General Relation Between Specific Heats

Consider a general homogeneous substance (not necessarily a gas) whose volume, $V$ , is the only relevant external parameter. Let us find the general relationship between this substance's molar specific heat at constant volume, $c_V$ , and its molar specific heat at constant pressure, $c_p$ .

The heat capacity at constant volume is given by

$$C_V=\left(\frac{{\mathchar'26\mkern-11mud}Q}{dT}\right)_V = T\left(\frac{\partial S}{\partial T}\right)_V.$$ (6.120)

Likewise, the heat capacity at constant pressure is written

$$C_p=\left(\frac{{\mathchar'26\mkern-11mud}Q}{dT}\right)_p = T\left(\frac{\partial S}{\partial T}\right)_p.$$ (6.121)

Experimentally, the parameters that are most easily controlled are the temperature, $T$ , and the pressure, $p$ . Let us consider these as the independent variables. Thus, $S=S(T,p)$ , which implies that

$${\mathchar'26\mkern-11mud}Q = T dS = T\left[\left(\frac{\partial... ...\partial T}\right)_p dT +\left(\frac{\partial S}{\partial p}\right)_T dp\right]$$ (6.122)

in an infinitesimal quasi-static process in which an amount of heat ${\mathchar'26\mkern-11mud}Q$ is absorbed. It follows from Equation (6.121) that

$${\mathchar'26\mkern-11mud}Q = T dS = C_p dT + T\left(\frac{\partial S}{\partial p}\right)_T dp.$$ (6.123)

Suppose that $p=p(T,V)$ . The previous equation can be written

$${\mathchar'26\mkern-11mud}Q = T dS = C_p dT + T\left(\frac{\par... ...partial T}\right)_V dT +\left(\frac{\partial p}{\partial V}\right)_T dV\right].$$ (6.124)

At constant volume, $dV=0$ . Hence, Equation (6.120) gives

$$C_V = T\left(\frac{\partial S}{\partial T}\right)_V =C_p + T\left... ...{\partial S}{\partial p}\right)_T \left(\frac{\partial p}{\partial T}\right)_V.$$ (6.125)

This is the general relationship between $C_V$ and $C_p$ . Unfortunately, it contains quantities on the right-hand side that are not readily measurable.

Consider $(\partial S/\partial p)_T$ . According to the Maxwell relation (6.119),

$$\left(\frac{\partial S}{\partial p}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_p.$$ (6.126)

Now, the quantity

$$\alpha_V\equiv \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p,$$ (6.127)

which is known as the volume coefficient of expansion, is easily measured experimentally. Hence, it is convenient to make the substitution

$$\left(\frac{\partial S}{\partial p}\right)_T =-V \alpha_V$$ (6.128)

in Equation (6.125).

Consider the quantity $(\partial p/\partial T)_V$ . Writing $V=V(T,p)$ , we obtain

$$dV = \left(\frac{\partial V}{\partial T}\right)_p dT +\left(\frac{\partial V}{\partial p}\right)_T dp.$$ (6.129)

At constant volume, $dV=0$ , so we obtain

$$\left(\frac{\partial p}{\partial T}\right)_V = -\left.\left(\frac... ...ial V}{\partial T}\right)_p\right/\left(\frac{\partial V}{\partial p}\right)_T.$$ (6.130)

The (usually positive) quantity

$$\kappa_T\equiv -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T,$$ (6.131)

which is known as the isothermal compressibility, is easily measured experimentally. Hence, it is convenient to make the substitution

$$\left(\frac{\partial p}{\partial T}\right)_V =\frac{\alpha_V}{\kappa_T}$$ (6.132)

in Equation (6.125). It follows that

$$C_p - C_V = V T \frac{\alpha_V^{ 2}}{\kappa_T},$$ (6.133)

and

$$c_p - c_V = v T \frac{\alpha_V^{ 2}}{\kappa_T},$$ (6.134)

where $v=V/\nu$ is the molar volume.

As an example, consider an ideal gas, for which

$$p V = \nu R T.$$ (6.135)

At constant $p$ , we have

$$p dV = \nu R dT.$$ (6.136)

Hence,

$$\left(\frac{dV}{dT}\right)_p = \frac{\nu R}{p} = \frac{V}{T},$$ (6.137)

and the expansion coefficient defined in Equation (6.127) becomes

$$\alpha_V = \frac{1}{T}.$$ (6.138)

At constant $T$ , we have

$$p dV + V dp = 0.$$ (6.139)

Hence,

$$\left(\frac{\partial V}{\partial p}\right)_T = - \frac{V}{p},$$ (6.140)

and the compressibility defined in Equation (6.131) becomes

$$\kappa_T = \frac{1}{p}.$$ (6.141)

Finally, the molar volume of an ideal gas is

$$v = \frac{V}{\nu} = \frac{R T}{p}.$$ (6.142)

Hence, Equations (6.134), (6.138), (6.141), and (6.142) yield

$$c_p - c_V=R,$$ (6.143)

which is identical to Equation (6.39).