A Summary of Thermodynamic Processes

Thermodynamic Processes

Calorimetry

Change of phase?Type of heatEquationTemperature

CalculationChange?

Nospecific heatdQ = mCdT [1]Yes

Yeslatent heatdQ = mL No

[1] Caution! Be careful of molar versus mass based specific heat constants.

Ideal Gas Law

PV = nRT,

P => pressure in Pascals (N/m2)

V => volume in m3

n => number of moles (dimensionless)

R => gas constant

T => temperature in Kelvin (not Celsius!)

Other Key Equations

dU = dQ – dW(first law of thermodynamics)

dQ = nCVT(ideal gas, specific heat at constant volume)

dQ = nCPT(ideal gas, specific heat at constant pressure)

dU = nCVT(ideal gas, derivation attached)

CP – CV = R(statistical mechanics)

Internal Energy of an Ideal Gas

The internal energy depends only on the endpoints. Pick a constant volume and constant pressure line segments to connect the endpoints. Using the first law:

U = nCV(T’ – T0) + nCP(Tf – T’) – 0 – Pf(Vf-V0) = nCV(Tf – T0), since

PfV0 = nRT’
Laws of Thermodynamics for Ideal Gases

processmeaningwork (W)heat (Q)entropy (S)

isobaricconstant pressureP0(VF – V0)nCP(TF – T0)nCP ln(TF/T0)

isochoricconstant volume0nCV(TF – T0)nCV ln(TF/T0)

isothermalconstant temperature(nRT0)ln(VF/V0)[1](nR)ln(VF/V0)

adiabatic [2]no heat exchange(PFVF- P0V0)/(1 - ) 00

[3]

[1]From the first law of thermodynamics, dU = dQ – dW; dU=0 for an isothermal process, so dQ = dW, or Q = W

[2]From the first law of thermodynamics, dU = dQ – dW; dQ=0 for an adiabatic process, so dU = -dW => nCVdT = -PdV;

from the ideal gas law, dT = d(PV/nR), so

n(CV/nR) d(PV) = -PdV

n(CV/nR) [PdV + VdP] = -PdV,

VdP = -PdV [1 + (CV/R)] / (CV/R);since R = CP - CV,  CP / CV

VdP = -PdV , or

P/P0 = (V/V0)-,

orPV = P0V0.

[3]dW = PdV, so W =  (P0V0)V-dV = (P0V0)V1-/(1 - ), V  [V0,VF]

W = (P0V0)V1-/(1 - ) = (PFVF- P0V0)/(1 - )

Examples follow

(1) a simple example

(2) Carnot cycle

(3) Otto cycle

(4) Diesel cycle

(5) Stirling cycle
Example 1: A Simple Example

Heat calculations:Work calculations:

Q12 = 8(CP/R)P0V0W12 = 8P0V0

Q23 = -9(CV/R)P0V0W23 = W41 = 0

Q34 = -2(CP/R)P0V0W34 = -2P0V0

Q41 = 3(CV/R)P0V0

Entropy calculations:Sums:

S12 = nCPln(3)Q = W = 6P0V0

S23 = -nCVln(4)U = Q - W = 0 (expected, closed cycle)

S34 = -nCPln(3)S = 0 (reversible process)

S41 = nCVln(4)

efficiency:

QH = Q12 + Q41 = (8CP + 3CV)P0V0/R(sum of positive heat results)

QC = |Q23 + Q34| = (2CP + 9CV)P0V0/R(sum of negative heat results)

e = 1 - (2CP + 9CV)/(8CP + 3CV) = 1 - (2 + 9)/(8 + 3);

for a monatomic gas,  = 5/3 and e = 0.24

Carnot efficiency:

TC = T4 = P0V0/(nR)

TH = T2 = 12P0V0/(nR)

e = 1 - TC/TH = 0.92 (notice that the actual efficiency is much lower)
Example 2: Carnot Cycle

STATET

aTH

bTH

cTC

dTC

______

STEPTYPEQWUS

a->bisothermalnRTHln(Vb/Va)Q0nRln(Vb/Va)

b->cadiabatic0UnCV(TC - TH)0

c->disothermalnRTCln(Vd/Vc)Q0nRln(Vd/Vc)

d->aadiabatic0UnCV(TH - TC)0

______

efficiency:

Qab = nRTHln(Vb/Va) > 0

Qcd = -nRTCln(Vd/Vc) < 0

|QC| / |QH | = (TC/TH)[| ln(Vd/Vc)/ln(Vb/Va) |]

TbVb-1 = TcVc-1|

| (adiabatic)=> Vb/Va = Vc/Vd=> |QC| / |QH | = (TC/TH)

TdVd-1 = TaVa-1|

e = 1 - (TC/TH)

entropy:

S = 0, see efficiency calculation. Reversible process.

