Problem 3.19 - Problem 3.19 [Difficulty: 2] Given: Data on...

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Problem 3.19 [Difficulty: 2] Given: Data on partitioned tank Find: Pressure of trapped air required to bring water and mercury levels equal if right air opening is sealed Solution: 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 (and therefore the mercury on the left) must move down distance x to bring the water and mercury levels equal. Then by mercury volume conservation, the mercury free surface (on the right) moves up (0.75/3.75) x = x /5. These two changes in level must cancel the original discrepancy 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 moves up 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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