Just a taste of thermodynamics
For a gas whose behavior depends only on pressure, p(t) (say), density, r(t),
and temperature, T(t), there are four physical dimensions that figure in derivations of
relations that govern p, r and T. With [q] to specify a fundamental or
derived dimension, the fundamental ones are
[t] = T, [length] = L, [mass] = M and [T] = D (degrees). (1)
(D rather than T in the last case to distinguish it from [t]=T.) The important derived
dimensions are
[speed] = L/T, [force] = ML/T2,
[p] = M/LT2, [r] = M/L3 and [p/r] =
(L/T)2. (2)
In the usual simple case, an ideal gas, the equation of state is
taken to be
p = rRT , pv = RT or pV = MRT , (3)
where p is pressure (force per unit area), r is density (mass per unit volume),
v (= 1/r) is specific-volume (volume per unit mass) and T is
absolute temperature (Kelvin or Rankine degrees). The first form of it is often seen
in fluid dynamics, the last two, in thermodynamics, and the first two use caps for the thermal
entities (e.g. [T] contains D) and lower case for mechanical entities (e.g. [v] contains M).
The last form, where V is volume and M is mass within V, serves to
show the difference between thinking of r(t) = M(t)/V (fluids) or
v(t) = V(t)/M (thermo).
With nothing said about R(p,r,T)
the relations are merely a definition of T, so at first eq(3) means only that
there is a thermal variable. An ideal gas is then said to be
one for which the empirical
relations, (p/r)=f(T) (Boyle's 'law') and rT=g(p) (Charles' 'law'),
also hold, no matter what. From the first, it follows that R is R(T); from the second,
that R is R(p); and thus R is a constant.
To see how the same result follows from dimensional analysis,
let p=rR0T/F(p,r,T), where R0 is any constant
with dimensions (L/T)2/D, and F is dimensionless ([F]=1).
To include the possibilities of exponential functions and fractional powers, introduce arbitrary
reference values, p0, r0, R0 and T0.
Only three of these are dimensionally independent, so we are free to choose
R0=p0/r0T0 (4)
Then
[p/p0] = [r/r0] = [T/T0]
= [r0R0T/p0] = [F] = 1 (5)
The easiest result comes from
rR0T/p =
F(p/p0 , r/r0 ,
T/T0 , k ), (6)
where k stands for any number of dimensionless constants. The (arbitrary) choices,
p0=p, r0=r and T0=T then imply
rR0T/p = F(1 , 1 , 1 , k ). (7)
In other words, F is a constant, and by the choice made for R0. it is '1'.
By essentially the same kind of argument, it follows from dimensional analysis that the most
general equation of state, say G(p,r,T,R,k)=0, that contains only the four variables
(with their dimensions) and dimensionless constants is RT=p/r, and R is a constant to boot.
Other coefficients that will appear presently are also constant, regardless of their dimensions,
and such a gas is called a perfect gas.
In other cases, physical constants can enter in modifications of eqs(3).
The gas may no longer be perfect, but similar arguments give the result that R
is constant in certain generalized versions of Boyle's and Charles' laws. The examples that follow are
familiar ones from the kinetic theory of gases.
Example 1: If m is the mass of a neutron, the density of a gas whose molecular
weight is M is r = mMN, where N(t) is the number of molecules per unit volume.
The equation of state becomes
p = mMNRMT, (8)
where it is now p0, N0
and T0 that are arbitrary and dimensionally independent. Now let
RM = R1/M, and cancel the 'M's
to evaluate R1. From that it follows that
pv = nRT with v=1/r , R = R1
and n º 1/M . (9)
R would be the gas constant for neutrons, and it can't be measured directly.
Example 2: Now consider
p1 = r1RT/M1
and p2 = r2RT/M2. (10)
From kinetic theory it follows when the N1 + N2 molecules are
all that is contained within
the volume, V, that p = p1 + p2 and
r = r1 + r2. From that and eqs(10) it follows that
p = rRT/M
where (1/M) = (1/M1) + (1/M2)
(or n = n1 + n2). (11)
This (sum the n's) applies for any number of components, provided
no chemical reactions are taking place,
no phase changes are taking place (e.g. no H2O is present),
no separation of lighter and heavier components is taking place.
Example 3:
p(v - nb) = nRT (12)
The parameter, b, is there to account for the volume that is not available in V because
the molecules fill some of the space. It depends on details of the structure of the molecules,
and there is no easy formula for it.
Example 4:
(p + n2a/v2)(v -
nb) = nRT (van der Waals) (13)
This one, with a correction for binary intermolecular interactions, is
a rough model that can describe solid, liquid and gaseous phases. The parameter, a, also
depends on molecular structure, and there is no easy formula for it.
To describe phases, consider the coefficient of thermal expansion
and the compressibility. For a perfect gas, v = RT/p, and they are
a º (¶v/¶T)/v = 1/T and
k º (¶v/¶p)/v = -1/p. (14)
In accord with the definition of partial derivatives, compressibility is sometimes called
isothermal compressibility, yet thermal expansion is almost never called isobaric.
