}dE)\vert {T=0}=\int_0^{E{fermi}}D(E)dE$
134 Probability of a Molecule not Collided over a distance $x$ $e^{-\frac{x}{\lambda}}$
  $\lambda =$ mean free path  
135 Number of Molecules haven’t yet collided over a distance $x$ $N=N_0 e^{-\frac{x}{\lambda}}$
136 Number of Molecules collided per distance $dx$ $N(x)-N(x+dx)=\frac{N_0}{\lambda}e^{-\frac{x}{\lambda}}dx$
137 Diffusion Distance $x_{diff}=\vert \vert \sum \vec{x}i\vert \vert =\sqrt{N{collision}}\lambda=\sqrt{\frac{t}{\lambda}}\lambda$
    $x_{diff}^2=\frac{\lambda^2}{\tau}t=\frac{\bar{v}}{n \sigma}t$
  $N_{collision}$ means number of collision a molecule suffered  
138 Diffusion Equation, $D$ is diffusion coefficient $D\vec{\nabla}^2n = \frac{dn}{dt}$
  Molecular Flux $\Gamma = \Phi$ $\Gamma =-D\vec{\nabla}n$
139 Average distance above a surface so that Molecules can diffuse through $\lang y\rang=\frac{\int \lambda \cos(\theta)\Delta \phi_{\theta, \phi, v}}{\int \Delta \phi_{\theta, \phi, v}}=\frac{2}{3}\lambda$
140 Electrical conductance $\sigma_c = \frac{1}{\rho_{res}}$, $n_e$=density of electron $\sigma_c = e^2 \frac{n_e}{m_e}\tau_e$
  $\lambda$ depends on density of nucleus, not electron $\vec{J}=\sigma \vec{E}=e(n_e \vec{v}_d)$
141 Heat Flux $Q_{flux}$ $Q_{flux}=\Phi \cdot E_{\text{heat per molecule}}$
  Assuming $E_{\text{heat per molecule}}=c_v T$ $Q_{flux}=-\frac{n\bar{v}}{3}c_v \lambda \frac{dT}{dy}=-\kappa \vec{\nabla}T$
  Thermal conductivity $\kappa$ $\kappa=\frac{n\bar{v}}{3}c_v \lambda=\frac{1}{3}\frac{\bar{v}c_v}{\sigma}$
142 Momentum Flux $\vec{G}$ and Viscosity $\vec{G}=\Phi \cdot P_{\text{momentum per molecule}}=\frac{n \bar{v}}{4}mu_x$
  $\vec{\nabla}v$ is velocity gradient $\vec{G}=-\frac{1}{3}n \bar{v}m\lambda \frac{du_x}{dy}=-\eta \vec{\nabla}v$
  $\eta$ is coefficient of viscosity $\eta=\frac{1}{3}n \bar{v}m\lambda=\frac{1}{3}\frac{m\bar{v}}{\sigma}$
Equation Number Title/Condition Equations