Elementary Principles 121

# Elementary Principles 121 - 4 3 5.23 Volume of balloon = 10...

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5- 9 5.23 Volume of balloon m m 3 == 4 3 10 4189 3 π bg Moles of gas in balloon nkmo l m 492 R atm 1 kmol 535 R 1 atm m STP kmol 3 3 = ° ° = 4189 3 22 4 5159 . . a. He in balloon: m kmol kg kmol kg He =⋅ = 4 003 2065 .. b g m kg m s N 1 kg m / s N g 22 = = 2065 9 807 1 20 250 . , b. PV n R T n R T n P P n atm 3 atm kmol kmol gas in balloon gas air displaced air air air gas gas di = = ⇒= ⋅= = 1 172 0 F buoyant F cable W total FW kmol 29.0 kg 9.807 m 1 kmol s N 1 N buoyant air displaced 2 kg m s 2 = 172 0 1 48 920 . , Since balloon is stationary, F 1 0 = FF W N kg 9.807 m s N 1 cable buoyant total 2 kg m s 2 =− = + = 48920 2065 150 1 27 20 , c. When cable is released, F = 27200 N M a net tot A = = a N1 k g m / s 2065+150 kg N ms 2 2 27200 12 3 . d. When mass of displaced air equals mass of balloon + helium the balloon stops rising. Need to know how density of air varies with altitude.
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## This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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