556_Physics ProblemsTechnical Physics

556_Physics ProblemsTechnical Physics - 558*P19.34 P19.35...

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558 Temperature *P19.34 Consider the air in the tank during one discharge process. We suppose that the process is slow enough that the temperature remains constant. Then as the pressure drops from 2.40 atm to 1.20 atm, the volume of the air doubles. During the first discharge, the air volume changes from 1 L to 2 L. Just 1 L of water is expelled and 3 L remains. In the second discharge, the air volume changes from 2 L to 4 L and 2 L of water is sprayed out. In the third discharge, only the last 1 L of water comes out. Were it not for male pattern dumbness, each person could more efficiently use his device by starting with the tank half full of water. P19.35 (a) PV nRT = n PV RT == × = 1013 10 100 8314 293 41 6 5 .. . . Pa m Jmo lK K mol 3 ej e j bg af (b) mn M = 416 289 120 . mol g mol kg , in agreement with the tabulated density of . k gm 3 at 20.0°C. *P19.36 The void volume is 0 765 0 765 0 765 1 27 10 0 2 7 75 10 22 2 5 ... . . . Vr total 3 m m m × = × −− ππ A . Now for the gas remaining
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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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