From thermodynamics,¹ the general equation of state of a pure substance is
$$\frac{\mathrm{d}V}{V}=\beta\mathrm{d}T-\kappa\mathrm{d}P\qquad(1)$$
where
$$\beta=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P=\ volume\ expansivity\qquad(2)$$
and
$$\kappa=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T=\ isothermal\ compressibility\qquad(3)$$
(Of course, V = volume, P = pressure, T = temperature.)
From my previous paper² (and Larson’s original work8),
$$V_L(T)=V_1+v_2+V_3\qquad(4)$$
where
VL = overall specific volume of liquid (cm³/g) (total volume/total mass)
V1 = specific volume increment at 0ºK and that due to the solid molecules in solution of the liquid (solid volume/total mass)
V2 = specific volume increment due to the liquid molecules of the substance, temperature above 0ºK (liquid volume/total mass)
V3 = specific volume increment due to the critical (gaseous or vapor) molecules in solution of the liquid (gaseous volume/total mass)
In this paper we will consider the effect of pressure on a liquid at temperatures below the liquid natural temperature unit, 510.8ºK. At low temperature, $V_3\approx 0$. Pressure has a different effect on V3 than it has on V2. Also, pressure has a different effect on a liquid at a temperature above, rather than below, 510.8ºK. These differences will be handled in another paper.
For a solid under pressure³, the volume is multiplied by $\sqrt{\frac{P_o}{P+P_o}}$, where Po is the internal pressure and P is the external pressure. For a liquid under pressure, the volume is multipled by the square of the solid factor, or simply $\frac{P_o}{P+P_o}$. So,
$$V_L(T,P)=V_1+V_2\left(\frac{P_o}{P+P_o}\right)\qquad(5)$$
It follows that isothermal compressibility is
$$\kappa=-\frac{1}{V_L}\left(\frac{\partial V_L}{\partial P}\right)_T=-\left(\frac{1}{V_1+V_2\left(\frac{P_o}{P+P_o}\right)}\right)(-V_2)(P_o)\left(\frac{1}{(P+P_o)^2}\right)\qquad(6)$$
It's often easier to work with the bulk modulus, B, which is the inverse of .
$$B=\frac{1}{\kappa}=\frac{\left(V_1+V_2\left(\frac{P_o}{P+P_o}\right)\right)(P+P_o)^2}{V_2P_o}\qquad(7)$$
From my previous paper,
$$V_1=V_{00}k_{s1}+\Delta s\approx V_{00}k_{s1} cm^3/g\qquad(8)$$
since $\Delta s$ is negligible for most liquids above the melting point.
$$ V_2=V_{00}k_{s2}\frac{T}{n_tT_{nu}}\ cm^3/g\qquad(9)$$
$$ V_{00}=\frac{10.5389n_v}{m}\ m^3/g$$
where $n_v$ is the number of volumetric units.
The internal pressure of a liquid is obviously different from that of a solid. The natural unit of pressure in the Reciprocal System is4 15,538,642 atm. To calculate the internal pressure of a solid we divide this number by the interregional ratio, 156.45. For a liquid, we divide by the square of the interregional ratio. Because liquid cohesion is two-dimensional rather than three-dimensional we must also multiply the expression by 2/3. Therefore,
$$P_{nu}=\frac{2}{3}\times\frac{15538642}{(156.45)^2}=423.2237\ _{atm}\qquad(11)$$
This expression is then multiplied by the number of pressure units, np, and divided by the ratio of the base volume to 1, raised to the 2/3 power. (The solid expression just uses volume, or so3.) Therefore,
$$P_o=\frac{423.22437n_p}{\left(\frac{V_{00}}{1}\right)^{\frac{2}{3}}}\ _{atm}\qquad(12)$$
Substituting eq. 10 in eq. 12, we get
$$P_o=88.045482m^{(2/3)}\frac{n_p}{n_v^{(2/3)}}\ _{atm}\qquad(13)$$
np is the number of atoms effectively acting against the external pressure. It is sometimes, but not usually, equal to the number of volumetric units, nv. Using eq. 8, 9, and 10, B can be expressed as
$$B=\frac{\kappa_{s1}n_tT_{nu}}{\kappa_{s2}T}\left(\frac{P^2}{P_o}+2P+P_o\right)+P+P_o\ _{atm}\qquad(14)$$
Now let's turn to calculating the volume expansivity.
$$\beta=\frac{1}{V_L}\left(\frac{\partial V_L}{\partial T}\right)_P=\left .\frac{\kappa_{s2}}{n_tT_{nu}\kappa_{s1}+\kappa_{s2}T\left(\frac{P_o}{P+P_o}\right)}+\beta_o\right._{\kappa^{-1}}\qquad(15)$$
where $\beta_o$ is the value of the expansivity at the end point of the solid.
