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Next: Free Expansion of Gas Up: Classical Thermodynamics Previous: Gibbs Free Energy

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

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

Likewise, the heat capacity at constant pressure is written

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

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

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

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

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

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

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

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

Now, the quantity

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

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

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

in Equation (6.125).

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

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

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

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

The (usually positive) quantity

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

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

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

in Equation (6.125). It follows that

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

and

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

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

As an example, consider an ideal gas, for which

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

At constant $ p$ , we have

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

Hence,

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

and the expansion coefficient defined in Equation (6.127) becomes

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

At constant $ T$ , we have

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

Hence,

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

and the compressibility defined in Equation (6.131) becomes

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

Finally, the molar volume of an ideal gas is

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

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

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

which is identical to Equation (6.39).


next up previous
Next: Free Expansion of Gas Up: Classical Thermodynamics Previous: Gibbs Free Energy
Richard Fitzpatrick 2016-01-25