# THE EQUATION OF STATE OF SOLID MATTER

For many years scientists and engineers have had available an excellent equation of state for gaseous matter. Now, at last, the Reciprocal System of Dewey B. Larson is able to give us an exact equation of state for solid matter. This paper will present a unified treatment of the subject, with Reference 1 as the starting point.

I. Volume of Solid as a Function of Temperature with Pressure Constant

From the material presented in Chapter 8 of Ref. 1, I have drawn a generalized plot of thermal expansion coefficient versus temperature, Figure l. The symbols are defined as follows:

ß = thermal expansion coefficient

T = temperature

TM = temperature of solid end point (at or close to melting point)

VM = volume at solid end point

T1 = first transition temperature

ß0 = initial value of thercnal expansion coefficient at absolute zero temperature

V0 = initial volume at absolute zero temperature

V1 = volume at tranSition temperature

ß0’ = initial vatue of thermal expansion coefficient based on second segment of curve

V0’ = initial volume based on second segment of curve

With the initial votume of the first segment of the curve included, eq. (8-4) of Ref. 1 becomes

 V = V0 + K/n³ T² (1)

where K is a constant and n is the number of rotational units that are themally vibrating.This equation can be put into a more usable form involving T, T1, ß1, end ß0 — all of which can be determined from theory. The thermal coefficient of expansion at temperature T is

 ß = 1/V dV/dT + ß0 = 2KT/n³ [1/V0 + KT²/n³ + ß0 (2) = 2KT/n³V0 + KT² + ß0

At T1,

 �1 = 2KT1/n³V0 + KT1² + ß0

Then,

 ß1�0/2T1 = K/n砎0 + T1�

So,

 2T1/�1-�0 = n砎0/K + t1�

 K/n³ = V0/2T1/�1-�0 - T1� (3)

Therefore,

 V = V0 + V0T²/2T1/�1-�0 - T1� (4)

This equation holds from T = 0 to T = T1. Larson has deduced the following values of ß1, ß0, and T1:

ß0 = 5.17 * 10-6/癒 for one unit

= 10.3 * 10-6/癒 for two units

= 15.5 * 10-6/癒 for three units

= 20.7 * 10-6/癒 for four units units

= units = [3576/TM]

ß0 = -2/7 * ß1 for electropositive elements

= -1/7 * ß1 for some 2lectronegative alements

Tl = 8.98 (a + z + y)°K

a, z, y are from Table 22 of Ref. 1

Thus given the volume of the solid at zero temperature, che rotational factcrs of the element, and the solid and point temperature, the volume V at any other temperature, (up to T1) can be easily determined.

The equation for the volume for temperatures above T1 has the same form as eq. (4):

 V0 T� V = V01 + ——— 2T1 (5) ——— - T� �1-�0

For this equation to be of use,V0’ and 0’ must be expressed in terms of known quantities such as VM and TM. Now,

 V0� TM� VM = V0� + ———– 2T1 ———– - T1� �1-�0�

 V0� T1� V1 = V0� + ———– 2T1 ———– - T1� �1-�0�

ln the equation for V1, solve for V0’ and put in equation for VM:

 V1 ( TM� ) VM = —————— 1 + ————— 1 +     T1� 2T1 ——––—– ——–  - T1� 2T1 �1�0� ———– - T1� �1-�0�

Or,

 TM� 1 + —————– 2T1 ——––  - T1� �1-�0� VM/V1 = ——————— T1� 1 + —————– 2T1 ——––  - T1� �1-�0�

Let,

 1 C2 = ————— 2T1 (6) ——–  - T1� �1-�0

Then,

 VM 1 + C2TM� —– = ————– V1 1 - C2T1�

Solve for C2:

 VM - V1 C2 = ——————— (7) V1TM� - VMT1�

From eqs. (6) and (7), ß0’, can be found:

 2T1 ß0’ = ß1 - V1TM² - VMT1² (8) + T1² VM - V1

This value of ß0’ can then be substituted into the equation for V0’:

 V1 VO’ = ———————– T1² 1 + —————— (9) 2T1 ———– - T1² ß1 - ßo’

With ß0’ and V0’ known, eq. (5) is ready for use. Larson has deduced the following values of TM and VM:

 TM = 1.80 * T1 for one rot. unit vibrating = 4.56 * T1 for two rot. units = 9.32 * T1 for three rot. units = 17.87 * T1 for four rot. units GM VM = 1.0625 V0 —– G0

where G0 is the initial crystal geometric constant and GM is the final one (some solids cnange crystalline structure as they expand).

