Concise Physical Chemistry

c05 JWBS043-Rogers September 13, 2010 11:25 Printer Name: Yet to Come

ENTROPY CHANGES 75

Hence

dSrev≡

dqrev

T

=

dH

T

=Cp

dT

T

Taking the integrals over the intervalT 1 toT 2 , we obtain

S=

∫T 2

T 1

Cp

dT

T

=Cp

∫T 2

T 1

dT

T

=Cpln

T 2

T 1

for the molar entropy change of heating at constant pressure. There is an analogous

equation for the energy change of heating at constant volume.

5.2.2 Expansion

For expansion at constant temperature,Uis constant. The first lawdU=dq+dw= 0

gives usdq=dwand the second law gives us

dSrev=

dqrev

T

=

dw

T

=

pdV

T

for reversible pressure–volume work. Taking the gas to be ideal for simplicity, we

obtain

dSrev=

pdV

T

=

RT

V dV

T

=R

dV

V

Integrating between limits as before, we obtain

Srev=R

∫V 2

V 1

dV

V

=Rln

V 2

V 1

for the reversible expansion of one mole of an ideal gas fromV 1 toV 2. The right-hand

side of the equation should be multiplied bynfor expansion ofnmoles of gas. If the

gas is not ideal, a real gas equation can be substituted for the ideal gas law in these

equations. Thus, the mathematical complexity will be increased, but the principle is

the same.

5.2.3 Heating and Expansion

BecausedSis an exact differential andS=f(T,p), we have

dS=

(

∂S

∂T

)

p

dT+

(

∂S

∂p

)

T

dp