Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

You have two containers each with 1 mole of xenon gas at \(15^{\circ} \mathrm{C} .\) Container \(\mathrm{A}\) has a volume of \(3.0 \mathrm{L},\) and container \(\mathrm{B}\) has a volume of 1.0 L. Explain how the following quantities compare between the two containers. a. the average kinetic energy of the Xe atoms b. the force with which the Xe atoms collide with the container walls c. the root mean square velocity of the Xe atoms d. the collision frequency of the Xe atoms (with other atoms) e. the pressure of the Xe sample

Short Answer

Expert verified
The average kinetic energy of the Xe atoms in both containers is the same. The force with which the Xe atoms collide with the container walls is three times greater in container A compared to container B. The root mean square velocities of the Xe atoms in both containers are the same. The collision frequency of the Xe atoms is three times greater in container B than in container A. The pressure in container A is three times greater than container B.

Step by step solution

01

a. The average kinetic energy of the Xe atoms:

The average kinetic energy (KE) of a gas is directly related to its temperature. Since both containers have the same temperature, the average kinetic energy of the xenon atoms in both the containers will be the same. Mathematically, KE = \( \frac{3}{2} kT \) where k is the Boltzmann constant and T is the temperature. Since the temperature of both containers is the same, their average kinetic energy will also be the same.
02

b. The force with which the Xe atoms collide with the container walls:

For a gas, the force exerted by its particles on the container wall is directly proportional to the pressure and the volume of the container. The ideal gas law states that: PV = nRT Where P is the pressure, V is the volume, n is moles of gas, R is the gas constant, and T is temperature. Since the temperature and the number of moles of gas are the same in both containers, we can compare their pressures: \( \frac{P_A}{P_B} = \frac{V_B}{V_A} = \frac{1.0}{3.0} \) So, the pressure in container A is three times greater than container B. Therefore, the force with which xenon atoms collide with the container wall is three times greater in container A compared to container B.
03

c. The root mean square velocity of the Xe atoms:

The root mean square velocity (v_rms) of a gas is given by: \( v_{rms} = \sqrt{\frac{3RT}{M}} \) Where R is the gas constant, T is temperature, and M is the molar mass of the gas. Since both containers have the same temperature and gas, their root mean square velocities will be the same.
04

d. The collision frequency of the Xe atoms (with other atoms):

The collision frequency of a gas depends on its concentration. A higher concentration of gas particles (in terms of particles per unit volume) results in more frequent collisions between the gas molecules. For an ideal gas, the concentration (C) is: C = n/V The concentration of xenon atoms in container A is: \( C_A = \frac{1}{3} \) In container B, the concentration of xenon atoms is: \( C_B = \frac{1}{1} \) So, the collision frequency of Xe atoms in container B is three times greater than in container A.
05

e. The pressure of the Xe sample:

We can use the ideal gas law to find the pressure of the xenon gas in both containers: PV = nRT For container A, \(P_A = \frac{nRT_A}{V_A} = \frac{1\cdot R\cdot (15+273)(K)}{3.0L}\) For container B, \(P_B = \frac{nRT_B}{V_B} = \frac{1\cdot R\cdot (15+273)(K)}{1.0L}\) As we found earlier, the pressure in container A is three times greater than container B.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Theory of Gases
The kinetic theory of gases provides a comprehensive framework to help us understand the behavior of gases. At its core, this theory presents gases as a collection of tiny particles, or molecules, that are in constant random motion. This movement of molecules is the primary source of force when they collide with the walls of their container.

These collisions are what we measure as gas pressure. A pivotal detail of the kinetic theory is that the temperature of a gas is actually a measure of the average kinetic energy of the gas molecules. Therefore, when we look at two containers of xenon gas at the same temperature, like in our exercise, the kinetic theory tells us that the average kinetic energy of the gas particles in both containers will be the same, even if the volumes are different.

