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A flask contains a mixture of neon (Ne), krypton (Kr), and radon (Rn) gases. Compare (a) the average kinetic energies of the three types of atoms and (b) the root-mean-square speeds. (Hint: Appendix D shows the molar mass (in \(\mathrm{g} / \mathrm{mol}\) ) of each element under the chemical symbol for that element.)

Short Answer

Expert verified
The average kinetic energies of Ne, Kr, and Rn are equal. The root-mean-square speed is highest for Ne, lower for Kr, and lowest for Rn, assuming they are at the same temperature.

Step by step solution

01

Evaluate the Average Kinetic Energy

We know that the average kinetic energy of a gas molecule is given by the equation: \[ KE_{avg} = \frac{3}{2} kT \]. Here, \(k\) represents Boltzmann's constant and \(T\) represents the temperature. If we assume that the three gases are at the same temperature, then their average kinetic energies will be the same since it is solely dependent on temperature and not on the type or molar mass of the gas. Therefore, the average kinetic energies of Ne, Kr, and Rn are equal.
02

Calculate the Root-Mean-Square Speeds

The root-mean-square speed is given by the formula: \[v_{rms} = \sqrt{\frac{3kT}{m}}\], where \(m\) is the molar mass of the gas. Smaller atoms are lighter, thus they move faster. Conversely, larger atoms are heavier and move slower. Therefore, the root-mean-square speed is inversely proportional to the square root of the molar mass. Appendix D should provide the necessary values for molar masses. Substituting the molar masses of Ne, Kr, and Rn into the equation will yield the root-mean-square speeds for each.
03

Compare the Root-Mean-Square Speeds

Using the values obtained from step 2, the root-mean-square speeds for each of the gases can be compared. The gas with the lightest molar mass (Ne) will have the highest root-mean-square speed, while the gas with the heaviest molar mass (Rn) will have the lowest root-mean-square speed.

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Key Concepts

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

Average Kinetic Energy
When we talk about the kinetic energy of gases, one of the fundamental concepts to grasp is the average kinetic energy. This is the energy that gas molecules have on average due to their motion. In a sample of gas, myriad molecules move at different speeds and in various directions. However, when we average their kinetic energies, we get a value that represents the whole gas.

The average kinetic energy of gas molecules can be calculated with the equation: \[ KE_{avg} = \frac{3}{2} kT \]Here, \(k\) is known as Boltzmann's constant, and \(T\) is the absolute temperature in Kelvin. Importantly, the average kinetic energy is directly proportional to the temperature. This means that regardless of the type of gas, if its temperature remains the same, its average kinetic energy stays constant as well. If we consider our given flask containing different gases at the same temperature, it becomes clear that the average kinetic energy is identical for neon (Ne), krypton (Kr), and radon (Rn).

Understanding the Implications

Imagine an invisible dance of molecules all moving and colliding. Although each type of molecule may dance to a different beat, the average kinetic energy reveals the overall intensity of the dance. This is a critical concept in thermodynamics and highlights the equality of energy distribution at a given temperature.
Root-Mean-Square Speed of Gas Molecules
Diving deeper into the behavior of gas molecules, we encounter the concept of root-mean-square speed (rms speed). This value gives us an indication of the molecules' speed in a gas. Unlike average kinetic energy, the rms speed does take into account the mass of the molecules.

The equation for the rms speed is: \[v_{rms} = \sqrt{\frac{3kT}{m}}\]where \(m\) is the molar mass of the gas molecule, and the other terms remain consistent with the kinetic energy equation. The square root is used to assure the speed comes out as a positive value, as it represents magnitude without direction.

Speed Varies with Molar Mass

As the equation shows, the rms speed is inversely proportional to the square root of the molar mass of a gas molecule. Consequently, lighter molecules, like those of neon, zip around faster compared to heavier molecules, like those of radon. Understanding the relationship between molar mass and rms speed is essential for predicting how different gases will behave under the same conditions. It's a dance of sorts where lighter partners can move swiftly, while heavier ones take a more measured pace.
Boltzmann's Constant
At the heart of the kinetic molecular theory lies Boltzmann's constant (\(k\)), a fundamental physical constant that relates the average kinetic energy of particles in a gas to the temperature of the gas. Its value is approximately \(1.38 \times 10^{-23} \) J/K (joules per kelvin).

