565_Physics ProblemsTechnical Physics

565_Physics ProblemsTechnical Physics - Chapter 19 FG nR IJ...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 19 567 P19.64 (a) From PV nRT = , the volume is: V nR P T = F H G I K J Therefore, when pressure is held constant, dV dT nR P V T == Thus, β F H G I K J = F H G I K J 11 V dV dT V V T , or = 1 T (b) At T =° = 02 7 3 C K, this predicts × −− 1 273 366 10 31 K K . Experimental values are: He K 3665 10 . and air K 367 10 . They agree within 0.06% and 0.2%, respectively. P19.65 For each gas alone, P NkT V 1 1 = and P V 2 2 = and P V 3 3 = , etc. For all gases PV N N N kT NNN k TP V 1 1 2 2 3 3 123 ++ = + + = …… bg and Also, VVV V === = , therefore PP P P =++ . P19.66 (a) Using the Periodic Table, we find the molecular masses of the air components to be M N u 2 = 28 01 ., M O u 2 = 32 00 M Ar u af = 39 95 . and M CO u 2 = 44 01 .. Thus, the number of moles of each gas in the sample is n n n nO N
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online