The ratio VM/V1 can be generalized to any pair of final to initial volumes:

 Vf 1 + C2Tf² — = ———— (10) Vi 1 + C2Ti²

Compare this with the equation for a gas:

 Vf Tf — = — (11) Vi Ti

II. Volume of Solid as a Function of Pressure with Temperature Constant

The comoression of a solid by hydrostatic pressure is discontinuous at certain aressures nere denoted as P1, P2, P3, P4 etc. At these pressures the internal pressure P0 can change to P01, P02, P03, P04, etc., thus altering the slope of the compression curve. Larson has shown that between the transitions, the volume naries as the inverse square root of the pressure. The most general way to express this is with the following aquation:

 V P0 + Pref —— = ———— (12) Vref P0 + P

For the given value of P, the vatues of Pref, Vref, and P0 must be detennined by theory (or empiricnlly if necessary) before V can be found. For a four transition solid we have the following:

 0 < P < P1 Vref = V0 Pref = 0 P0 = P0 P1 < P < P2 Vref = V1 Pref = P1 P0 P01 P2 < P < P3 Vref = V2 Pref = P2 P0 = P02 P3 < P < P4 Vref = V3 Pref = P3 P0 = P03

Now, in the MKS system,

 aZy KN P0 = 4.177 * 10-23 —– —– (13) S0³ m²

where a, Z, y are the rotational compression values (simitar to the thermat values) and so is the base interatomic spacing. At each transition a, Z, y can change (and possible S0), thus causing P0 to change.

Before continuing the discussion of the equation of state I will discuss some subsidiary properties of matter: the bulk modulus, the modulus of elasticity, and Poisson’s ratio. Larson has derived the equation for compressibility; the solid bulk modulus is the inverse of this:

 B = 2 * P0 (14)

(at zero external pressure and zero temperature for a pure substance). I witl not derive the equation for the modulus of elasticity, E. In eq. (13) let the constants of the equation be written as J and generalize s0 (for the moment) to s. Then the initial internal stress is

 s = -P = -Js-3 (15)

By definition,

 ds E = —– de

And,

 ds ds ds — = — — de ds de

where e is the strain:

 s - s0 e = ——– s0

So,

 s = s0e + s0

thus,

 ds — = s0 de

Since

 ds — = 3Js-4 ds

then

 ds — = 3Js-4 s0 de

and when s = s0 and T = 0 癒 for a pure substance,

 ds — s0,T = 0 = 3Js0-3 = -3P0 de

 E = -3P0 (16)

(stress and pressure are in opposite directions)

Poisson’s ratio can be determined from the well-known equation

 3P0 n = .5 - ——– (17) 6 2P0

Thus at zero temperature and pressure for a pure substance,

 3P0 n = .5 - ——– = .25 (18) 6 2P0

This is in the “ball park” for most solids; however, most substances used in construction are impure and at other than zero temperature dnd, in addition, may contain a proportion of tiquid molecules — thus drastically changing the values of Poisson’s ratio and the modulus of elasticity. These considerations will be left to another paper.

Going back to eq. (12) we can generalize to the ratio of final to initial volume within a segment:

 Vf P0 + Pi — = ———– (19) Vi P0 + Pf

This compares with the equation for a gas:

 Vf Pi — = — (20) Vi Pf

III.Volume of Solid as a Function of Both Temperature and Pressure

The solid can be considered to undergo a aressure change at zero temperature and then a temperature change from the new volume. Let P < P1 Then

 P0 V0NEW = V0 ——–– P0 + P

Let T < T1. Then,

 V0NEW T² V = V0NEW + ———————— 2T1NEW (21) ———— - T1NEW2 ß1 - ß0

The value of T1 is not the same as before. To get to the original value of V1 the new value of T1 must be higher

 2 T1NEW = ——————— ß1 - ß0 ——— + ß1 - ß0 (22) V1 —————– V0 (P0/ P0 + P)½

where V1 is calculated from the original T1. I am assuming here that. is as before.

If P < P1 and T > T1, then the value of V0’ has to be modified, since T1 and TM are different. I assume that ß0’ is the same. Then the second term on the right in eq. (8) is the same and the new value of TM can be found:

 ( { T1NEW [ V1TM2 - VMT12 ] } VM-V1 +VM ) ½ TMNEW = ———– ——————— + T12 - T1NEW² — T1NEW T1 VM - V1 V1
(23)

Eq. (5) becomes

 V0NEWT² V = V0NEW + ———– 2T1NEW (24) ———– - T1NEW2 ß1 - ß0

Equations (21) and (24j (combined) represent the complete equation of the solid state.

IV. Exampte Calculations

As an example, consider one volume unit of silver at zero degrets K and zero external pressure. Whnt is the volume at temperature T and pressure P?

