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

This paper provides a step-by-step procedure for the calculation of liquid specific volume as a function of composition and temperature, based on the Reciprocal System of D. B. Larson1.  In this theory, each individual molecule may be in the solid, liquid, or gaseous (or vapor) state, regardless of the state of the majority of molecules of the substance.

First let's define some terms:

$V_L$ = overall specific volume of liquid (cm3/g) (total volume/total mass)

$V_1$ = specific volume increment at 0 oK 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 0 oK (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)

Then,

$$V_L=V_1+V_2+V_3\qquad(1)$$

The initial values of these three components are designated $V_{01}$, $V_{02}$, $V_{03}$.  These differ only by a geometric factor (designated $k_{s1}$, $k_{s2}$, $k_{s3}$) applied to a base initial value, $V_{00}$, determined as follows.

Just as the volume of a gas is determined by the number of molecules, so the volume of a liquid is determined by the number of volumetric groups which it contains.  In an organic compound, for instance, each of the common interior groups, such as CH2, CH, or CO, constitutes one volumetric group.  The CH3 groups in the end positions of the aliphatic chains occupy two units each.  So hexane, represented as CH3CH2CH2CH2CH2CH3, has 8 volumetric groups.  Let $n_v$ be the number of volumetric groups and recall that the factor .707 expresses the geometric reduction obtained by the close-packed arrangement of the liquid groups because of their flexibility of movement.  Then, in natural units, the base initial volume is directly proportional to the number of volumetric units, reduced by close-packing:

$$V_{00}=.707n_v \qquad(2)$$

Let $m$ be the molecular weight (non-dimensional) of the molecule of the substance, $m_u$ be the value of the natural unit of atomic mass in g, and $V_{nu}$ be the value of the natural unit of liquid volume expressed in cm3. Then, in conventional units, the basic initial value is

$V_{00}=\frac{.707n_vV_{nu}}{mm_u}\ cm^3/g \qquad(3)$

$V_{nu}$ is not the cube of the natural unit of space in the time-space region, which is applicable only to the gaseous state.  Rather, $V_{nu}$ is the cube of the natural unit of space in the time region, which is 1/156.45 (the inter-regional ratio) of that in time-space region, or 2.9139 x 10-8 cm.  Cubing this we get

$$V_{nu}=2.47417\times 10^{-23}\ cm^3$$

The natural unit of mass is 1 atomic mass unit, so $m_u$ is 1.65979x10-24 g.  Putting these values in eq. 3, we get

$$V_{00}=\frac{10.5389n_v}{m}\ cm^3/g \qquad(4)$$

For hexane, $n_v$ is 8 and the molecular weight is 86.18.  Therefore,

$$V_{00}=.9783\ cm^3/g$$

For the critical (gaseous or vapor) specific volume increment, the geometric factor $k_{s3}$ is always 1.00.  For the solid specific volume increment, the geometric factor $k_{s1}$ is .891 (the cube root of .707) where close-packing in the solid state can be achieved.  Where such packing cannot be achieved, the geometric factor $k_{s1}$ is 1.000.  The same applies to the geometric factor for the liquid specific volume increment, $k_{s2}$.  Therefore, the initial values of the three volume components may be expressed as

$$V_{01}=V_{00}k_{s1}\qquad(6)$$

$$V_{02}=V_{00}k_{s2}\qquad(7)$$

$$V_{03}=V_{00}k_{s3}=V_{00}\qquad(8)$$

In a multi-group molecule, the value of the geometric factors $k_{s1}$ and $k_{s2}$ represent averages, since some groups may be at .891 while others at 1.000.  Let $n_{s1}$ = the number of close-packed groups per molecule in the solid state, and let $n_{s2}$ = the number of close-packed groups per molecule in the liquid state.  Then

$$k_{s1}=\frac{n_{s1}\times.891+(n_v-n_{s1})\times1.000}{n_v}\qquad(9)$$

$$k_{s2}=\frac{n_{s2}\times.891+(n_v-n_{s2})\times1.000}{n_v}\qquad(10)$$

For hexane, for instance,$k_{s1}$ is .9864 (with 1 group at .891 and 7 groups at 1.0000, the average is 7.891/8 or .9864) and $k_{s2}$ is .9728 (with 2 groups at .891 and 6 groups at 1.0000).  Therefore, for hexane, the initial values of the specific volume increments are

$$V_{01}=.9783\times.9864=.9650\ cm^3/g$$

$$V_{02}=.9783\times.9728=.9517\ cm^3/g$$

$$V_{03}=.9783\times1.0000=.9783\ cm^3/g$$

From eq. 10 it's clear that ordinarily $n_v\ge n_{s1},n_v\ge n_{s2}$.  However, for lower group elements, hyrdrogen through fluorine, closer packing than normal can be achieved because of inactive dimensions of the gravitational repulsion force.  This means that, in effect, for lower group elements the geometric factors can be less than .891.  We can still use eq. 10, though, if we allow the value of the number of solid groups to exceed the number of volumetric units.

