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

Short Answer

Expert verified
First, convert the given Celsius temperatures to Kelvin: -50°C = 223.15 K, and 0°C = 273.15 K. Then note the molar mass of each gas: Molar mass of Krypton (Kr) = 83.798 g/mol, and Molar mass of Argon (Ar) = 39.948 g/mol. Make observations comparing gas properties and temperature. Finally, sketch the molecular speed distributions on a single plot: a curve for Krypton at -50°C (slowest distribution, lower peak, wider base), a curve for Krypton at 0°C (higher distribution than Krypton at -50°C, higher peak, narrower base), and a curve for Argon at 0°C (highest distribution among the three, with the highest peak and narrowest base).

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given Celsius temperatures to Kelvin since the Maxwell-Boltzmann distribution uses Kelvin temperatures. We can do this by adding 273.15 to each Celsius temperature: -50°C = -50 + 273.15 = 223.15 K 0°C = 0 + 273.15 = 273.15 K
02

Note the properties of the gases

Now we need to take into account the properties of both gases. For this exercise, we can focus on the molar mass of each gas since it will have the most impact on the molecular speed distribution: Molar mass of Krypton (Kr) = 83.798 g/mol Molar mass of Argon (Ar) = 39.948 g/mol
03

Compare the gas properties and temperature

When comparing the molar masses and temperatures of each situation, we can make some observations: 1. Krypton at -50°C (223.15 K) will have a lower molecular speed distribution than Krypton at 0°C (273.15 K) because its molecules will have less kinetic energy at a lower temperature. 2. Krypton at 0°C (273.15 K) will have a lower molecular speed distribution than Argon at 0°C (273.15 K) because the molar mass of Krypton is greater than the molar mass of Argon. This means that the kinetic energy of Argon's lighter molecules will be higher than Krypton's heavier molecules at the same temperature.
04

Sketch the distributions

Based on the observations made in step 3, we can now sketch the molecular speed distributions on a single plot: 1. Sketch a curve for Krypton at -50°C (223.15 K) with the slowest molecular speed distribution because of its low temperature. This curve should have a lower peak and wider base compared to the other distributions. 2. Sketch a curve for Krypton at 0°C (273.15 K) with a higher molecular speed distribution compared to Krypton at -50°C. However, Krypton at 0°C will still have a slower distribution than Argon at 0°C due to its heavier molecules. This curve should have a higher peak and slightly narrower base than Krypton at -50°C. 3. Sketch a curve for Argon at 0°C (273.15 K) with the highest molecular speed distribution compared to the two Krypton distributions. This is because Argon has lighter molecules and a higher temperature than Krypton at -50°C. The curve should have the highest peak and narrowest base among the three distributions. Now the single plot should have three qualitatively sketched molecular speed distributions for the different gases at varying temperatures.

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Key Concepts

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

Maxwell-Boltzmann Distribution
The Maxwell-Boltzmann distribution is a fundamental concept describing the statistical distribution of speeds among particles in a gas. It is vitally important in the field of thermodynamics and provides insights into how particles' speeds vary at different temperatures.

The distribution curve, when plotted, represents the probability of particles possessing a certain speed at a given temperature. It is characterized by its peak, where the most probable speed is located, and its tails, which indicate the presence of both slow and fast-moving particles. As the temperature of the gas increases, the peak shifts to higher speeds, illustrating that particles have greater kinetic energy and therefore move faster. Conversely, lower temperatures result in slower particle speeds, with the peak shifting to the left.

When we compare the distribution of molecular speeds at different conditions – say, for Krypton gas at two distinct temperatures – we can predict that the curve for the gas at the higher temperature will peak further to the right, signifying higher average speeds due to increased kinetic energy.
Kinetic Molecular Theory
Kinetic Molecular Theory (KMT) provides insight into the behavior of particles in gases and underpins the Maxwell-Boltzmann distribution concept. This theory is based on several key assumptions: gases are composed of a large number of small particles that are far apart relative to their size, and these particles are in constant, random motion. The theory postulates that pressure is due to particle collisions with the container walls and that these collisions are perfectly elastic, meaning no kinetic energy is lost.

Under the KMT framework, the temperature of a gas is directly related to the average kinetic energy of its particles. When the temperature rises, the gas particles gain kinetic energy, leading to an increase in their speed, which is reflected in a widening and shifting of the Maxwell-Boltzmann distribution curve. Importantly, the theory also explains how at any given temperature, lighter gas molecules will travel faster on average than heavier ones, due to their lower mass and the influence of kinetic energy distribution.
Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It has a significant role in determining the behavior of gas particles, as it directly correlates to the kinetic energy distribution among those particles. For instance, argon has a lower molar mass than krypton, meaning that, at a given temperature, argon's molecules will generally move faster than those of krypton.

The influence of molar mass becomes clear when considering its impact on gas behavior under the same environmental conditions. Lower molar mass gases will have a distribution curve that peaks at higher speeds and is narrower, indicating a higher average speed and a tighter speed range. On the other hand, gases with higher molar masses will exhibit a broader distribution, with a peak at slower speeds, as seen with krypton in the given exercise.

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Most popular questions from this chapter

Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.452 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27{ }^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

You have an evacuated container of fixed volume and known mass and introduce a known mass of a gas sample. Measuring the pressure at constant temperature over time, you are surprised to see it slowly dropping. You measure the mass of the gas-filled container and find that the mass is what it should be-gas plus container-and the mass does not change over time, so you do not have a leak. Suggest an explanation for your observations.

After the large eruption of Mount St. Helens in 1980 , gas samples from the volcano were taken by sampling the downwind gas plume. The unfiltered gas samples were passed over a goldcoated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it and then analyzed. In one particular set of experiments scientists found a mercury vapor level of \(1800 \mathrm{ng}\) of Hg per cubic meter in the plume at a gas temperature of \(10^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of Hg vapor in the plume, (b) the number of \(\mathrm{Hg}\) atoms per cubic meter in the gas, \((\mathrm{c})\) the total mass of Hg emitted per day by the volcano if the daily plume volume was \(1600 \mathrm{~km}^{3}\).

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise \(10.55 .\) The unknown vapor had a mass of \(0.846 \mathrm{~g} ;\) the volume of the bulb was \(354 \mathrm{~cm}^{3}\), pressure 752 torr, and temperature \(100{ }^{\circ} \mathrm{C}\). Calculate the molar mass of the unknown vapor.

(a) What are the mole fractions of each component in a mixture of \(15.08 \mathrm{~g}\) of \(\mathrm{O}_{2}, 8.17 \mathrm{~g}\) of \(\mathrm{N}_{2}\), and \(2.64 \mathrm{~g}\) of \(\mathrm{H}_{2} ?\) (b) What is the partial pressure in atm of each component of this mixture if it is held in a 15.50-L vessel at \(15^{\circ} \mathrm{C}\) ?

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