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

On a single plot, qualitatively sketch the distribution of molecular speeds for (a) \(\mathrm{Kr}(\mathrm{g})\) at \(-50^{\circ} \mathrm{C}\), (b) \(\mathrm{Kr}(\mathrm{g})\) at \(0^{\circ} \mathrm{C}\), (c) \(\operatorname{Ar}(g)\) at \(0^{\circ} \mathrm{C}\). [Section 10.7]

Short Answer

Expert verified
The molecular speed distributions for the given gases can be qualitatively sketched as follows: 1. \(\mathrm{Kr}(g)\) at 223 K: This distribution will be the narrowest and shifted to the lowest speeds due to its low temperature and high molar mass. 2. \(\mathrm{Kr}(g)\) at 273 K: This distribution will be broader and shifted to slightly higher speeds because of its higher temperature, but it will still be narrower and shifted to lower speeds than \(\operatorname{Ar}(g)\) at 273 K due to its higher molar mass. 3. \(\operatorname{Ar}(g)\) at 273 K: This distribution will be the broadest and shifted to the highest speeds due to its lower molar mass. The final plot should show these distributions qualitatively, illustrating the effects of temperature and molar mass on the Maxwell-Boltzmann distribution of molecular speeds.

Step by step solution

01

Recall Maxwell-Boltzmann Distribution

The Maxwell-Boltzmann distribution is used to describe the distribution of molecule speeds in a given gas. The distribution depends on the gas's temperature and molar mass, where higher temperatures lead to broader distributions and a shift to higher speeds, while heavier gases have a narrower distribution shifted to lower speeds.
02

Convert Temperatures to Kelvin

When working with temperature in gas behaviors, it is important to use the Kelvin scale. Convert the given temperatures to Kelvin. \(a) -50^{\circ}C = 223 K\) \(b) 0^{\circ}C = 273 K\) \(c) 0^{\circ}C = 273 K\)
03

Temperature Effects on Distribution

Let's consider the effect of temperature on the speed distribution for the gases. Higher temperatures cause the distribution to broaden and shift to higher speeds. In this exercise, we have two different temperatures: 223 K and 273 K. Thus, the distributions for \(\mathrm{Kr}(g)\) at 223 K will be narrower and shifted to lower speeds than the distribution for \(\mathrm{Kr}(g)\) at 273 K.
04

Molar Mass Effects on Distribution

Now let's consider the effect of molar mass on the distribution of molecular speeds. Heavier gases have a narrower distribution that shifts to lower speeds. In this exercise, we have two different gases, \(\mathrm{Kr}(g)\) and \(\operatorname{Ar}(g)\). Since \(\mathrm{Kr}\) is heavier than \(\operatorname{Ar}\) (molar mass of \(\mathrm{Kr} = 83.8 g/mol\) and molar mass of \(\operatorname{Ar} = 39.9 g/mol\)), the distribution of molecular speeds for \(\mathrm{Kr}(g)\) will be narrower and shifted to lower speeds compared to \(\operatorname{Ar}(g)\) at the same temperature.
05

Qualitatively Sketch Distribution

With the understanding of how temperature and molar mass affect the distribution of molecular speeds, we can now qualitatively sketch the molecular speed distributions for these gases in a single plot. Be sure to follow the rules discussed regarding the influence of temperature and molar mass. 1. \(\mathrm{Kr}(g)\) at 223 K will have the narrowest distribution, shifted to the lowest speeds (due to low temperature and high molar mass). 2. \(\mathrm{Kr}(g)\) at 273 K will have a broader distribution and shifted to slightly higher speeds (due to higher temperature) but still narrower and shifted to lower speeds than \(\operatorname{Ar}(g)\) at 273 K (due to higher molar mass). 3. \(\operatorname{Ar}(g)\) at 273 K will have the broadest distribution, shifted to the highest speeds (due to lower molar mass). The final plot should show these three distributions qualitatively and help visualize the effects of temperature and molar mass in the Maxwell-Boltzmann distribution of molecular speeds.

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.

Molecular Speed Distribution
Understanding how gas particle speeds are distributed is crucial in grasping the behavior of gases. The Maxwell-Boltzmann distribution represents this spread of velocities among the molecules in a gas. By visualizing this distribution, we can observe that most molecules move at moderate speeds, with fewer molecules at the extremes of very slow or very fast speeds.

The shape of this distribution curve is not static; it shifts and changes based on various factors such as temperature and the molar mass of the gas, which we'll explore in the following sections. In essence, knowing the molecular speed distribution for a gas at given conditions can provide insights into many properties of the gas, from diffusion rates to kinetic energy.
Temperature Effects on Gas Behavior
Temperature exerts a significant influence on how gas particles behave. As temperature rises, particles gain kinetic energy, which increases their speed. This relationship is directly reflected in the Maxwell-Boltzmann distribution. At higher temperatures, the peak of the distribution curve flattens and broadens, indicating a greater proportion of molecules are moving at higher speeds.

When comparing gases such as krypton (Kr) at different temperatures, the distribution curve for the higher temperature will always be broader and shifted towards the right, implying faster average speeds. This concept explains why gases expand when heated—the increased molecular speeds cause the particles to collide more forcefully and frequently with the container's walls.
Molar Mass Impact on Gas Particles
The molar mass of gas particles has a pronounced effect on the speed distribution of those particles. Heavier molecules, with higher molar masses, move slower on average than lighter molecules due to having more mass. Consequently, the Maxwell-Boltzmann distribution for a heavier gas like krypton would be narrower and peak at a lower speed compared to a lighter gas like argon at the same temperature.

Considering the molar mass in our examination of gases provides a deeper understanding of diffusion and effusion rates. A heavier gas like Kr will diffuse more slowly compared to a lighter gas like Ar because its molecular speed is inherently lower due to its higher molar mass. Grasping these relationships between molar mass, molecular speed, and temperature is instrumental in predicting gas behavior in various physical and chemical processes.

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

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In \(1 \mathrm{hr}\) the average cockroach running at \(0.08 \mathrm{~km} / \mathrm{hr}\) consumed \(0.8 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at 1 atm pressure and \(24^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in \(1 \mathrm{hr}\) by a \(5.2-\mathrm{g}\) cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 1-qt fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, will the cockroach consume more than \(20 \%\) of the available \(\mathrm{O}_{2}\) in a 48-hr period? (Air is \(21 \mathrm{~mol} \% \mathrm{O}_{2}\).)

Hurricane Wilma of 2005 is the most intense hurricane on record in the Atlantic basin, with a low-pressure reading of 882 mbar (millibars). Convert this reading into (a) atmospheres, (b) torr, and (c) inches of \(\mathrm{Hg}\) -

Consider a mixture of two gases, \(A\) and \(B\), confined in a closed vessel. A quantity of a third gas, \(C\), is added to the same vessel at the same temperature. How does the addition of gas \(C\) affect the following: (a) the partial pressure of gas A, (b) the total pressure in the vessel, (c) the mole fraction of gas B?

\(10.73\) A quantity of \(\mathrm{N}_{2}\) gas originally held at \(5.25 \mathrm{~atm}\) pressure in a \(1.00\) - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5\) - \(\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25\) atm and \(26^{\circ} \mathrm{C}\) in a \(5.00\)-L container is transferred to this same container. What is the total pressure in the new container?

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L}\); in other words, rate is the amount that diffuses over the time it takes to diffuse.)

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