Compressibility

When the density $\rho$ changes, both the pressure p and the temperature T will change, in general. The usual way to describe these changes in thermodynamics is to change the volume V occupied by a fixed number N of particles, so that  

$$dV=\left( \frac{\partial V}{\partial p}\right) _{T}dp+\left( \frac{\partial V}{\partial T}\right) _{p}dT$$ (2)

It is convenient to consider the fractional volume change dV/V and to define the isothermal compressibility  

$$\kappa _{T}=-\frac{1}{V}\left( \frac{\partial V}{\partial p}\right) _{T}$$ (3)

and the thermal expansion coefficient

$$\beta =\frac{1}{V}\left( \frac{\partial V}{\partial T}\right) _{p}$$

which are positive quantities characteristic of the material (fluid or solid). To convert from dV/V to a change of density, we note that a fixed N implies that $\rho V$ is a constant (it is the mass contained in V). Then $\rho dV+Vd\rho =0$ and we get  

$$\frac{d\rho }{\rho }=-\frac{dV}{V}=\kappa _{T}dp-\beta dT$$ (4)

As an example, consider an ideal gas. From pV=Nk_BT, get pdV+Vdp=Nk_BdT, hence  

$$\kappa _{T}=1/p\quad \quad \quad \quad {\rm (ideal\;gas)}$$ (5)

$$\beta =1/T\quad \quad \quad \quad {\rm (ideal\;gas)}$$ (6)

As we said already, there are two limiting cases where heat transfer and fluid motions are effectively decoupled, and then the dynamics of the fluid is fully determined by the appropriate compressibility coefficient:

• If the density changes slowly, so that heat conduction keeps the temperature T constant, eq. (5.4) reduces to $d\rho /\rho =\kappa _{T}dp$, where $\kappa _{T}$ is the isothermal compressibility (5.3). As we have already noted, $\kappa _{T}$ is positive (because a compression corresponds to a positive dp and a negative dV), and does not depend on the size of V (because for a larger V we get a correspondingly larger dV for the same dp). However, $\kappa _{T}$ can depend on p and on the temperature T. For example, in an ideal gas, $\kappa _{T}=1/p$.
• If the density changes rapidly, so that no heat can flow, we have instead $d\rho /\rho =\kappa _{S}dp$, where  

  $$\kappa _{S}=-\frac{1}{V}\left( \frac{\partial V}{\partial p}\right) _{S}$$ (7)

  is the adiabatic compressibility, which we now discuss. The subscript S denotes a process at constant entropy, in accordance with the fact that an adiabatic volume change (dQ=0, no heat transfer) is also isentropic (dS=0). In such a process all the work done in changing the volume goes into the internal energy E of the fluid and eq. (5.1) reduces to

  $$dE=-pdV\quad \quad \quad {\rm (adiabatic\;process)}$$

  This condition allows us to compute $\kappa _{S}$. For example, in an ideal gas dE=C_VdT, hence for an adiabatic process C_VdT = -pdV. Using this result in eq. (5.5), we find (1 + Nk_B)pdV + Vdp = 0 and by the definition (5.8):

  $$\kappa _{S}=\frac{1}{p}\,\frac{C_{V}}{C_{V}+Nk_{B}}\quad \quad \quad \quad {\rm (ideal\;gas),}$$

  Recalling that C_p=C_V+Nk_B is the specific heat at constant pressure and comparing with equation (5.5), we see that  

  $$\frac{\kappa _{S}}{\kappa _{T}}=\frac{C_{V}}{C_{p}}$$ (8)

  This relation is true in general (as can be proved by thermodynamics). It is almost always true that C_p>C_V, so that $\kappa _{S}<\kappa _{T}$: it is harder to compress a fluid if heat does not flow out of it.

In a typical liquid, there is little difference between $\kappa _{S}$ and $\kappa _{T}$, but in a monatomic ideal gas the difference is large: C_V=(3/2)Nk_B and C_p=(5/2)Nk_B. The ratio C_p/C_V appears often in the physics of gases and is called $\gamma$. For a monatomic ideal gas, $\gamma =5/3=1.667$; for air near room temperature $\gamma$ is very close to 1.4, which is the value for a diatomic ideal gas when the molecules can be regarded as rigid (at higher temperatures, molecular vibrations increase C_V, so that $\gamma$ decreases).

The inverse of the compressibility coefficient is called the bulk modulus B. We can define $B_{T}=1/\kappa _{T}$ and $B_{S}=1/\kappa _{S}$, but we can use simply B and let the context distinguish what we mean (keeping in mind that for liquids, and solids, the distinction is unimportant). We also want to relate pressure changes to density changes, rather than volume changes. Using rdV + Vdr = 0 as in the derivation of eq. (5.4), we see that  

$$dp=\frac{B}{\rho }\,d\rho$$ (9)

Values of 1/B are given in Table 7.16 of PQRG. We see that for common organic liquids B is about 1 GigaPascal (10$^{9}\,$N/m²), increasing slowly with pressure. Water is a little less compressible: $B=2.17\,\,$GPa at $T=22{{}^\circ}$C and $p=0.1\,$MPa. Mercury is a lot less compressible: with $B=25\,\,$GPa it is getting close to the range of solid metals.