A formula analogous to that for CP-CVrelates the isothermal and isentropic compressibilities of a material:

κT=κS+TVβ2CP.

(Here κS=-(1/V)(V/P)Sis the reciprocal of the adiabatic bulk modulus considered in Problem 1.39.) Derive this formula. Also check that it is true for an ideal gas.

Short Answer

Expert verified

κT=κS+β2VTCP

Step by step solution

01

To  prove

κT=κS+β2VTCP

02

Explanation

The isothermal and isentropic compressibilities can be calculated as follows:

κT=-1VVPT.....(1)κS=-1VVPS.....(2)

If S is a function of P and T, then V=V(P,T)is obtained from the definition of the derivative:

role="math" localid="1648438079130" dS=SPTdP+STPdT....(3)

If V is a function of P and S, then V=V(P,S)is obtained from the definition of the derivative:

dV=SPTdP+VSTdS

Substitute from (3) to get:

dV=SPTdP+VSTSPTdP+STPdTdV=SPT+VSTSPTdP+STPdT

At constant temperature dT=0we get:

(dV)T=SPT+VSTSPTdPVPT=SPT+VSTSPT

03

Further continuation for the proof

substitute from (1) and (2) to get:

-VκT=-VκS+VSTSPT......(4)

From the Gibbs energy and the Maxwell relation, we can conclude:

SPT=VTP

The thermal expansion is determined by:

β=1VVTP

Add these two equations together to get the following result:

SPT=-βV

substitute into (4) to get:

-VκT=-VκS-βVVST

04

Further continuation for the proof 

At constant pressure, the volume changes owing to a temperature change, and it is given by:

dV=βVdT....(6)

and the entropy change is given by:

dS=dQT=CPdTT....(7)

divide (6) over (7) to get:

VSP=βVTCP

substitute into (5) to get:

-VκT=-VκS-β2V2TCP

κT=κS+β2VTCP.......(8)

05

Further continuation for the proof 

For an ideal gas we have:

β=1VVTP=1VTNkTPP=NkPV=1T......(9)

κT=-1VVPT=1VPNkTPT=NkTP2V....(10)

substitute from (9), (10) and (11) into (8) to get:

κS=κT-β2VTCPκS=NkTP2V-β2VTNk1+f2-1κS=NkTP2V-β2VTNk2+f2-1κS=NkTP2V-VNkT22+f

To use PV=NKTwe get:

κS=PVP2V-VPV22+fκS=1Pf2+f......(12)

06

Further continuation for the proof 

Now we must verify this relationship: in an ideal gas, in an isentropic (adiabatic) process, we have:

PVγ=K

whereγ=(f+2)/fandKis constant, write forVto get:

V=KP1/γ

κS=1VPKP1/γκS=1VγKP(1/γ)1KP2κS=1VγKP(1/γ)PKKP2κS=1VγVPKKP2κS=1PγκS=1Pf2+f

which is equivalent to (12)

Therefore we proved thatκT=κS+β2VTCP

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