Problem 3.19[Difficulty: 2]Given:Data on partitioned tankFind:Pressure of trapped air required to bring water and mercury levels equal if right air opening is sealedSolution:First we need to determine how far each free surface moves.In the tank of Problem 3.18, the ratio of cross section areas of the partitions is 0.75/3.75 or 1:5. Suppose the water surface (andtherefore the mercury on the left) must move down distance xto bring the water and mercury levels equal. Then by mercury volumeconservation, the mercury free surface (on the right) moves up (0.75/3.75)x= x/5. These two changes in level must cancel the originaldiscrepancy in free surface levels, of (1m + 2.9m) - 3 m = 0.9 m. Hence x+ x/5 = 0.9 m, or x= 0.75 m. The mercury level thus movesup x/5 = 0.15 m.Assuming the air (an ideal gas, pV=RT) in the right behaves isothermally, the new pressure there will be
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