_v_ = _c_ (267 + _t_)/(_p_),
in which _v_ is this volume, _t_ the temperature, _p_ the pressure, and
_c_ a constant quantity depending on the weight of the vapor and on the
units chosen. We give here the table of the volumes occupied by a gramme
of vapor formed at different temperatures, and consequently under
different pressures.
───────────────────────┬───────────────────────┬───────────────────────
_t_ │ _p_ │ _v_
or degrees Centigrade. │or tension of the vapor│ or volume of a gramme
│ expressed in │ of vapor expressed in
│millimetres of mercury.│ litres.
───────────────────────┼───────────────────────┼───────────────────────
° │ mm. │ lit.
0│ 5.060 │ 185.0
20│ 17.32 │ 58.2
40│ 53.00 │ 20.4
60│ 144.6 │ 7.96
80│ 352.1 │ 3.47
100│ 760.0 │ 1.70
───────────────────────┴───────────────────────┴───────────────────────
The first two columns of this table are taken from the _Traité de
Physique_ of M. Biot (vol. i., p. 272 and 531). The third is calculated
by means of the above formula, and in accordance with the result of
experiment, indicating that water vaporized under atmospheric pressure
occupies a space 1700 times as great as in the liquid state.
By using three numbers of the first column and three corresponding
numbers of the third column, we can easily determine the constants of
our equation
_t_ = (_A_ + _B_ log _v_)/(_A′_ + _B′_ log _v_).
We will not enter into the details of the calculation necessary to
determine these quantities. It is sufficient to say that the following
values,
_A_ = 2268, _A′_ = 19.64,
_B_ = −1000, _B′_ = 3.30,
satisfy fairly well the prescribed conditions, so that the equation
_t_ = (2268 − 1000 log _v_)/(19.64 + 3.30 log _v_)
expresses very nearly the relation which exists between the volume of
the vapor and its temperature. We may remark here that the quantity _B′_
is positive and very small, which tends to confirm this proposition—that
the specific heat of an elastic fluid increases with the volume, but
follows a slow progression.
NOTE E.—Were we to admit the constancy of the specific heat of a gas
when its volume does not change, but when its temperature varies,
analysis would show a relation between the motive power and the
thermometric degree. We will show how this is, and this will also give
us occasion to show how some of the propositions established above
should be expressed in algebraic language.
Let _r_ be the quantity of motive power produced by the expansion of a
given quantity of air passing from the volume of one litre to the volume
of _v_ litres under constant temperature. If _v_ increases by the
infinitely small quantity _dv_, _r_ will increase by the quantity _dr_,
which, according to the nature of motive power, will be equal to the
increase _dv_ of volume multiplied by the expansive force which the
elastic fluid then possesses; let _p_ be this expansive force. We should
have the equation
_dr_ = _pdv_. (1)
Let us suppose the constant temperature under which the dilatation takes
place equal to _t_ degrees Centigrade. If we call _q_ the elastic force
of the air occupying the volume 1 litre at the same temperature _t_, we
shall have, according to the law of Mariotte,
(_v_)/(1) = (_q_)/(_p_), whence _p_ = (_q_)/(_v_).
If now _P_ is the elastic force of this same air at the constant volume
1, but at the temperature zero, we shall have, according to the rule of
M. Gay-Lussac,
_q_ = _P_ + _P_ (_t_)/(267) = (_P_)/(267)(267 + _t_);
whence
_q_ = _p_ = (_P_)/(267) (267 + _t_)/(_v_).
If, to abridge, we call _N_ the quantity (_P_)/(267), the equation would
become
_p_ = _N_ (_t_ + 267)/(_v_);
whence we deduce, according to equation (1),
_dr_ = _N_ (_t_ + 267)/(_v_)_dv_.
Regarding _t_ as constant, and taking the integral of the two numbers,
we shall have
_r_ = _N_(_t_ + 267) log _v_ + _C_.