Thermodynamics and Chemistry

CHAPTER 4 THE SECOND LAW

4.3 CONCEPTSDEVELOPED WITHCARNOTENGINES 112

Thus, the efficiency of a Carnot engine must depend only on the values ofTcandTh
and not on the properties of the working substance. Since the efficiency is given byD
1 Cqc=qh, the ratioqc=qhmust be a unique function ofTcandThonly. To find this function
for temperatures on the ideal-gas temperature scale, it is simplest to choose as the working
substance an ideal gas.
An ideal gas has the equation of statepV D nRT. Its internal energy change in a
closed system is given by dU DCVdT (Eq.3.5.3), whereCV (a function only ofT) is
the heat capacity at constant volume. Reversible expansion work is given by∂wDpdV,
which for an ideal gas becomes∂wD .nRT=V /dV. Substituting these expressions for
dUand∂win the first law, dUD∂qC∂w, and solving for∂q, we obtain

∂qDCVdTC

nRT

V

dV (4.3.4)

(ideal gas, reversible

expansion work only)

Dividing both sides byTgives

∂q

T

D

CVdT

T

CnR

dV

V

(4.3.5)

(ideal gas, reversible

expansion work only)

In the two adiabatic steps of the Carnot cycle,∂qis zero. We obtain a relation among the
volumes of the four labeled states shown in Fig.4.3by integrating Eq.4.3.5over these steps
and setting the integrals equal to zero:

Path B!C:

Z

∂q

T

D

ZTc

Th

CVdT

T

CnRln

VC

VB

D 0 (4.3.6)

Path D!A:

Z

∂q

T

D

ZTh

Tc

CVdT

T

CnRln

VA

VD

D 0 (4.3.7)

Adding these two equations (the integrals shown with limits cancel) gives the relation

nRln

VAVC

VBVD

D 0 (4.3.8)

which we can rearrange to

ln.VB=VA/Dln.VD=VC/ (4.3.9)

(ideal gas, Carnot cycle)

We obtain expressions for the heat in the two isothermal steps by integrating Eq.4.3.4with
dTset equal to 0.

Path A!BW qhDnRThln.VB=VA/ (4.3.10)

Path C!DW qcDnRTcln.VD=VC/ (4.3.11)

The ratio ofqcandqhobtained from these expressions is

qc

qh

D

Tc

Th



ln.VD=VC/

ln.VB=VA/

(4.3.12)