One could plug (or 1/B) and into eq. 1 and integrate, but the resulting equation is more complex than eq. 5 and thus not useful.
In summary, to calculate bulk modulus and volume expansivity of a liquid, it is necessary to determine
$m$, the molecular weight
$n_v$, the number of volumetric units
$s1$, geometric factor
$s2$, geometric factor
$n_t$, the number of temperature units
$n_p$, the number of pressure units
Example Calculations and Comparisons with Experiment5,6
I selected four important liquids: acetic acid, carbon tetrachloride, ethyl acetate, and water. Here are the results, in table format.
|
Chemical |
Formula |
M |
$\kappa_{s1}$ |
$\kappa_{s2}$ |
Nv |
Nt |
Np |
P atm |
T ok |
Balc atm |
Bobs atm |
|
Acetic Acit |
CH3CO2H |
60.05 |
.9046 |
.7820 |
4 |
1.0 |
7 |
1 |
288.16 |
11441.503 |
11279.014 |
|
Carbon Tetrachloride |
CCl4 |
153.81 |
1.0 |
.9183 |
6 |
1.0 |
5 |
1 |
250.26 |
12334.317 |
11878.218 |
|
Ethyl Acetate |
CH3CO2C2H5 |
88.10 |
.9818 |
.9818 |
6 |
1.0 |
6 |
1 |
293.16 |
8687.0274 |
8733.6283 |
|
Water |
H2O |
18.0153 |
.8707 |
.8707 |
1.5 |
2.0 |
9 |
1 |
273.16 |
19697.992 |
19698.877 |
|
Chemical |
calc k1 |
obsK-1 |
|
acetit acid |
1.1377x10-3 |
1.269x10-3 |
|
Carbon |
1.240x10-3 |
1.2987x10-3 |
|
Ethyl Acetate |
1.24398x10-3 |
1.304x10-3 |
|
Water |
7.72383x10-4 |
7.992x10-4 |
(The values of $\beta_o$ have not yet been determined, which explains the descrepancy between calc and obs.)
The np values are easy to understand. In acetic acid, the CH3 contributes 3 units and the CO2H contributes 4. In carbon tetrachloride, each atom contributes 1 unit. In ethyl acetate, each volumetric group contributes a unit. In water, 3 molecules of 3 atoms each act against the external pressure, for a total of 9. All values of nv, nt, and np are integral or half-integral, as required by the nature of the Reciprocal System. This is very different from the empirical correlations used by other investigators.7
In the coming years I hope some member of ISUS will calculate the results for thousands of liquids following the equations given here.
References:
1. M. Abbott, H. Van Ness, Thermodynamics (New York: McGraw-Hill Book Company, 1972), p. 105.
2. R. Satz, "The Liquid State in the Reciprocal System: The Volume/Temperature Relation, a Contemporary Mathematical Treatement," Reciprocity, Vol. XXIII, No. 2, Autumn 1994. Incidentally, the normal function should have been denoted by $\Phi$, not erf.
$\Phi(z)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^ze^{-t^2}\mathrm{d}t$
$erf(z)=\frac{1}{\sqrt{\pi}}\int_{-\infty}^ze^{-t^2}\mathrm{d}t$
(The numerical results of the paper do not change, because $\Phi$ was actually used.)
3. R. Satz, "The Equation of State of Solid Matter," Reciprocity, Vol. X, No. 2, Spring-Summer 1980.
4. D. Larson, Nothing But Motion (Portland, Oregon: North Pacific Publishers, 1979), p. 160.
5. Handbook of Chemistry and Physics, 72nd Edition (Cleveland: The Chemical Rubber Company, 1991-1992), pp. 6-108 to 6-110.
6. American Institute of Physics Handbook (New York: McGraw-Hill Book Company, 1972). values are difficult to find. If you know the volume at temperature i and temperature f (and the pressure is constant), then from equation 1, $\beta\approx\frac{\ln\left(\frac{V_f}{V_i}\right)}{T_f-T_i}$.
7. R. Reid, J. Prausnitz, B. Poling, The Properties of Gases and Liquids, 4th Edition (New York: McGraw-Hill, Inc., 1987).
8. D. Larson, The Liquid State, privately circulated series of papers on the liquid state, circa. 1960-1964. Note: for this work, I made use of his paper IV. Larson used the semi-empirical value 415.84 atm for the liquid natural pressure unit. My derivation of Plnu is unique.