ISOTHERMAL AND ISENTROPIC COMPRESSIBILITIES

Link to: physicspages home page.To leave a comment or report an error, please use the auxiliary blog.Reference: Daniel V. Schroeder, An Introduction to Thermal Physics,

(Addison-Wesley, 2000) - Problem 5.16.An expression similar to that relating the heat capacities can be derived

to relate the isothermal and isentropic compressibilities κT and κS, definedas

κT ≡ − 1V

(∂V∂P

)T

(1)

κS ≡ − 1V

(∂V∂P

)S

(2)

These quantities measure the fractional change in volume of a substancein response to a change in pressure. To obtain the relation between them,we use a method similar to that for heat capacities CV and CP.

If we write S = S (P,T ) then

dS =

(∂S∂P

)T

dP+

(∂S∂T

)P

dT (3)

Also, starting with V =V (P,S) we have

dV =

(∂V∂P

)S

dP+

(∂V∂S

)P

dS (4)

Substituting 3 into 4 we get

dV =

[(∂V∂S

)P

(∂S∂P

)T+

(∂V∂P

)S

]dP+

(∂V∂T

)P

dT (5)

At constant temperature dT = 0 and we get(∂V∂P

)T

=

(∂V∂S

)P

(∂S∂P

)T+

(∂V∂P

)S

(6)

−V κT =

(∂V∂S

)P

(∂S∂P

)T−V κS (7)

From the Maxwell relation from the Gibbs energy1

ISOTHERMAL AND ISENTROPIC COMPRESSIBILITIES 2

(∂S∂P

)T=−

(∂V∂T

)P

(8)

Also, from the definition of the thermal expansion coefficient β

β ≡ 1V

(∂V∂T

)P

(9)

Combining these last two equations gives

−V κT =−βV(

∂V∂S

)P−V κS (10)

To get rid of the last partial derivative, we observe that the volume changedV due to a temperature change dT at constant pressure is

dV = βV dV (11)

The entropy change due to an influx of heat dQ at constant pressure attemperature T is

dS =dQT

(12)

= CPdTT

(13)

Dividing these two relations gives

(∂V∂S

)P=

TV β

CP(14)

Inserting this into 10 and cancelling off a factor of −V gives the finalresult

κT = κS +TV β 2

CP(15)

For an ideal gas, we can use this equation to work out κS:

ISOTHERMAL AND ISENTROPIC COMPRESSIBILITIES 3

β =1V

(∂V∂T

)P=

NkPV

=1T

(16)

κT = − 1V

(∂V∂P

)T=

NkTP2V

=1P

(17)

CP = CV +Nk (18)

= Nk(

1+f2

)(19)

κS =1P− V

NkT(

1+ f2

) (20)

=1P

f( f +2)

(21)

where in the third line, we’ve used Schroeder’s equation 1.48, and f isthe number of degrees of freedom of each gas molecule.

To check this, recall that for an isentropic (adiabatic) process in an idealgas

PV γ = K (22)

V =

(KP

)1/γ

(23)

where γ = ( f +2)/ f and K is a constant. So

κS = − 1V

(∂V∂P

)S

(24)

= −(

PK

)1/γ (−1

γ

)(KP

)1/γ 1P

(25)

=1

Pγ=

1P

f( f +2)

(26)

which is the same as 21, so equation 15 checks out for an ideal gas.

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