# THE LIQUID STATE IN THE RECIPROCAL SYSTEM: THE VOLUME/PRESSURE RELATION, A CONTEMPORARY MATHEMATICAL TREATMENT, PART II

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 Tetrachloride 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 CH3contributes 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.

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