Example 3: Otto Cycle

STATEPVT

aPaVa = rVbTa

bPb = Pa rVbTb = Ta r-1

cPc = Pb(Tc/Tb)VbTc = Td r-1

dPd = Pc(1/r)Va = rVbTd

______

STEPTYPEQWUS

a->badiabatic0nCV(Tb – Ta)-W0

b->cisochoricnCV(Tc – Tb)0QnCVln(Tc/Tb)

c->dadiabatic0nCV(Td – Tc)-W0

d->aisochoric nCV(Ta – Td)0QnCVln(Ta/Td)______

efficiency:

Qbc = nCV(Tc – Tb) > 0

Qcd = nCV(Ta – Td) < 0

|QC| / |QH | = (Td – Ta) / (Tc – Tb) = (Td – Ta)/[r-1(Td – Ta)]

= 1/r-1, or e = 1 - 1/r-1

NOTE:Tc > Tb > Ta (since Pc>Pb);Td/Ta = Tc/Tb > 1 => Td > Ta;

so that Tc = TH and Ta = TCOLD; using these temperatures,

the Carnot efficiency is e = 1 – (1/r-1)(Ta/Td) > Otto efficiency

entropy:

S = nCVln(Tc/Tb) + nCVln(Ta/Td)

= nCVln[(Tc/Tb)(Ta/Td)]

= nCVln[(Td/Ta)(Ta/Td)]

= nCVln (1) = 0

S = 0. Reversible process.

Example 4: Diesel Cycle

STATEPVT

aPaVa = rVbTa

bPb = Pa rVbTb = Ta r-1

cPbVc= rcVbTc = (Vc/Vb)Tb= (Vc/Vb)Ta r-1

dPd = Pa (Vc/Vb)Va = rVbTd= (Vc/Vb)(1/r)-1Tb= (Vc/Vb)Ta

______

STEPTYPEQWUS

a->badiabatic0nCV(Tb – Ta)-W0

b->cisobaricnCP(Tc – Tb)0QnCPln(Tc/Tb)

c->dadiabatic0nCV(Td – Tc)-W0

d->aisochoric nCV(Ta – Td)0QnCVln(Ta/Td)______

efficiency:

Qbc = nCP(Tc – Tb) > 0

Qcd = nCV(Ta – Td) < 0

|QC| / |QH | = (1/) (Td – Ta)/ (Tc – Tb)

= (1/) [(Vc/Vb)– 1)Ta] / [ (Vc/Vb) – 1]Tb

= (1/) [(Vc/Vb)– 1)] / [ (Vc/Vb) – 1](1/r-1)

e = 1 –{[rc– 1] / [rc– 1]} (1/r-1)

NOTE: this is indeterminate for rc= 1;

the efficiency at this point is 1 – (-1)/(r-1) => 1 – 2/(5r2/3) for monatomic

NOTE: for rc > 1, e -> 1 – (1/)(rc/r)-1=> 1 – 3(r/rc)-2/3 for monatomic

entropy:

S = nCPln(Tc/Tb) + nCVln(Ta/Td)

= nCV [ ln(Tc/Tb) + ln(Ta/Td)]

= nCV ln(Tc/Tb)(Ta/Td)

= nCV ln(Vc/Vb)(Vb/Vc)

= nCV ln(1)= 0

S = 0. Reversible process.

Example 5: Stirling Cycle

STATEPVT

aPaVa = rVbTC

bPbVbTC

cPcVbTH

dPdVa= rVbTH

______

STEPTYPEQWUS

a->bisothermalW-nRTCln(r)0nRln(Vb/Va)

b->cisochoricnCV(TH – TC)0QnCVln(Tc/Tb)

c->disothermalWnRTH ln(r)0nRln(Vd/Vc)

d->aisochoric -nCV(TH – TC)0QnCVln(Ta/Td) ______

efficiency:

Qcd = nRTH ln(r) + nCV(TH – TC) > 0

Qab= -nRTC ln(r) - nCV(TH – TC) < 0

|QC| / |QH | = [TH ln(r) + (CV/R)(TH – TC)] / [TC ln(r) + (CV/R)(TH – TC)]

= [TH (ln(r) + CV/R) – (CV/R))TC)] / [TC (ln(r) - CV/R) + (CV/R)TH]

e = 1 - 1/(r-1)

entropy:

S = nRln(Vb/Va) + nRln(Vd/Vc) + nCVln(Tc/Tb) + nCVln(Ta/Td)

= nRln[(Vb/Va)(Vd/Vc)] + nCVln[(Tc/Tb)(Ta/Td)]

= nRln[(1/r)(r)] + nCVln[(TH/TC)(TC/TH)]

= nRln(1) + nCVln(1

S = 0. Reversible process.