558 Temperature*P19.34Consider the air in the tank during one discharge process. We suppose that the process is slowenough that the temperature remains constant. Then as the pressure drops from 2.40 atm to1.20 atm, the volume of the air doubles. During the first discharge, the air volume changes from 1 Lto 2 L. Just 1 L of water is expelled and 3 L remains. In the second discharge, the air volume changesfrom 2 L to 4 L and 2 L of water is sprayed out. In the third discharge, only the last 1 L of watercomes out. Were it not for male pattern dumbness, each person could more efficiently use his deviceby starting with the tank half full of water.P19.35(a)PVnRT=nPVRT==×⋅=1013 10100831429341 65....PamJmolKKmol3ejejbgaf(b)mnM=416289120.molg molkg, in agreement with the tabulated density of. kgm3at 20.0°C.*P19.36The void volume is 0 7650 7650 7651 27 100 27 75 102225......Vrtotal3m mm×=×−−ππA. Now forthe 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 .