6 OUTLINES OF PHYSICAL CHEMISTEY
If we introduce this value into the equation, we get : pv =p v [l + ±(t - 273)]=p v (l + £L-1)
T PqVq
For any given mass of gas, the value of =^-P is a con-
stant usually represented by r, so that we can give the
general equation the simple form :
N.B. — The numerical value of r depends, on the unit of volume and unit of pressure chosen. We shall return to this subject.
Determination of the density of a Gas or Vapour
Having given a certain weight g of a gas, We have to
observe :
v, its volume in cubic centimetres ;
p, its pressure in millimetres oi mercury ;
t, its temperature in degrees Centigrade.
We have then all the data necessary for the calculation of the density of the gas, relatively to air.
Since p v = p v (1 + a t),
therefore v = — -J-—-
��Po(l + at)
will be the volume of the gas at 0° and under normal pressure. The same volume of air would weigh
0-00129 P v , x grams,
since one cubic centimetre of air at 0° and 760 millimetres pressure weighs 0*00129 grams.
�� �
