Thermodynamics

Zeroth Law of Thermodynamics

An isolated system will evolve to equilibrium.

1st Law of Thermodynamics

Conservation of energy. dU = 0 for isolated system.

Internal energy of a system increases by heat added and decreases by work done by the system.

dU = δQ-δW

Isochoric and relate to first law

Constant Volume process

Since dW = pdV, dW = 0, therefore dU = dQ

Isobaric and relate to first law

Constant Pressure process

dU = δQ - δW and dQ = dU + pdV = d(U+pV) = dH (enthalpy)

Adiabatic and relate to first law

δQ = 0 so dU = -δW

Enthalpy

isobaric: dQ = dU + pdV = d(U+pV) = dH

H = U +pV

Specific Heat

C is the ratio of heat add to a system and the corrisponding change in temperature. C = dQ/dT

C_v = (dQ/dT)_v = (dU/dT)_v

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

Isothermal

Temperature is constant. dU = 0; dQ = dW

p = nRT/V = const/V

W = ∫pdV = ∫nRT/V dV = nRT*ln(V2/V1)

Second Law of Thermodynamics

States that the entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium -- the state of maximum entropy.

The total entropy of any system will not decrease other than by increasing the entropy of some other system.

Thomson's postulate, Clausius's postulate

Entropy

The measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.

dS ≥ δQ/T

= for reversible, > for irreversible.

if reversible and adiabatic dS = 0, so Entropy stays constant during this process.

Entropy and 1st law and 2nd law

TdS ≥ dU + dW

and dW ≤ -dU + TdS ≤ -d(U - TS)

Ideal Gas: law, internal energy equ, Cp and Cr relation, isothermal Work, isobaric Work, adiabatic Work

pV = nRT

(dU/dV)_t = 0 so dU = dQ = n*C_v*dT

C_p = C_v + R

W_t = nRT*ln(V2/V1)

W_p = p∆V = nR∆T

W_q = -∆U = -n ∫ C_v*dT

Ideal gas: γ, and adiabatic Work

γ = Cp/Cv

C=pV^γ

W_q = (pV - pV) / (γ-1)

Hess's Law

Regardless of the steps in a reaction, total enthalpy change encompasses the sum of all changes. ∆H is a state function that only needs the initial and final conditions.

v_i

stoichiometric coefficients

Standard Enthalpy of formation

∆H⁰₂₉₈ = ∑v_i*∆H⁰₂₉₈(i) - ∑v_i*∆H⁰₂₉₈(i)

Standard Enthalpy of reaction

∆H⁰_T = ∆H⁰₂₉₈ + ∫₂₉₈ ∆Cp*dT

Enthalpy, H, of an Ideal gas

and heat of reaction at constant pressure

H = U + pV = U + nRT

∆H⁰_T = ∆U⁰_T + ∆v_gas*RT

Draw Heat engine P-V diagram

good job

Efficiency of a heat engine and why.

η = W / |Q_h| = (|Q_h| - |Q_c|) / |Q_h|

dU of a cycle equals 0 because its a state function,

therefore dQ = dW or W = |Q_h| - |Q_c| and W cannot be greater then dQ so W cannot be greater than Q_h.

result: η ≤ 1

Thomson's postulate

There cannot be a closed path through p_V space where all Q_h results in Work.

Clausius's postulate

Heat cannot flow spontaneously from a cooler to a hotter system

Reversible process

A process that can go from point A to B and back to A with no change in the universe.

Carnot Cycle

A reversible process in which the system is alternately in contact with two heat reservoirs at constant temperature. paths are isothermal, adiabatic, isothermal, adiabatic.

The entire change in internal energy of the carnot cycle is 0.

W = W₁₂ +W₂₃ + W₃₄ + W₄₁ = |Q_h| - |Q_c|

Carnot Efficiency

since |Q_h| / T_h = |Q_c| / T_c

so η = 1 - T_c / T_h

Second Carnot theorem

the efficiency of any machine cannot be larger then the efficiency of the Carnot machine with same heater and cooler

Entropy: isobaric, isothermal, ∆S_phase transition , isothermal and ideal gas

dS = ∫Cp*dT/T ; ∆S = ∆Q/T ; ∆S_p.t. = ;

dU = 0 so ∆S =∫pdV/T = nR*ln(V/V) = nR*ln(p_1/p_2)

Entropy of mixing two gases.

∆S = n₁R*ln(V/V₁) + n₂R*ln(V/V₂)

and V_i / V = n_i / n = x_i (molar fraction)

so ∆S = -nR [x₁ln(x₁) + x₂ln(x₂)]

Third law of thermodynamics

The heat capacity of any substance goes to 0 as T→0 K. The entropy of equilibrium of any system will got to 0 as T→0 K. @ T=0 K, all processes occur with no change of entropy.

