Relations between heat capacities

Short description: Equations definiting head capacities in thermodynamics

In thermodynamics, the heat capacity at constant volume, CV, and the heat capacity at constant pressure, CP, are extensive properties that have the magnitude of energy divided by temperature.

## Contents

* 1 Relations
* 2 Derivation
* 3 Ideal gas
* 4 Implications
* 5 See also
* 6 References

## Relations

The laws of thermodynamics imply the following relations between these two heat capacities (Gaskell 2003:23):

CP−CV=VTα2βT

CPCV=βTβS

Here α is the thermal expansion coefficient:

α=1V(∂V∂T)P

βT is the isothermal compressibility (the inverse of the bulk modulus):

βT=−1V(∂V∂P)T

and βS is the isentropic compressibility:

βS=−1V(∂V∂P)S

A corresponding expression for the difference in specific heat capacities (intensive properties) at constant volume and constant pressure is:

cp−cv=Tα2ρβT

where ρ is the density of the substance under the applicable conditions.

The corresponding expression for the ratio of specific heat capacities remains the same since the thermodynamic system size-dependent quantities, whether on a per mass or per mole basis, cancel out in the ratio because specific heat capacities are intensive properties. Thus:

cpcv=βTβS

The difference relation allows one to obtain the heat capacity for solids at constant volume which is not readily measured in terms of quantities that are more easily measured. The ratio relation allows one to express the isentropic compressibility in terms of the heat capacity ratio.

## Derivation

If an infinitesimally small amount of heat δQ is supplied to a system in a reversible way then, according to the second law of thermodynamics, the entropy change of the system is given by:

dS=δQT

Since

δQ=CdT

where C is the heat capacity, it follows that:

TdS=CdT

The heat capacity depends on how the external variables of the system are changed when the heat is supplied. If the only external variable of the system is the volume, then we can write:

dS=(∂S∂T)VdT+(∂S∂V)TdV

From this follows:

CV=T(∂S∂T)V

Expressing dS in terms of dT and dP similarly as above leads to the expression:

CP=T(∂S∂T)P

One can find the above expression for CP−CV by expressing dV in terms of dP and dT in the above expression for dS.

dV=(∂V∂T)PdT+(∂V∂P)TdP

results in

dS=[(∂S∂T)V+(∂S∂V)T(∂V∂T)P]dT+(∂S∂V)T(∂V∂P)TdP

and it follows:

(∂S∂T)P=(∂S∂T)V+(∂S∂V)T(∂V∂T)P

Therefore,

CP−CV=T(∂S∂V)T(∂V∂T)P=VTα(∂S∂V)T

The partial derivative (∂S∂V)T can be rewritten in terms of variables that do not involve the entropy using a suitable Maxwell relation. These relations follow from the fundamental thermodynamic relation:

dE=TdS−PdV

It follows from this that the differential of the Helmholtz free energy F=E−TS is:

dF=−SdT−PdV

This means that

−S=(∂F∂T)V

and

−P=(∂F∂V)T

The symmetry of second derivatives of F with respect to T and V then implies

(∂S∂V)T=(∂P∂T)V

allowing one to write:

CP−CV=VTα(∂P∂T)V

The r.h.s. contains a derivative at constant volume, which can be difficult to measure. It can be rewritten as follows. In general,

dV=(∂V∂P)TdP+(∂V∂T)PdT

Since the partial derivative (∂P∂T)V is just the ratio of dP and dT for dV = 0, one can obtain this by putting dV = 0 in the above equation and solving for this ratio:

(∂P∂T)V=−(∂V∂T)P(∂V∂P)T=αβT

which yields the expression:

CP−CV=VTα2βT

The expression for the ratio of the heat capacities can be obtained as follows:

CPCV=(∂S∂T)P(∂S∂T)V

The partial derivative in the numerator can be expressed as a ratio of partial derivatives of the pressure w.r.t. temperature and entropy. If in the relation

dP=(∂P∂S)TdS+(∂P∂T)SdT

we put dP=0 and solve for the ratio dSdT we obtain (∂S∂T)P. Doing so gives:

(∂S∂T)P=−(∂P∂T)S(∂P∂S)T

One can similarly rewrite the partial derivative (∂S∂T)V by expressing dV in terms of dS and dT, putting dV equal to zero and solving for the ratio dSdT. When one substitutes that expression in the heat capacity ratio expressed as the ratio of the partial derivatives of the entropy above, it follows:

CPCV=(∂P∂T)S(∂P∂S)T(∂V∂S)T(∂V∂T)S

Taking together the two derivatives at constant S:

(∂P∂T)S(∂V∂T)S=(∂P∂T)S(∂T∂V)S=(∂P∂V)S

Taking together the two derivatives at constant T:

(∂V∂S)T(∂P∂S)T=(∂V∂S)T(∂S∂P)T=(∂V∂P)T

From this one can write:

CPCV=(∂P∂V)S(∂V∂P)T=βTβS

## Ideal gas

This is a derivation to obtain an expression for CP−CV for an ideal gas.

An ideal gas has the equation of state: PV=nRT

where

* P = pressure
* V = volume
* n = number of moles
* R = universal gas constant
* T = temperature

The ideal gas equation of state can be arranged to give:

V=nRT/P or nR=PV/T

The following partial derivatives are obtained from the above equation of state:

(∂V∂T)P =nRP =(VPT)(1P)=VT

(∂V∂P)T =−nRTP2 =−PVP2 =−VP

The following simple expressions are obtained for thermal expansion coefficient α:

α=1V(∂V∂T)P =1V(VT)

α=1/T

and for isothermal compressibility βT:

βT=−1V(∂V∂P)T =−1V(−VP)

βT=1/P

One can now calculate CP−CV for ideal gases from the previously obtained general formula:

CP−CV=VTα2βT =VT(1/T)21/P=VPT

Substituting from the ideal gas equation gives finally:

CP−CV=nR

where n = number of moles of gas in the thermodynamic system under consideration and R = universal gas constant. On a per mole basis, the expression for difference in molar heat capacities becomes simply R for ideal gases as follows:

CP,m−CV,m=CP−CVn=nRn=R

This result would be consistent if the specific difference were derived directly from the general expression for cp−cv.

## Implications

A state of stable equilibrium implies the non-negativity of the heat capacities and the compressibilities.[1] The non-negativity of the heat capacity is a requirement for thermal stability. For if the heat capacity were negative, a small fluctuation of heat added to a subsystem would result in a decrease of temperature relative to its surroundings, which would lead to even more heat added to the subsystem. Similarly, the non-negativity of the compressibility is a requirement for mechanical stability. For if the compressibility were negative, a small fluctuation of volume added to a subsystem would result in an increase of pressure relative to its surroundings, which would lead to even more volume added to the subsystem. Hence we have the inequality: CP≥CV≥0

A relation between βT and βS can be derived similarly as, or derived from, the relation between CP and CV:[2] CP(βT−βS)=VTα2

This implies: βT≥βS≥0

## See also

* Heat capacity ratio

## References

1. ↑ Reichl 2016, p. 55
2. ↑ Reichl 2016, p. 49

* Gaskell, David R. (2008). Introduction to the thermodynamics of materials (Fifth ed.). Taylor & Francis. ISBN 1-59169-043-9.
* Reichl, Linda E. (2016). A Modern Course in Statistical Physics (Fourth ed.). WILEY-VCH.

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