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Chapter 5: Problem 95

Discuss the following phenomena in terms of the gas laws: (a) the pressure increase in an automobile tire on a hot day; (b) the "popping" of a paper bag; (c) the expansion of a weather balloon as it rises in the air; (d) the loud noise heard when a lightbulb shatters.

Short Answer

Expert verified
Situation A uses Gay-Lussac's Law to explain how temperature affects pressure in an automobile tire. Situation B uses Boyle's Law to explain how volume and pressure relate in the 'popping' of a paper bag. Situation C also uses Boyle's Law, demonstrating how a decrease in pressure can increase volume in a weather balloon. Situation D, like B, applies Boyle's law, showing how a sudden change in pressure, when a lightbulb shatters, causes a loud noise.

Step by step solution

01

Situation A: Automobile tire on a hot day

The pressure inside a tire increases on a hot day due to Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature when volume is held constant. In this case, as the temperature increases (from the day becoming hot), so does the pressure inside the tire.
02

Situation B: Popping of a paper bag

The 'popping' of a paper bag is a result of a rapid change in volume and therefore pressure - related to Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when the temperature is held constant. The rapid increase in volume when the bag pops causes a decrease in pressure, producing the 'pop' sound.
03

Situation C: Expansion of a weather balloon

The expansion of a weather balloon as it rises in the air can be explained by Boyle's Law - as the balloon rises, the outside air pressure decreases, causing the volume of the gas inside the balloon to expand. Because the law says that pressure and volume are inversely proportional (assuming a constant temperature), a decrease in one will cause an increase in the other.
04

Situation D: Noise when a lightbulb shatters

The loud noise heard when a lightbulb shatters is due to a sudden change in pressure. When the lightbulb shatters, the gas inside the bulb escapes rapidly, creating a change in pressure - this can be again related to Boyle's law.

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Most popular questions from this chapter

The following apparatus can be used to measure atomic and molecular speed. Suppose that a beam of metal atoms is directed at a rotating cylinder in a vacuum. A small opening in the cylinder allows the atoms to strike a target area. Because the cylinder is rotating, atoms traveling at different speeds will strike the target at different positions. In time, a layer of the metal will deposit on the target area, and the variation in its thickness is found to correspond to Maxwell's speed distribution. In one experiment it is found that at \(850^{\circ} \mathrm{C}\) some bismuth (Bi) atoms struck the target at a point \(2.80 \mathrm{~cm}\) from the spot directly opposite the slit. The diameter of the cylinder is \(15.0 \mathrm{~cm}\) and it is rotating at 130 revolutions per second. (a) Calculate the speed (m/s) at which the target is moving. (Hint: The circumference of a circle is given by \(2 \pi r,\) where \(r\) is the radius. \()\) (b) Calculate the time (in seconds) it takes for the target to travel \(2.80 \mathrm{~cm} .\) (c) Determine the speed of the Bi atoms. Compare your result in (c) with the \(u_{\mathrm{rms}}\) of \(\mathrm{Bi}\) at \(850^{\circ} \mathrm{C}\). Comment on the difference.

Cite two pieces of evidence to show that gases do not behave ideally under all conditions.

A relation known as the barometric formula is useful for estimating the change in atmospheric pressure with altitude. The formula is given by \(P=P_{0} e^{-g .1 h / R T},\) where \(P\) and \(P_{0}\) are the pressures at height \(h\) and sea level, respectively; \(g\) is the acceleration due to gravity \(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) ; \mathscr{M}\) is the average molar mass of air \((29.0 \mathrm{~g} / \mathrm{mol}) ;\) and \(R\) is the gas constant. Calculate the atmospheric pressure in atm at a height of \(5.0 \mathrm{~km}\), assuming the temperature is constant at \(5^{\circ} \mathrm{C}\) and \(P_{0}=1.0 \mathrm{~atm} .\)

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.

The boiling point of liquid nitrogen is \(-196^{\circ} \mathrm{C}\). On the basis of this information alone, do you think nitrogen is an ideal gas?
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