How Volume and Pressure Interact

The kinetic theory also implies that gas molecules will have more room to move in a larger volume, leading to fewer collisions with the container walls per unit time compared to a gas in a smaller volume at the same temperature. However, the average kinetic energy—and thus the temperature—remains consistent across different volumes when no energy is added or removed.

This understanding is crucial in explaining why, in our exercise, container A, with a larger volume, exerts the same average kinetic energy as container B but with different pressures and collision forces against the container walls.
Ideal Gas Law
The ideal gas law, represented by the equation PV = nRT, provides the relationship between pressure (P), volume (V), gas moles (n), the gas constant (R), and temperature (T). This equation can be a powerful tool in predicting the behavior of an ideal gas under various conditions.

The term 'ideal gas' refers to a hypothetical gas whose molecules exhibit no intermolecular attraction and occupy no volume. While no gas is truly ideal, many real gases behave in a manner closely approximating this at high temperatures and low pressures. The ideal gas law is derived from combining several empirical gas laws, such as Boyle's Law, Charles’s Law, and Avogadro's Law. Each of these describes gas behavior in response to changes in variable pairs (such as pressure and volume, volume and temperature, or number of moles and volume).

Applying the Ideal Gas Law

In our exercise, we use the ideal gas law to determine the pressure and force exerted by the gas molecules within the two containers. Despite container A having a larger volume, the pressure in container B is found to be three times greater due to its smaller volume. This is because, according to the ideal gas law, pressure is inversely proportional to volume when temperature and moles of gas remain constant.
Root Mean Square Velocity
Root mean square velocity (RMS velocity) is a way to quantify the speed of gas particles. Statistically, it reflects the average speed of a collection of gas particles at a given temperature, regardless of their individual directions of travel.

The RMS velocity is calculated using the formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \),where R is the universal gas constant, T is the absolute temperature in Kelvin, and M is the molar mass of the gas. This equation arises from the kinetic theory of gases and plays an indispensable role in understanding gas dynamics.

Significance in Different Conditions

Given our textbook exercise, this calculation shows that the root mean square velocity of the xenon atoms in both containers remains consistent because the temperature and the molar mass of xenon do not change. It's essential to note that volume and the number of moles don't affect the RMS velocity directly; it's solely dependent on the type of gas (its molar mass) and the temperature of the system.

So, even though container A and container B have different volumes and pressures, the RMS velocity of the xenon gas in both scenarios stays the same, illustrating the uniformity of particle speed within the ideal gas model at a constant temperature.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be produced by the following reaction: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$ Hydrogen at STP flows into a reactor at a rate of \(16.0 \mathrm{L} / \mathrm{min.}\) Carbon monoxide at STP flows into the reactor at a rate of \(25.0 \mathrm{L} / \mathrm{min.}\). If \(5.30\) \(\mathrm{g}\) methanol is produced per minute, what is the percent yield of the reaction?

Consider the following chemical equation. $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ If \(25.0 \mathrm{mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to \(400^{\circ} \mathrm{C}\) in a nickel reaction vessel. A \(100.0\) -\(\mathrm{mL}\) nickel container is filled with xenon and fluorine, giving partial pressures of \(1.24\) atm and \(10.10\) atm, respectively, at a temperature of \(25^{\circ} \mathrm{C}\). The reaction vessel is heated to \(400^{\circ} \mathrm{C}\) to cause a reaction to occur and then cooled to a temperature at which \(\mathrm{F}_{2}\) is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining \(\mathrm{F}_{2}\) gas is transferred to another A \(100.0\) -\(\mathrm{mL}\) nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(7.62\) atm. Assuming all of the xenon has reacted, what is the formula of the product?

An organic compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) ii. A 65.2 -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ), giving \(35.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) at \(740 .\) torr and \(25^{\circ} \mathrm{C}\) iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{mUmin.}\) The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{mL} / \mathrm{min.}\) What is the molecular formula of the compound?

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free