Assuming that the temperature remains constant, what will be the volume of the balloon when it is taken to a mountain resort at an altitude of 2500 m, where the atmospheric pressure is 557 Torr? If not possible, explain why.Įxercise 5.3B Use principles from this section to explain (a) how a beverage is transferred from glass to mouth when one uses a soda straw and (b) why an old-fashioned hand-operated pump cannot raise water from a well if the water level is more than about 30 feet below the pump.Įxample 5.4 A helium-filled party balloon has a volume of 4.50 L at sea level, where the atmospheric pressure is 748 Torr. If possible, also place a barometric pressure of (f) 103 kPa in the order. (d) PNe 735 mmHgĮxample 5.3 continued Exercise 5.3A Place a barometric pressure of (e) 101 kPa into the order of increasing pressures established in Example 5.3. Finally, in the closed-end manometer in (d), the difference in the two mercury levels, 735 mm, is the actual gas pressure. Although there is much less mercury in (c) than in (a), the pressure of the mercury column in (c) is greater than in (a) because of its greater height: 75.0 cm = 750 mm. The pressure of helium in the open-end manometer (b) is slightly greater than Pbar, which is 762 Torr = 762 mmHg and, therefore, greater than the pressure in (a). What is the pressure in atmospheres that is exerted by this depth of water? What is the total pressure the diver experiences at this depth? Explain.Įxample 5.3 A Conceptual Example Without doing calculations, arrange the drawings in Figure 5.5 so that the pressures denoted in red are in increasing order.Īnalysis and Conclusions The pressure in (a) is expressed as the depth of the liquid mercury, that is, 745 mmHg. To measure normal atmospheric pressure with a water-filled barometer, we would have to use one that is more than 10 m tall-as tall as a threestory building!Įxercise 5.2A Calculate the height of a column of carbon tetrachloride, CCl 4 (d = 1.59 g/cm3), that exerts the same pressure as a column of mercury (d = 13.6 g/cm3) 760 mm high.Įxercise 5.2B A diver reaches a depth of 30.0 m. Then we solve the expression forĮxample 5.2 continued Assessment Because water has a much lower density than mercury (about 1/14), it should take a much taller water column to produce the same pressure as a mercury column, and that is what we found. Next we cancel the factor g and substitute the known numerical values into the resulting expression. Solution We start by equating the two liquid pressures. When we equate these two pressures, we will find that there is only one unknown, for which we can solve. Strategy With the above equation, we can establish the pressures exerted by the columns of mercury and water: Hg to Torr (b) 98.2 kPa to Torr (d) 768 Torr to atmĮxercise 5.1B What is the pressure, in kilopascals and in atmospheres, if a force of 1.00 x 102 N is exerted on an area of 5.00 cm2?Įxample 5.2 Calculate the height of a column of water (d = 1.00 g/cm3) that exerts the same pressure as a column of mercury (d = 13.6 g/cm3) 760 mm high. With this conversion factor, we find thatĮxample 5.1 continued Assessment Note that the units kPa cancel and that the pressure expressed in the unit torr is a larger number (by a factor of about 7.5) than the pressure in kilopascals, as expected from the form of the conversion factor.Įxercise 5.1A Carry out the following conversions of pressure units. Solution To convert from kPa to Torr, we need this relationship from Table 5.2: 1 atm = 760 Torr = 101.325 kPa.
Generally we can find such a relationship in Table 5.2.
Strategy To convert a pressure from one unit to another, we need a relationship between the two units that we can use as a conversion factor. What is the pressure expressed in the unit torr? Example 5.1 A Canadian weather report gives the atmospheric pressure as 100.2 kPa.
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