Statistical Physics: The Laws of Thermodynamics (unfinished)

What is a thermodynamic system

Any microscopic system, often consider N particles where N ∼ Na ∼ 6×10²³ (avorgos number.)

What are equations of state?

Relationships between thermodynamic paramaters for a system in equilibrium.

What is the general equation for the work done on a system?

dW = (intensive)d(extensive)

What is an extensive variable? What is an intensive variable?

Extensive: A variable which scales with the system size

Intensive: A variable which doesn’t scale with the system size.

What is the ideal gas equation?

PV = NRT

What is the zeroth law in thermodynamics? What does this law imply about the system?

If two systems A and B are in equilibrium with a system C, then they are in equilibrium with each other.

This law implies that there will always exist a state variable such that systems in equilibrium will have the same T.

If we considered two systems A and B being brought into equilibrium, what relationship is satisfied and what does it mean?

Ⓗ_a (A1, A2, …, ) = Ⓗ_b (B1, B2, …,)

This implies there exists some f₀ of the thermodynamic variable that characterises the system.

(Derivation in notes)

What is the first law of thermodynamics? How do we write this law if its an infinitesimal process? 

ΔE = ΔQ-ΔW 

For an infinitesimal process we write it as: 𝑑E = 𝑑Q + 𝑑W

What is the formula for the Heat capacity at constant volume?

(dQ/dT)_v = (dE/dT)v = Cv

Derive the Heat Capacity at constant volume from the first law

1. Set up system: We consider a gas with equations of state. E = E(P,V)

2. Partial Derivative: dQ = (dE/dP)_v dP + (dE/dV)_p dV

3. Consider other state variable: Now we consider E = E(T,V)

4. Partial Derivative: dQ = (dE/dT)_v dT + [(dE/dV)_T + P]dV

5. Solve: (dQ/dT)_v = (dE/dT)v = Cv

What is the formula for the Heat capacity at constant pressure?

(dQ/dT)_p = (dH/dT)_p = Cp

Derive the Heat Capacity at constant pressure the first law

1. Set up system: We consider a gas with equations of state. E = E(P,V)

2. Partial Derivative: dQ = (dE/dP)_v dP + (dE/dV)_p dV

3. Consider other state variable: Now we consider E = E(T,P)

4. Partial Derivative: dQ = [(dE/dT)_v + P(dV/dT)_p] dT + [(dE/dP)_T + P(dV/dP)_T]dP 

5. Solve using Enthalpy: H = E + PV => (dQ/dT)_v = (dE/dT)v = Cp

What is the second law of thermodynamics? If B is reversiably accessible from A i.e. Qrev, what does this imply?

There is an extensive thermodynamic variable S(E,..,V,…,), the ebtropy and if state B is adibatically accessible from A then S_B ≥ S_A.

If B is reversiably accessible from A, then S_b = S_a for reversible adiabatic processes. 

Using the second law, how can we show that two bodies in equilibrium have the same temperature

1. We consider the change in entropy dS

2. Let dS = 0 as dQrev = 0

3. T ≡ (dE/dS)_x ≥ 0, obtain maxwell equation for internal energy

4. dE₁ = dE₂ ≠ 0 as E and X are fixed.

5. Solve for dS, everything cancels

6. Therefore T1 = T2 holds for all dE.

(Full derivation in notes… its long!)

Based on the second law, if T1 > T2, what does this imply about dE?

T₁ > T₂ => dE < 0 

Based on the second law, if T1 < T2, what does this imply about dE?

T₁ < T₂ => dE > 0