Now that we have the initial values as a function of composition, we can determine the values of the three components as a function of temperature.  The solid specific volume increment not only includes the initial volume at 0 oK but also a factor proportional to the number of solid molecules in the substance at any temperature, $\Delta s$, which can be determined by probability considerations.

$$V_1=V_{01}+\Delta s\qquad(11)$$

To use the normal probability function or table we need to know the value of the normal random variable, $z_s$, applicable.  It should be proportional to the difference between the liquid temperature $T$ and the melting point $T_M$, in degrees K, divided by the melting point. The coefficient and the intercept  have unfortunately not been worked out theoretically, but are given empirically by Larson (Ref. 1) as follows:

$$z_s=\frac{4(T-T_M)}{T_M}+.40\qquad(12)$$

We want the right tail of the distribution, so we subtract the value of the normal function, denoted by erf(zs), from 1 and then multiply by the average difference in specific volume between solid and liquid molecules, denoted by $\Delta\nu$:

$$\Delta s=(1-erf(z_s))\Delta\nu\qquad(13)$$

Larson uses an average value of $\Delta \nu$ of .080 for paraffin hydrocarbons (C14 and below) and .084 for paraffins above C14 (rather than computing the individual values).  For hexane, $T_M$ = 178 K (-95 oC).  At $T$ = -50 oC, $z_s$=1.41 and from the normal probability table, erf(zs)$\approx$.9207.  Subtracting this from 1.0000, we get .0793, which means that 7.93 % of the molecules in the liquid hexane aggregate at -50 oC are in the solid state.  Multiplying this figure by the approximate difference in specific volume between solid and liquid molecules, .080, we get .0063 cm3/g for the value of $\Delta s$.

The thermal motion beyond the initial point of the liquid (considered as starting at 0 oK) is the one-dimensional equivalent of the thermal motion of a gas, and thus the volume generated is directly proportional to the temperature, $T$.  Let $T_{nu}$ be the natural unit of temperature in the time region (for the condensed states of matter) and $n_t$ be the temperature factor.  Then

$$V_2=\frac{T}{n_tT_{nu}}V_{02}\qquad(14)$$

In Ref. 2, Larson derived the value of $T_{nu}$ to be 510.8 K.  For simple substances, $n_t$ is 1.  More complex or more electropositive substances have values of $n_t$ of 2 up to 16.  Hexane has a value of 1; water, 2; silver, 16.  Compounds of electropositive and electronegative elements have intermediate values (some with half-integral values, which are averages), as would be expected.

The gaseous or vapor increment of specific volume depends on the proportion of critical molecules existing in the aggregate at each temperature, which can be computed from probability considerations.  Larson uses two random variables for this computation, both a function of the critical temperature, $T_c$:

$$z_{c1}=\frac{9(T_c-T)}{Tc+T_{nu}/2}\qquad(15)$$

$$z_{c2}=\frac{27(T_c-T)}{T_c+T_{nu}/2}\qquad(16)$$

Then the specific volume increment due to critical molecules in the substance is

$$V_3=(2-(erf(z_{c1})+erf(z_{c2})))V_{03}\qquad(17)$$

For hexane, $T_c$ = 508 K.  At $T$ = 210 oC, $z_{c1}$= .2947 and $z_{c2}$ =.8106.  The corresponding values of the normal probability function are .6144 and .8109.  Then, from eq. 17,

$$V_3=(2-(.6144+.8109))(.9783)=(.5747)(.9783)=.5622\ cm^3/g$$

The .5747 factor means that 57.47% of the molecules at this temperature are in the critical state.

Having determined $V_1$, $V_2$ and $V_3$ we can now calculate $V_L$ from eq. 1.

To automate the task of comparing the theoretical values with those observed, I've prepared a computer program and run it on most of the same liquids Larson used in the original series of papers:  hexane, hexadecane, benzene, acetic acid, ethyl acetate, ethyl chloride, ethanethiol, fluorine, hydrochloric acid, sulfur dioxide, carbon tetrachloride, and water.  Printouts from the program for all of these liquids follow.  The observed values come from the same sources Larson used:  Timmermans' Physico-chemical Constants of Pure Organic Compounds, the American Petroleum Institute, and the International Critical Tables.

Most of the computer results are in harmony with Larson's manual calculations.  The two seeming exceptions are for acetic acid and water.  For acetic acid, Larson used a value of initial liquid specific volume of .5469, which is .7795 that of his base initial volume, .7016; but .891 is supposedly the smallest allowed fraction.  For water, Larson used a value of .7640 for both the initial solid and liquid specific volumes, but this is only .8713 that of his base initial volume, .8769, not .891.  Actually. these differences are due to "hydrogen bonding", which can allow closer packing than normal.  In a second calculation for water, I input 1.78 for $n_{s1}$ and $n_{s2}$ so as to get the initial volumes to be .7640.  The theoretical results computed came out to be much closer to the experimental ones than the previous run.

To compute the specific volume for any liquid of your choice, follow these steps:

1.  Determine the formula of the compound and its molecular weight.

2.  From the formula, determine the number of volumetric units and number of temperature units.

3.  Use equation 4 to obtain the base initial volume.

4.  Use equations 9 and 10 to compute the geometric factors; some iteration here may be required to the get the right values.

5.  Compute the initial volumes with equations 6, 7, and 8.

6.  Using equations 12 and 13, compute the solid specific volume increment, equation 11.

7.  Use equation 14 to compute the liquid specific volume increment.

8.  Using equations 15 and 16, compute the critical specific volume increment, equation 17.

9.  Sum the results to get the final value, from equation 1.

References:

1.  D. Larson, The Liquid State , privately circulated series of papers on the liquid state, circa. 1960-1964.  Note:  I made use of the papers numbered I, II, II-supplement, and III.  I've reorganized all of the equations and changed some of the symbols for the sake of clarity.  I've also used the latest values of the conversion constants.  The computer program is entirely original.

2.  D. Larson, Basic Properties of Matter (Salt Lake City, UT:  International Society of Unified Science, 1959-1988), pp. 59-60.

Appendix:  The Computer Program

The following pages show the input screens of the program.  The data base language is filePro Plus and the computation language is TrueBasic.  This is the first of what will be a comprehensive series of programs for the calculation of all properties of matter based on the Reciprocal System of theory.  Eventually the programs will be made available for purchase.

International Society of  Unified Science
Reciprocal System Research Society

Salt Lake City, UT 84106
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