Page:International Library of Technology, Volume 93.djvu/52

 — Using formula 3 and substituting,

44. Mixture of Two Quantities of Air Having: Unequal Pressures, Volumes, and Temperatures. If a body of air having a temperature $$\scriptstyle t_1$$, a pressure $$\scriptstyle p_1$$, and a volume $$\scriptstyle v_1$$ is mixed with another volume of air having a temperature $$\scriptstyle t_2$$, a pressure $$\scriptstyle p_2$$, and a volume $$\scriptstyle v_2$$ to form a volume $$\scriptstyle V$$ having a pressure $$\scriptstyle P$$ and a temperature $$\scriptstyle t$$, either the new temperature $$\scriptstyle t$$, the new volume $$\scriptstyle V$$, or the new pressure $$\scriptstyle P$$ may be found, if the other two quantities are known, by the following formula, in which $$\scriptstyle T_1$$, $$\scriptstyle T_2$$, and $$\scriptstyle T$$ are the absolute temperatures corresponding to $$\scriptstyle t_1$$, $$\scriptstyle t_2$$, and $$\scriptstyle t$$:

It follows from this formula that

— Five cubic feet of air having a pressure of 30 pounds per square inch and a temperature of 80° F. is to be compressed, together with 11 cubic feet of air having a pressure of 21 pounds per square inch and a temperature of 45° F., in a vessel whose cubical contents is 8 cubic feet; the new pressure is required to be 45 pounds per square inch. What is the temperature of the mixture?

. — Substituting the given values in formula 4,

. — Fourteen cubic feet of air at 80 pounds pressure, and at a temperature of 100° F., is compressed with 26 cubic feet of air at 60 pounds pressure and 60° F. into a volume of 20 cubic feet; what is the resultant pressure, if the final temperature is 140° F.?