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Chapter 5: Problem 83

The average distance traveled by a molecule between successive collisions is called mean free path. For a given amount of a gas, how does the mean free path of a gas depend on (a) density, (b) temperature at constant volume, (c) pressure at constant temperature, (d) volume at constant temperature, and (e) size of the atoms?

Short Answer

Expert verified
The mean free path depends inverse proportionally on density, pressure at constant temperature and size of the atoms, but is directly proportional on volume at constant temperature. There's no significant effect of temperature at constant volume on the mean free path.

Step by step solution

01

(a) Dependency on Density

The mean free path of a gas is inversely proportional to its density. When the density of a gas increases, the number of molecules per unit volume also increases, resulting in more frequent collisions and hence, a shorter mean free path.
02

(b) Dependency on Temperature at Constant Volume

The mean free path is not directly affected by the temperature of a gas at constant volume. This is due to the fact that a change in temperature at constant volume does not significantly alter the density of the gas, thus it does not affect the average distance that molecules travel between collisions.
03

(c) Dependency on Pressure at Constant Temperature

The mean free path of a gas is inversely proportional to its pressure at constant temperature. When the pressure of a gas increases at constant temperature, the density of the gas also increases, leading to more frequent collisions and hence, a shorter mean free path.
04

(d) Dependency on Volume at Constant Temperature

The mean free path of a gas is directly proportional to its volume at constant temperature. When the volume of a gas increases at constant temperature, the density of the gas decreases, leading to less frequent collisions and hence, a longer mean free path.
05

(e) Dependency on Size of the Atoms

The mean free path of a gas is inversely proportional to the size of its atoms. The larger the size of the atoms, the greater the chances of collisions, and thus, the shorter the mean free path.

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Most popular questions from this chapter

A gas evolved during the fermentation of glucose (wine making) has a volume of \(0.78 \mathrm{~L}\) at \(20.1^{\circ} \mathrm{C}\) and 1.00 atm. What was the volume of this gas at the fermentation temperature of \(36.5^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) pressure?

A compound of \(\mathrm{P}\) and \(\mathrm{F}\) was analyzed as follows: Heating \(0.2324 \mathrm{~g}\) of the compound in a \(378-\mathrm{cm}^{3}\) container turned all of it to gas, which had a pressure of \(97.3 \mathrm{mmHg}\) at \(77^{\circ} \mathrm{C}\). Then the gas was mixed with calcium chloride solution, which turned all of the \(\mathrm{F}\) to \(0.2631 \mathrm{~g}\) of \(\mathrm{CaF}_{2}\). Determine the molecular formula of the compound.

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

A barometer having a cross-sectional area of \(1.00 \mathrm{~cm}^{2}\) at sea level measures a pressure of \(76.0 \mathrm{~cm}\) of mercury. The pressure exerted by this column of mercury is equal to the pressure exerted by all the air on \(1 \mathrm{~cm}^{2}\) of Earth's surface. Given that the density of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\) and the average radius of Earth is \(6371 \mathrm{~km},\) calculate the total mass of Earth's atmosphere in kilograms. (Hint: The surface area of a sphere is \(4 \pi r^{2},\) where \(r\) is the radius of the sphere.)

Apply your knowledge of the kinetic theory of gases to the following situations. (a) Two flasks of volumes \(V_{1}\) and \(V_{2}\left(V_{2}>V_{1}\right)\) contain the same number of helium atoms at the same temperature. (i) Compare the root-mean-square (rms) speeds and average kinetic energies of the helium (He) atoms in the flasks. (ii) Compare the frequency and the force with which the He atoms collide with the walls of their containers. (b) Equal numbers of He atoms are placed in two flasks of the same volume at temperatures \(T_{1}\) and \(T_{2}\left(T_{2}>T_{1}\right) .\) (i) Compare the rms speeds of the atoms in the two flasks. (ii) Compare the frequency and the force with which the He atoms collide with the walls of their containers. (c) Equal numbers of He and neon (Ne) atoms are placed in two flasks of the same volume, and the temperature of both gases is \(74^{\circ} \mathrm{C}\). Comment on the validity of the following statements: (i) The rms speed of He is equal to that of Ne. (ii) The average kinetic energies of the two gases are equal. (iii) The rms speed of each He atom is \(1.47 \times 10^{3} \mathrm{~m} / \mathrm{s}\)
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