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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 1 1 V dV dT V V T , or β = 1 T (b) At T = ° = 0 273 C K , this predicts β = = × 1 273 3 66 10 3 1 K K . Experimental values are: β He K = × 3 665 10 3 1 . and β air K = × 3 67 10 3 1 . They agree within 0.06% and 0.2%, respectively. P19.65 For each gas alone, P N kT V 1 1 = and P N kT V 2 2 = and P N kT V 3 3 = , etc. For all gases P V P V P V N N N kT N N N kT PV 1 1 2 2 3 3 1 2 3 1 2 3 + + = + + + + = b g b g and Also, V V V V 1 2 3 = = = = , therefore P P P P = + + 1 2 3 . P19.66 (a) Using the Periodic Table, we find the molecular masses of the air components to be M N u 2 b g = 28 01 . , M O u 2 b g = 32 00 . , M Ar u a f = 39 95 . and M CO u 2 b g = 44 01 . . Thus, the number of moles of each gas in the sample is
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