Helmholtz free energy

δW ≤ -(dU - TdS) ≤ -d(U-TS)

and U-TS = A....

so δW ≤ -dA

and δW_nonmechanical ≤ -dA - pdV

Gibbs free energy

δW_nonmechanical ≤ -dA - pdV ≤ -d(A + pV) ≤ -dG

G = A + pV = U - TdS + pV = H - TS

Will a spontaneous process occur, and work

If W_nonmechanical > W_mechanical then it will occur.

ie. W_nonmechanical ≥ 0

or dA ≤ 0

dG ≤ 0

Free energy Conditions of equilibrium.

dA = 0 at const T,V

dG = 0 at const T,P

Entropy and Helmholtz and Gibbs free energies

dH = TdS + Vdp

dA = -SdT - pdV

dG = -SdT - Vdp

S = -(dG/dT)_p

V = (dG/dp)_T

[d(G/T)/dT]_p = -H/T²

dependence of Gibbs free energy on pressure for an ideal gas

G_T = G⁰_T + nRT*ln(p)

Internal energy of an open system and Chemical Potential. Define chemical potential.

dU = TdS - pdV + ∑ µ_i * dn_i

µ_i = (dU/dn_i)_S,V,n

Chemical potential is the work necessary to remove one mole of component i from the system.

H, G, and A with respect to chemical potential.

dH = TdS + Vdp + ∑ µ_i * dn_i

dA = -SdT - pdV + ∑ µ_i * dn_i

dG = -SdT + Vdp + ∑ µ_i * dn_i

Chemical potential of pure materials.

µ = G/n

Chemical potential of component i in an ideal gas mixture.

µ_i = µ⁰_i + RTln(p_i)

µ⁰_i is the chemical potential of pure i at given temp

p_i is the partial pressure of component i

dimensions of err thing

Q

U

W

S

H

A

G

µ

Conditions for equilibrium.

T₁ = T₂

p₁ = p₂

µ₁ = µ₂ (chemical equilibrium)

Triple point

existence of all three phases

Critical Point

End of liquid-vapor line where where the liquid and vapor have same structural properties. Beyond is single phase region.

Degrees of Freedom, F

Number of independent parameters that must be set to fully determine the state of a system in equilibrium.

Number of components, C, of a system

The number of substances minus the number of independent equations describing the chemical reactions relating the substances. ie CO, CO₂, O₂, Fe, FeO (5)

2Fe + O₂ → 2FeO ,

FeO + CO → Fe + CO₂,

so C = 5 - 2 = 3

Gibbs Phase Rule

F = C + 2 - P (the 2 is for temp and pressure)

& F ≥ 0

Clausius-Clapeyron equation for phase transition

dp/dT = ∆S/∆V

dp/dT = ∆H_p.t. / T∆V

ideal gas d(ln(p))/dT = dH/RT²

Solution

Homogeneous mixture of two or more components.

Molar Fraction

x_i = n_i / n

Weight fraction

ratio of mass's

[i] = w_i / w

[%w] = 100*[i]

Molarity

c_i = n_i / V

#moles per liter of solution

Molality

m_i = n_i / (1kg of solvent)

Partial molar values of Extensive properties(F)

(with bar over the top)F_i = dF/dn_i

Excess values or ∆F_mix

F^M = F - ∑n_i * F⁰_i

Relative partial molar value

(with bar over top)(F_i)^M = d∆F_mix / dn_i

= (with bar over top)F_i - F⁰_i

Gibbs-Duhem relation: general and binary solu.

∑[(dF_i(bar) / dn_j) *n_i] = 0

in binary solution: x₁(dF₁(bar) / dx₁) = x₂(dF₂(bar) / x₂) @constant temp, press

Ideal Dilute Solutions

Solution where the solvent molar fraction approaches unity.

x₁→1

Henry's law and constant

In ideal dilute solution the saturated vapor pressure of the solute is proportional to the solute concentration in the solution. p₂ = h*c₂ and h = k↑ / k↓.

Sieverts law and how different from Henry's

Applies to diatomic gases in metals.

p₂ = s*c₂^2

ie. p_N₂ = s*c_[N] ^2

where c_[N] is concentration of nitrogen ATOMS (not N₂)

Condition for vapor in equilibrium with solution.

µ₂ = µ₂_vapor

Chemical potential of solute vapor in a solution.

µ = µ⁰ + RT*ln(p)

so µ₂ = µ⁰₂_vapor + RT*ln(p₂)

= µ⁰₂_vapor +RT*lnh +RT*lnc₂

-or- phi₂ + RT*lnc₂

phi₂ is a concentration independent constant

Standard state of the Solute

if c₂ = 1 then RT*ln(c₂) = 0 and µ₂ = phi₂

phi₂ = µ⁰₂_vapor +RT*ln(h)

Chemical potential of solute of vapor in an IDEAL DILUTE solution.

µ₂ = phi_x + RT*lnx₂

Chemical potential of solvent

µ₁ = µ⁰₁ + RT*lnx₁

Raoult's law

p₁ = p⁰₁*x₁

The partial pressure of a solvent in the saturated vapor above a solution is proportional to its molar fraction in the solution.

Ideal dilute solution, partial molar:

entropy, enthalpy, and volume

S₁(bar) = S⁰₁ - R*lnx₁

S₂(bar) = S₂(dot) - R*lnc₂ (like µ₂)

H₁(bar) = H⁰₁

H₂(bar) = H₂(dot)

V₁(bar) = V⁰₁

V₂(bar) = V₂(dot)