For a very rough description of phases: a gas expands to fill V, and a and k are
of equal importance; a is much the more important of the two in a liquid; and neither of them
matters as much in a solid, which has other physical properties.
Now comes thermal energy of a perfect gas, and this is not a standard treatment of it. The equations of
fluid dynamics provide a relation between kinetic energy, gravitational potential energy, and
other kinds of energy that are neither of the first two. The first law of thermodynamics is
the assertion that there is a way to define a thermal potential energy that can be written as
I(p,v,T,c), where the 'c' stands for any number of variables that are not present in a
perfect gas (e.g. concentrations of reacting components).
In general the dimensions of energy are [work]=ML2/T2, and the quantity
that traditionally appears in thermodynamics is energy per unit mass, so (for a perfect gas)
[I]=L2/T2
and I = I0 + lRT + mpv, (15)
where l and m are dimensionless constants. The inclusion of I0 serves to
emphasize that I is potential energy per unit mass. Like g(z-z0), it has no independent value,
and it appears only in relations that describe how p, v, and T are related to one another in situations
where they vary with time. Since pv=RT, we can start with I=I0+CT
where C=(l+m)R and proceed with a specification of how
I(t) = I0 + CT(t-t0) does that.
To see how work works, consider the time derivative of the
equation of state,
pdv + vdp = RdT (16)
In this, 'd' signifies the operation (d/dt), the time derivative, and the dimensions of
the terms in eq(16) are [(work/dt)/M] = [power per unit mass].
What eq(16) indicates is that there are three kinds of work in play; one, thermal, and two, mechanical.
When dT = 0, the mechanical doppelgangers are balanced, with vdp = -pdv.
It is pdv that is the more familiar mechanical work (power per unit mass here), and the first
formulation of the first law of thermodynamics is
dI = CvdT = -pdv + Q. (17)
In that, -pdv is the rate at which mechanical work changes the specific internal energy.
(Note, -pdv decreases I when dv>0 -- think of expansion.)
Q(t) contains the net rate of absorbtion (possibly negative) of radiant energy,
effects of friction (necessarily positive), and other heat sources (e.g. latent heats).
In parallel with the result above, the other result about work and thermal energy is
dH = CpdT = vdp + Q. (18)
Eqs(17 and 18) are definitions of specific heat at constant volume and
specific heat at constant pressure. For a perfect gas they are constants and it follows
from eqs(16, 17 and 18) that
(Cp - Cv)dT = RdT. (19)
Hence Cp = R + Cv, and H and I are doppelgangers too. H is called
the enthalpy.
In fact, H = I + pv in general, and H = I + RT as well for a perfect gas, so
d(I + pv) = (Cv+R)dT =
d(pv) - pdv + Q, (20)
and eq(18) is just an alternate formulation of the first law of thermodynamics. The point of doing both
ways is merely to explain that there are different uses for the C's, but there are no
contradictions. The results from kinetic theory are that Cv is near (3/2)R for spherical
molucules like He, near (5/2)R for rod-shaped ones like N2 and O2, and
near 3R for polyatomic molecules like CO2, CH4 and H2O.
To complete this review of results for a perfect gas, note that the use of p/T = R/v in eq(17) implies
CvdT/T + Rdv/v
= d(Cv ln(T/T0)
+ R ln(v/v0)) = Q/T, (21)
and the use of v/T = R/p in eq(18) implies
CpdT/T - Rdp/p
= d(Cp ln(T/T0)
- R ln(p/p0)) = Q/T. (22)
Both ways, we are lead to define entropy as
S - S0 =
óõt
t0Q/T dt' . (23)
Then, for a perfect gas,
Cv ln(T/T0) + R ln(v/v0) =
Cp ln(T/T0) - R ln(p/p0)
= S - S0 , (24)
Note that S, once evaluated from the definite integral, is S(x,t) when 'x' is present
as well as 't', and the d's in
dI = CvdT = TdS - pdv
and
dH = CpdT = TdS + vdp (25)
will not be restricted to being time-derivatives after coordinates x, y and z have been introduced.
At this point it would be shameful not to introduce the other two thermal potentials, so here they are:
| The four thermal potentials |
| name | args | first law | Maxwell | perfect |
| Internal energy | I(v,S) | dI=TdS-pdv |
(¶T/rv)S=-(¶p/¶S)v | I=CvT |
| Enthalpy | H(p,S)=I+pv | dH=TdS+vdp |
(¶T/¶p)S=(¶v/¶S)p | H=CpT |
| Free energy | F(v,T)=I-TS | dF=-SdT-pdv |
(¶S/¶v)T=(¶p/¶T)v | F=(Cv-S)T |
| Gibbs free energy | G(p,T)=H-TS | dG=-SdT+vdp |
(¶S/¶p)T=-(¶v/¶T)p | G=(Cp-S)T |