First the thermal rotational factors, a-Z-y, from Tabte 22 of Ref, 1 are found; they are 4-3-l. With these, the temperature of the first transition point, T1, can be calcutated:

 T1 = 8.98 (a+Z+y) = 8.98 (8) = 71.84oK

Silver has a maximum of four magnetit rotational units vibrating, so the solid end point is

 TM = 17.87 * T1 = 17.87 * 71.84 = 1283.78oK

In this case the endgoint appears to be somewhat higher than the empirical melting point, 1234 oK Thus it would seem that the thermal factors at the end point are towered by one to 3-3-1, so that

 TM = 17.87 * 8.98 (3+3+1) = 1123.31oK

Now the number of units to use in selecting ß1 is

 [3576 / 1123.31] = 3

and therefore

 ß1 = 15.5 * 10-6 / oK

Since siiver is etectronegative,

 s0 = - 1/7 ß1 = 1/7 * 15.5 * 10-6 = -2.214 * 10-6

Then from eq. (4),

 V T² — = 1 + ———————— V0 2 * 71.84 ———————– - 71.84² (15.5 + 2.214) * 10-6

 V — = 1 + 1.234 * 10-7 T� T < T1, P = 0 V0

This equation holds good up to T = T1 at which point

 V1 —  = 1 + 1.234 * 10-7 T1² = 1.0006369 V0

For temperatures above T1, the values of V0’ and ß0’ are needed. To calculate ß0’ I am going to use the empirical value of TM pending theoretical clarification.
From eqs. (8) and (9),

 2 * 71.84 ß0’ = 15.5 * 10-6 - ———————————— 1.0006369 * 1234² - 1.0625 * 71.84² ——————————————— + 7184² 1.0625 - 1.0006369

 = 9.647 * 10-6

 1.0006369 V0’ = —————– 1 + 71.84² ———— = 1.0004265 2 * 71.84² ———————–     - 71.84² (15.5 - 9.647) * 10-6

Thus from eq. (5),

 1.0004265 T² V = 1.0004265 + —————— 2 * 7184 ————————       - 71.84² (15.5 - 9.647) * 10-6

 V = 1.0004265 + 4.07623 * 10-8 T² T > T1, P = 0, V0 = 1

(Note: no crystal change from FCC is assumed here).

Now we’11 go on to look at the pressure relations. Assume that P is less than the first transition pressure P1 (which is approx. 107 KN/m² ) so that the initial compressibility factors from Table 14 of Ref. 1 can be used: a-Z-y = 4-8-1. From Table 4 of Ref. 1,
s0 = 2.87 x 10-10 m. Then from eq. (13),

 P0 = 4.177 * 10-23 (4 * 8 * 1) / (2.87 * 10-10)³ = 5.654 * 107 KN/m²

Since P c p , P = 0, dnd V = V . Then eq. (12) is

 V ( 5.554 * 107 ) — = ——————– ½            P < 1.0 * 107 KN/m² V0 5.554 * 107 + P T = 0

If P = .001 P ,

 V/V0 = (P0/1.001 P0)½ = .999500

The bulk modulus B, modulus of elasticity E, and Poisson’s ratio can now be calculated for a pure sample of silver at zero temperature:

 B = 2 * P0 = 2 * 5.654 * 107 = 1.1305 * 108 KN/m² E = 13 * P01 = 3 * 5.654 * 107 = 1.1692 * 108 KN/m² V = .25

 V0 5.654 * 107 ——————– T² V = V0 5.654 * 107 5.654 * 107 + P ——————– + ———————— 5.654 * 107 + P 2 T1 ——————— - T12 (15.5 + 2.214) * 10-6

The value of T1 to be used here comes from eq. (22):

 T1 = 2 ———————————————————— 15.5 * 10-6 + 2.214 * 10-6      + (15.5 + 2.214) * 10-6 ——————————— 1.0006369 ————–

If P = .001 P0, then

 5.654 * 10T - 1 ——————– 5.65 * 107 1P

 T1 = 128.24 癒

Putting this value of T1 into the above gives:

 V — = .99950 + 6.91095 * 10-8 T² T < T1 V0 P = .001 P0

The new value of T1 gives the new value of V0’:

 V0’ = 1.0006369 ————– 1 +      128.224 —————————— = 1.0002614 2 * 128.24 —————— - 128.242 (15.5 - 9.647) * 10-6

Thus, for temperatures above T1

 V = 1.0002614 + 1,0002614 T² ——————–– 2 * 128.24 ————–— - 128.24² (15.5 - 9.647) * 10-6 V = 1.0002614 + 2.28350 * 10-8 T² T > T1, P = .001P0 V0 = 1

Finally, from eq. (23) I find that the new melting temperature is:

 TMNEW = 1650.88 癒

(I have assumed, however, that this does not affect the original value of ß1) .

Basically the same procedure could be used with other elements, atloys, and compounds. Corresponding equations do not exist in quantum mechanics. A solution in “principle only” is not a true solution. A true solution is based on principle and works in practice.

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Reference

1. Dewey B. Larson, Solid Matter, prepublication version of second volume of the revised edition of The Structure of the Physical Universe (Portland, Oregon: North Pacific Publishers, 1980).

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