(4) V = v/(1 + 0.00366t). The following problem will serve as an illustration of the application of this equation. The volume of a gas at 20 deg. is 750 cc.; find the volume it will occupy at 0 deg., the pressure remaining constant. In this case, v = 750 cc. and t = 20. By substituting these values, equation (4) becomes V = 750/(1 + 0.00366 x 20) = 698.9 cc. ~Law of Boyle.~ This law expresses the relation between the volume occupied by a gas and the pressure to which it is subjected. It may be stated as follows: _The volume of a gas is inversely proportional to the pressure under which it is measured, provided the temperature of the gas remains constant._ If V represents the volume when subjected to a pressure P and v represents its volume when the pressure is changed to p, then, in accordance with the above law, V : v :: p : P, or VP = vp. In other words, for a given weight of a gas the product of the numbers representing its volume and the pressure to which it is subjected is a constant. Since the pressure of the atmosphere at any point is indicated by the barometric reading, it is convenient in the solution of the problems to substitute the latter for the pressure measured in grams per square centimeter. The average reading of the barometer at the sea level is 760 mm., which corresponds to a pressure of 1033.3 g. per square centimeter. The following problem will serve as an illustration of the application of Boyle's law. A gas occupies a volume of 500 cc. in a laboratory where the barometric reading is 740 mm. What volume would it occupy if the atmospheric pressure changed so that the reading became 750 mm.? Substituting the values in the equation VP = vp, we have 500 x 740 = v x 750, or v = 493.3 cc. ~Variations in the volume of a gas due to changes both in temperature and pressure.~ Inasmuch as corrections must be made as a rule for both temperature and pressure, it is convenient to combine the equations given above for the corrections for each, so that the two corrections may be made in one operation. The following equation is thus obtained: (5) V_{s} = vp/(760(1 + 0.00366t)), in which V_{s} represents the volume of a gas under standard conditions and v, p, and t the volume, pressure, and temperature respectively at which the gas was actually measured. The following problem will serve to illustrate the application of this equation. A gas having a temperature of 20 deg. occupies a volume of 500 cc. when subjected to a pressure indicated by a barometric reading of 740 mm. What volume would this gas occupy under standard conditions? In this problem v = 500, p = 740, and t = 20. Substituting these values in the above equation, we get V_{s} = (500 x 740)/(760 (1 + 0.00366 x 20)) = 453.6 cc. [Illustration: Fig. 8]