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Chapter 5: Problem 135

Under the same conditions of temperature and pressure, why does \(1 \mathrm{~L}\) of moist air weigh less than \(1 \mathrm{~L}\) of dry air? In weather forecasts, an oncoming lowpressure front usually means imminent rainfall. Explain.

Short Answer

Expert verified
Moist air weighs less than dry air because it contains more of the lighter water vapor molecules and less of the heavier nitrogen and oxygen molecules. A low-pressure front usually means imminent rainfall because as the less dense air in these systems rise and cools, it creates clouds from which precipitation can fall.

Step by step solution

01

Understand why moist air weighs less than dry air

When air is dry, it is primarily composed of nitrogen (molecular weight 28) and oxygen (molecular weight 32). When air is moist, it has a significant amount of water vapor (molecular weight 18), which is lighter than both nitrogen and oxygen. Taking this into account, 1 liter of moist air will weigh less than 1 liter of dry air because it contains more water vapor and less of the heavier gases. This simple relationship is due to the ideal gas law principle where at a given temperature and pressure, equal volumes of different gases contain the same number of molecules.
02

Explain why a low pressure front usually means imminent rainfall

Low-pressure systems are characterized by relatively less dense air masses that rise. As this air rises, it cools. Cooler air cannot hold as much water vapor as warmer air, so the excess water vapor condenses around tiny particles in the air, forming clouds. From these clouds, precipitation, such as rain, often fall. Hence, a low pressure front usually signifies the possibility for rain.

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Most popular questions from this chapter

(a) Show that the pressure exerted by a fluid \(P\) (in pascals) is given by \(P=h d g,\) where \(h\) is the column of the fluid in meters, \(d\) is the density in \(\mathrm{kg} / \mathrm{m}^{3},\) and \(g\) is the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) (Hint: See Appendix 1.) (b) The volume of an air bubble that starts at the bottom of a lake at \(5.24^{\circ} \mathrm{C}\) increases by a factor of 6 as it rises to the surface of water where the temperature is \(18.73^{\circ} \mathrm{C}\) and the air pressure is 0.973 atm. The density of the lake water is \(1.02 \mathrm{~g} / \mathrm{cm}^{3}\). Use the equation in (a) to determine the depth of the lake in meters.

Describe how you would measure, by either chemical or physical means, the partial pressures of a mixture of gases of the following composition: (a) \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2,}(\mathrm{~b}) \mathrm{He}\) and \(\mathrm{N}_{2}\)

Helium is mixed with oxygen gas for deep-sea divers. Calculate the percent by volume of oxygen gas in the mixture if the diver has to submerge to a depth where the total pressure is 4.2 atm. The partial pressure of oxygen is maintained at 0.20 atm at this depth.

One way to gain a physical understanding of \(b\) in the van der Waals equation is to calculate the "excluded volume." Assume that the distance of closest approach between two similar atoms is the sum of their radii \((2 r) .\) (a) Calculate the volume around each atom into which the center of another atom cannot penetrate. (b) From your result in (a), calculate the excluded volume for 1 mole of the atoms, which is the constant \(b\). How does this volume compare with the sum of the volumes of 1 mole of the atoms?

A certain anesthetic contains 64.9 percent \(\mathrm{C}, 13.5\) percent \(\mathrm{H},\) and 21.6 percent \(\mathrm{O}\) by mass. At \(120^{\circ} \mathrm{C}\) and \(750 \mathrm{mmHg}, 1.00 \mathrm{~L}\) of the gaseous compound weighs \(2.30 \mathrm{~g}\). What is the molecular formula of the compound?
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