Page:Scientific Papers of Josiah Willard Gibbs.djvu/417

Rh work done by the conversion of a unit of N2O4 into NO2 under constant pressure is $$a_{2}t$$. Therefore, the ratio of the heat absorbed to the external work done by the conversion of N2O4 into NO2 is $$7181 \div t$$, or 23 at the temperature of 40° centigrade. Let us next consider how much more rapidly this vapor expands with increase of temperature at constant pressure than air. From the necessary relation where $$m$$ denotes the weight of the vapor, and $$k$$ a constant, we obtain  where the suffix p indicates that the differential coefficients are for constant pressure. The last term of this expression evidently denotes the part of the expansion which is due to the conversion of N2O4 into NO2, and the preceding term the expansion which would take place if there were no such conversion, and which is identical with the expansion of the same volume of air under the same circumstances. The ratio of the two terms is $$-\frac{t}{\text{D}}\left( \frac{d\text{D}}{dt} \right)_{p}$$, the numerical value of which for the temperature of 40° is 2.42, as may be found by differentiating equation (10), or, with less precision, from the numbers in the third column of Table I. Let us now suppose that equal volumes of peroxide of nitrogen and of air at the temperature of 40° and the pressure of one atmosphere receive equal infinitesimal increments of temperature under constant pressure. The heat absorbed by the peroxide of nitrogen on account of the conversion of N2O4 into NO2 is 23 times the external work due to the same cause, and this work is 2.42 times the external work done by the expansion of the air. But the heat absorbed by the air in expanding under constant pressure is well known to be 3.5 times the work done. Therefore the heat absorbed on account of the conversion of N2O4 into NO2 is (23 x 2.42 ÷ 3.5 =) 15.9 times the heat absorbed by the air. To obtain the whole heat absorbed by the vapor we must add that which would be required if no conversion took place. At 40° the vapor of peroxide of nitrogen contains about 54 molecules of N2O4 to 46 of NO2, as may easily be calculated from its density. The specific heat for constant pressure of a mixture in such proportions of gases of such molecular formulae, if no chemical action could take place, would be about twice that of the same volume of air. Adding this to the heat absorbed by the chemical action we obtain the final result,—that at 40° and the pressure of the atmosphere the specific