Test 3 study notes

Notes to study for the third test of this course

Definitions

• Latent heat: Heat required to convert a solid into liquid or liquid into gas without a change in temperature.
• Macrostate: A state of the system specified by macroscopic quantities
• Microstate: An exact microscopic description of the system, for example, a list of the positions and velocities of all particles in a gas.
• Multiplicity: $\Omega$ is the number of microstates per macrostate
  • $S = k \ln\Omega$
• Fundamental Assumption of Statistical Mechanics: In an isolated system with fixed internal energy $U$, fixed volume $V$, and fixed number of atoms $N$, then all microstates are equally likely.
  • This imples that an isolated system always evolve towards a macrostate with the largest multiplicity
• Quasistatic: Keeping the system in equilibrium with itself as the process happens
  • This is the same as keeping the system close to maximum entropy for a given macrostate at all times
  • Is generally slow and careful
• Temperature: Part of system that tells how much heat it is throwing off.
  • Is related to how much entropy it loses when it throws off heat
  • $T = \left(\frac{\partial S}{\partial U}|_{V,N}\right)^{-1} = \frac{\partial U}{\partial S}|_{V,N}$
• Pressure: Related to how much entropy system gains when volume is increased
  • $P = T\frac{\partial S}{\partial V}|_{U,N}$
• Thermodynamic identity: $dU = -PdV + TdS$
• Hemlholtz free energy: $F = U - TS$
  • Extensive state function
  • For reversible processes, $Q = T\Delta S$
    • F must stay the same if process is reversible
  • For irreversible processes, $Q < T\Delta S$ since new entropy is created spontaneously
    • F must decrease if the process is irreversible
    • Implies that increase in free energy F cannot be larger than work done on system
  • $\Delta F = \Delta U - T\Delta S$
  • $\Delta U = Q + W$
  • $\Delta F = Q + W - T\Delta S$
• Enthalpy: $H = U + PV$
  • $\Delta H = \Delta U + \Delta (PV)$
  • $\Delta H = Q - W + \Delta P V + P\Delta V$
  • $\Delta H = Q - P \Delta V + \Delta P V + P\Delta V$
  • $\Delta H = Q + \Delta P V$
• Gibb's free energy: $G = H - TS$
  • $G = U + PV - TS$
  • $G = F + PV$
  • Phase with lower gibb's free energy is more stable
• Chemical Potential: Chemical potential $\mu \equiv -T \frac{\partial S}{\partial N} |_{U,V}$
  • Particles flow from high potential to low potential
  • $\mu$ is an intensive quantity
  • Relevant when N is not constant
  • $G = N\mu$
• Clausius-Clapeyron equation: $\frac{dP}{dT} = \frac{L}{T\Delta V}$
  • Can predict change in melting pressure when applying extra pressure if we know volume and latent heat of melting.
    • When one of the substance is gas, a useful approximation is $\Delta V \approx V(gas) \approx \frac{RT}{P}$
      • Further, assuming that $L$ is a constant, we get that $\frac{dP}{dT} = \frac{LP}{RT^2}$
  • Vapour equation: $T - T_0 = \frac{RT_0^2}{P_0L}(P-P_0)$
    • Derivation:
      • $P = \text{const. }e^{-L/RT} = P_0e^{L/RT_0 - L/RT}$
      • $P = P_0e^{L(T-T_0)/RT_0T} \approx P_0e^{L(T-T_0)/RT_0^2}\approx P_0+\frac{P_0L}{RT_0^2}(T-T_0)$
    • Partial pressures for each component are proportional to number of particles of that component
      • Ex: $\frac{P_{\text{partial}}(H_2O)}{P_{\text{total}}} =\frac{N(H_2O)}{N_{\text{total}}}$
      • At 39 degrees celcius and 0.07atm, at 100% humidity, air is about 7% vapor by volume

Heat engines

• Carnot cycle efficiency $e \equiv \frac{\text{desirable output}}{\text{cost}} = \frac{W}{Q_h} = \frac{Q_h - Q_c}{Q_h}$ where $Q_h$ is $Q_{in}$ and $Q_c$is $Q_{out}$
  • Requires that the cycle must be reversible and involves no change in entropy
• efficiency has a bound: $e \leq 1- \frac{T_c^{\text{reservoir}}}{T_h^{\text{reservoir}}}$
• efficiency max: $e_{max} = e_{carnot} = 1 - \frac{T_c}{T_h}$
• First law (conservation of energy) states that $e\leq 1$ (cannot win)
• Second law states that $e \leq 1-\frac{T_c}{T_h}$ (cannot break even)

if the whole process is reversible, efficiency: $e = 1 - \frac{T_c^{\text{reservoir}}}{T_h^{\text{reservoir}}}$ exactly.

Refrigerators

A refrigerator moves heat $Q_c$ from the cold reservoir and heat $Q_h$ to the hot reservoir. The entropy of the universe changes like $Q_h/T_h^{\text{reservoir}} - Q_c/T_c^{\text{reservoir}}$

Instead of efficiency, we use the coefficient of performance (COP)
$\text{COP} \equiv \frac{\text{desirable output}}{\text{the cost}} = \frac{Q_c}{W}$

where $W + Q_{\text{in}} = Q_{\text{out}}$, $Q_{\text{in}} = Q_c$, $Q_{\text{out}} = Q_h$

$\text{COP}_{\text{max}} = \frac{T_c}{T_h - T_c}$ for refrigerators

Throttling

A throttling process is a process where there is no enthalpy change between states, no work is done, and the process is adiabatic. It is fundamentally irreversible

Ex. 1kg of liquid water at its boiling point at 1atm is throttled so that its temperature is lowered to $20^o\text{C}$. What fraction of water is vaporized?

Ans: Knowing that enthalpy of water $H_{water} = 419$ at 100 degrees, then the enthalpy must be the same since it is a throttled process. Let $x$ be the unknown vaporized water. $H_{water} = (1-x)H_{water} + H_{steam}$

Battery voltage

Given $\Delta G$ and $\Delta H$, one can easily calculate the battery voltage with:

$Voltage = \frac{work}{charge} = \frac{-\Delta G}{\# electrons \times e \times N_A}$

where $N_A$ is avagadro's number and e is the charge of an electron

Reversible batteries

For reversible operation of batteries in the base case scenario, $Q = T\Delta S = \Delta H - \Delta G$ where a $+$ sign for $Q$ means heat flows into the battery.

Relationships of formulas

$H = U + PV$
$G = F + PV$
$F = U - TS$
$G = H - TS$

Thermodynamic identities

$dG = -SdT + VdP + \mu dN$	$G = G(T,P,N)$
$dU = TdS - PdV + \mu dN$	$U = U(S,V,N)$
$dF = -SdT - PdV + \mu dN$	$F = F(T,V,N)$
$dH = TdS + VdP + \mu dN$	$H = H(S,P,N)$

Fundamental theorems

$\Delta S_{\text{universe}} \ge 0$

For ideal gas:

$H = U + PV = \frac{f}{2}NkT + NkT = \frac{f+2}{2}NkT$