Lecture 3

Question 1

the heat capacity at constant volume , Cv (formula)

Answer 1

Cv = lim(ΔT->0) (δQ/ΔT)v = (δU/δT)v

Question 2

the heat capacity at constant pressure, Cp (derivation)

Answer 2

dU = (δU/δT)p dT + (δU/δP)T dP

dV = (δV/δT)p dT + (δV/δP)T dP

substituting into first law

δQ = dU + PdV then taking a factor out of δ/δT)p

gives H = U + PV

with differential form

dH = d(U+PV) = dU + PdV + VdP = δQ + VdP

Cp = lim(ΔT->0) (δQ/ΔT)P = (δH/δT)p

Question 3

functions of state

Answer 3

U, P, V and H

Question 4

the enthalpy

Answer 4

H is an energy or a thermodynamic potential. Commonly used for isobaric conditions (dP = 0).

Question 5

specific enthalpy (derivation)

Answer 5

U2 - U1 = Q - Wd + P1V1 - P2V2

(U2 + P2V2) - (U1 + P1V1) = Q - Wd

ΔH = H2 - H1 = Q - Wd

(h1 - h2) dm/dt + dQ/dt = dWd/dt = Power

Question 6

Bernoulli’s equation (derivation for comparison)

Answer 6

ΔH = Q - Wd

including kinetic and potential energies

ΔH + Δ(εk + εp)bulk = Q - W

wd = h1 - h2 + 1/2(v1^2 -v2^2) + g(z1 - z2) + q

comparing with bernoulli’s eq

0 = 1/p (P1-P2) + 1/2(v1^2 - v2^2) + g(z1 - z2)

identical no work no heat and no change in internal energy