565_Physics ProblemsTechnical Physics

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

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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
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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