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 paper^{² }(and Larson’s original work^{8}),

$$V_L(T)=V_1+v_2+V_3\qquad(4)$$

where

*V** _{L }*= overall specific volume of liquid (cm

^{³}/g) (total volume/total mass)

*V** _{1 }*= specific volume increment at 0

^{º}K and that due to the solid molecules in solution of the liquid (solid volume/total mass)

*V*_{2}_{}= specific volume increment due to the liquid molecules of the substance, temperature above 0ºK (liquid volume/total mass)

*V** _{3}* = 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 *V** _{3}* than it has on

*V*

*. 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.*

_{2}For a solid under pressure^{³}, the volume is multiplied by $\sqrt{\frac{P_o}{P+P_o}}$, where *P _{o}* 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 is^{4 }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, *n _{p}*, and divided by the ratio of the base volume to 1, raised to the 2/3 power. (The solid expression just uses volume, or s

_{o}

^{3}.) 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

$$P_o=88.045482m^{(2/3)}\frac{n_p}{n_v^{(2/3)}}\ _{atm}\qquad(13)$$

*n _{p}* is the number of atoms effectively acting against the external pressure. It is sometimes, but not usually, equal to the number of volumetric units,

*n*. Using eq. 8, 9, and 10, B can be expressed as

_{v}$$B=\frac{\kappa_{s1}n_tT_{nu}}{\kappa_{s2}T}\left(\frac{P^2}{P_o}+2P+P_o\right)+P+P_o\ _{atm}\qquad(14)$$

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 Experiment ^{5,6}**

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 |
CH |
60.05 |
.9046 |
.7820 |
4 |
1.0 |
7 |
1 |
288.16 |
11441.503 |
11279.014 |

Carbon Tetrachloride |
CCl |
153.81 |
1.0 |
.9183 |
6 |
1.0 |
5 |
1 |
250.26 |
12334.317 |
11878.218 |

Ethyl Acetate |
CH |
88.10 |
.9818 |
.9818 |
6 |
1.0 |
6 |
1 |
293.16 |
8687.0274 |
8733.6283 |

Water |
H |
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 |
1.269x10 |

Carbon |
1.240x10 |
1.2987x10 |

Ethyl Acetate |
1.24398x10 |
1.304x10 |

Water |
7.72383x10 |
7.992x10 |

(The values of $\beta_o$ have not yet been determined, which explains the descrepancy between _{calc} and _{obs}.)

The *n _{p}* values are easy to understand. In acetic acid, the CH

_{3}

_{}contributes 3 units and the CO

_{2}H 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

*n*,

_{v}*n*, and

_{t}*n*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.

_{p}^{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 P_{lnu} is unique.