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Chapter 10: Problem 84

At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, \(\lambda\) is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it \(R_{\mathrm{mfp}}\), like the ideal-gas constant) and define units for \(R_{\mathrm{mfp}^{-}}\)

Short Answer

Expert verified
The formula for the mean free path (λ) of a gas molecule, taking into account its direct proportionality to temperature (T), inverse proportionality to pressure (P), and inverse proportionality to the square of the diameter of the gas molecules (d), is given by: \[λ = R_{\mathrm{mfp}} \times T \times \frac{1}{P} \times \frac{1}{d^2}\] where the proportionality constant R_mfp has the units of \(L/(K \times Pa)\).

Step by step solution

01

Relation between λ and Temperature

According to the problem, at constant pressure, the mean free path (λ) of a gas molecule is directly proportional to temperature (T). So, we can write this as: \[λ ∝ T\]
02

Relation between λ and Pressure

Next, at constant temperature, the problem states that λ is inversely proportional to pressure (P). So, we can write this as: \[λ ∝ \frac{1}{P}\]
03

Relation between λ and Diameter of Gas Molecules

Finally, if we compare two different gas molecules at the same temperature and pressure, λ is inversely proportional to the square of the diameter of the gas molecules (d). Therefore, we can write this as: \[λ ∝ \frac{1}{d^2}\]
04

Combining the Relations

Now, we will combine the three proportional relationships to form a single formula for λ: \[λ ∝ T \times \frac{1}{P} \times \frac{1}{d^2}\]
05

Introducing the Proportionality Constant

To convert this proportionality relation into an equation, we need to introduce a proportionality constant, called R_mfp, which will have the same functionality as the ideal-gas constant in the ideal gas equation. So the final formula will look like this: \[λ = R_{\mathrm{mfp}} \times T \times \frac{1}{P} \times \frac{1}{d^2}\]
06

Defining the Units for R_mfp

Since the units for λ should be length (L), for temperature (T) it's Kelvin (K), for pressure (P) it's Pascal (Pa), and for the diameter of gas molecules (d) it's also length (L), we need to define the units for R_mfp in such a way that the units in the formula balance out properly. Therefore R_mfp should have units: \[[R_{\mathrm{mfp}}] = \frac{L^3}{K \times Pa \times L^2} = \frac{L}{K \times Pa}\] So, the required formula for the mean free path of a gas molecule is given by: \[λ = R_{\mathrm{mfp}} \times T \times \frac{1}{P} \times \frac{1}{d^2}\] where R_mfp has the units of \(L/(K \times Pa)\).

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Most popular questions from this chapter

(a) The compound 1 -iodododecane is a nonvolatile liquid with a density of \(1.20 \mathrm{~g} / \mathrm{mL}\). The density of mercury is $13.6 \mathrm{~g} / \mathrm{mL}$. What do you predict for the height of a barometer column based on 1 -iodododecane, when the atmospheric pressure is 749 torr? \((\mathbf{b})\) What is the pressure, in atmospheres, on the body of a diver if he is $21 \mathrm{ft}$ below the surface of the water when the atmospheric pressure is 742 torr?

In the contact process, sulfur dioxide and oxygen gas react to form sulfur trioxide as follows: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ At a certain temperature and pressure, \(50 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) reacts with \(25 \mathrm{~L}\) of \(\mathrm{O}_{2}\). If all the \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are consumed, what volume of \(\mathrm{SO}_{3}\), at the same temperature and pressure, will be produced?

In a "Kipp generator", hydrogen gas is produced when zinc flakes react with hydrochloric acid: $$2 \mathrm{HCl}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ If \(30.0 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(20^{\circ} \mathrm{C}\) and a barometric pressure of \(101.33 \mathrm{kPa}\), how many grams of \(\mathrm{Zn}\) have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

A scuba diver's tank contains \(2.50 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(11.0 \mathrm{~L}\). (a) Calculate the gas pressure inside the tank at \(10^{\circ} \mathrm{C}\). (b) What volume would this oxygen occupy at \(25^{\circ} \mathrm{C}\) and \(101.33 \mathrm{kPa} ?\)

A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(202.7 \mathrm{kPa}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.
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