Stat Mech Flashcards
What does classical thermodynamics describe?
Macroscopic systems ~in equilibrium in terms of a few measurable variables
What are state variables?
Variables that describe the current state of a system. Importantly, they do not depend on the history of the system.
Includes temperature, pressure, volume, composition, internal energy, and entropy.
Cannot all be varied independently, equations linking them are equations of state.
What is quasi-equilibrium?
Changes to conditions (eg pressure) can be made sufficiently slowly that the system remains in equilibrium at all times.
What is a reversible process?
A process where an infinitesimal change in the external conditions is enough to reverse the direction of the process;
e.g. compression of a gas in a cylinder with a frictionless piston by exerting an external force only just sufficient to move the piston one way or the other.
What is a macrostate?
The state of a macroscopic system in equilibrium specified by a handful of macroscopically-manipulable variables. It completely ignores what is going on with the individual atoms that comprise the system.
What is a microstate?
A description of the positions and momenta of all atoms (or of their combined quantum state) of a system.
The Zeroth Law of Thermodymamics
If two bodies are separately in thermal equilibrium with a third body, they are also in thermal equilibriums with one another. All three are then said to be at the same temperature.
Absolute temperature
Measured in Kelvin, defined to be 273.16 K at the triple point of water:
T = lim(P→0) PV/(PV)ₜᵣᵢₚₗₑ × 273.16 K
Low pressure limit taken so real gases approach ideal behaviour.
Zero of the Kelvin scale is absolute zero. At this temperature, pressure of an ideal gas would vanish, because (classically) the motion of its molecules would cease.
First Law of Thermodynamics
Any change in internal energy, E, of a system is due to the amount of heat added to it and the work done on it.
∆E = Q + W
dE = đQ + đW
Energy and work in a reversible process.
dE = đQʳᵉᵛ + đWʳᵉᵛ
Adiabatic processes
No heat transfer
Q = 0
System in a cycle
Returns to its initial state
∆E = 0
Work done in reversible processes
Compression of a fluid:
đWʳᵉᵛ = −P dV
Stretching a wire of tension Γ by dl:
đWʳᵉᵛ = Γ dl
Increasing the area of a film of surface tension γ by dA:
đWʳᵉᵛ = γ dA
Increasing the magnetic field B imposed upon a paramagnetic sample:
đWʳᵉᵛ = −m · dB = −VM · dB
where M is magnetisation per unit volume, and m is total magnetic moment of the sample.
To calculate the internal energy change for irreversible processes (such as free expansion of a gas), it is necessary to find a reversible process linking the same initial and final states of the system.
Second law of thermodynamics
No process which would decrease the entropy of a system will happen spontaneously.
OR
If left alone, systems always head towards the same equilibrium as determined by external constraints.
OR
In an isolated system, entropy can only increase (or stay the same)
Kelvin and Planck statement
“It is impossible to construct an engine which, operating in a cycle, will produce no other effect than the extraction of heat from a reservoir and the performance of an equivalent amount of work”
Clausius statement
“It is impossible to construct an refrigerator which, operating in a cycle, will produce no other effect than the transfer of heat from a cooler body to a hotter one.”
Entropy
dS = đQʳᵉᵛ/T
For an isolated system, during any spontaneous change
dS ≥ 0
Fundamental thermodynamic
relation
dE = TdS − PdV
(dE = đQʳᵉᵛ + đWʳᵉᵛ)
E : energy
T : absolute temperature
S : entropy
P : pressure
V : volume
Carnot engine
Qₕ is removed from a hot reservoir and Q꜀ is discarded to a cold one in such a way that the combined entropy change is zero.
Qₕ/Tₕ = Q꜀/T꜀
The difference is available to do work
W = Qₕ − Q꜀
From the fundamental thermodynamic relation, derive functions for fixed properties.
dE = TdS − PdV
⇒E(S, V) with
(∂E/∂S)|ᵥ = T and (∂E/∂V)|ₛ = −P
dS = 1/T dE + P/T dV
⇒S(E, V) with
(∂S/∂E)|ᵥ = 1/T and (∂S/∂V)|ₑ = P/T
Helmholtz free energy
System in contact with heat bath, spontaneous change at constant volume, Q absorbed.
∆E = Q.
System starts and ends at temperature T. Total change in entropy has two parts, ∆S for system and −Q/T for surroundings.
∆Sₜₒₜ = ∆S − Q/T = 1/T (T∆S − ∆E) ≥ 0
⇒ ∆(TS − E) ≥ 0
In approach to equilibrium, quantity TS − E is maximised.
Helmholtz free energy, F, is negative of this, so it is minimised.
F = E − TS
⇒ dF = −SdT − PdV
(∂F/∂T)|ᵥ = −S and (∂F/∂V)|ₜ = −P
Gibbs free energy
Approach to equilibrium minimises Gibbs free energy, G, of a system.
G = E − TS + PV
⇒ dG = −SdT + VdP
(∂G/∂T)|ₚ = −S and (∂G/∂P)|ₜ = V
Gibbs free energy and chemical potential
µ = (∂G/∂N)|ₜ,ₚ
G = ΣμᵢNᵢ
μᵢ is chemical potential of each species
Nᵢ is number of molecules of each species
Entropy for an ideal gas as a function of temperature and pressure
S(T, P) = S(T₀, P₀) + Cₚ ln(T/T₀) − nRln(P/P₀)