Boltzmann's constant acts as a bridge between the macroscopic world that we can observe (like temperature) and the microscopic world of atoms and molecules in motion. By using this constant, we can link the temperature we feel to the unseen agitation of a sea of particles.

The Universal Gas Constant

It's also noteworthy to mention the universal gas constant (\(R\)), which is used in the ideal gas law. It's related to Boltzmann's constant by the number of molecules in a mole (Avogadro's number), specifically \(R = k \times N_A\). Through understanding Boltzmann's constant, we gather insight not only into energy per particle but also the larger thermodynamic properties of gases.
Molar Mass and Gas Properties
Gas properties can vary widely depending on their molar mass, which is the mass of one mole of a substance usually expressed in g/mol. For gases, the molar mass is critical in determining both the rms speed and how they will conduct themselves under various conditions.

In our earlier discussion on rms speeds, we learned that the speed is inversely related to the square root of the molar mass. This concept is integral when examining how gases diffuse or how quickly a gas will effuse through a small opening. Gases of lower molar mass spread more rapidly due to their higher speeds.

Understanding Molar Mass Effects

Reflect on how different balloons filled with helium (a very light gas) and carbon dioxide (a heavier gas) will behave. The helium balloon rises and diffuses quickly into the atmosphere, while the carbon dioxide-filled balloon rises more slowly and diffuses less rapidly. This illustrates the tangible effects of molar mass on gas properties in the world around us.

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Most popular questions from this chapter

(a) Calculate the mass of nitrogen present in a volume of \(3000 \mathrm{~cm}^{3}\) if the gas is at \(22.0^{\circ} \mathrm{C}\) and the absolute pressure of \(2.00 \times 10^{-13}\) atm is a partial vacuum easily obtained in laboratories. (b) What is the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the \(\mathrm{N}_{2}\) ?

A diver observes a bubble of air rising from the bottom of a lake (where the absolute pressure is \(3.50 \mathrm{~atm}\) ) to the surface (where the pressure is 1.00 atm \() .\) The temperature at the bottom is \(4.0^{\circ} \mathrm{C},\) and the temperature at the surface is \(23.0^{\circ} \mathrm{C}\). (a) What is the ratio of the volume of the bubble as it reaches the surface to its volume at the bottom? (b) Would it be safe for the diver to hold his breath while ascending from the bottom of the lake to the surface? Why or why not?

A parcel of air over a campfire feels an upward buoyant force because the heated air is less dense than the surrounding air. By estimating the acceleration of the air immediately above a fire, one can estimate the fire's temperature. The mass of a volume \(V\) of air is \(n M_{\text {air }},\) where \(n\) is the number of moles of air molecules in the volume and \(M_{\text {air }}\) is the molar mass of air. The net upward force on a parcel of air above a fire is roughly given by \(\left(m_{\text {out }}-m_{\text {in }}\right) g,\) where \(m_{\text {out }}\) is the mass of a volume of ambient air and \(m_{\text {in }}\) is the mass of a similar volume of air in the hot zone. (a) Use the ideal-gas law, along with the knowledge that the pressure of the air above the fire is the same as that of the ambient air, to derive an expression for the acceleration \(a\) of an air parcel as a function of \(\left(T_{\text {out }} / T_{\text {in }}\right),\) where \(T_{\text {in }}\) is the absolute temperature of the air above the fire and \(T_{\text {out }}\) is the absolute temperature of the ambient air. (b) Rearrange your formula from part (a) to obtain an expression for \(T_{\text {in }}\) as a function of \(T_{\text {out }}\) and \(a\). (c) Based on your experience with campfires, estimate the acceleration of the air above the fire by comparing in your mind the upward trajectory of sparks with the acceleration of falling objects. Thus you can estimate \(a\) as a multiple of \(g .\) (d) Assuming an ambient temperature of \(15^{\circ} \mathrm{C}\), use your formula and your estimate of \(a\) to estimate the temperature of the fire.

A cylindrical diving bell has a radius of \(750 \mathrm{~cm}\) and a height of \(2.50 \mathrm{~m}\). The bell includes a top compartment that holds an undersea adventurer. A bottom compartment separated from the top by a sturdy grating holds a tank of compressed air with a valve to release air into the bell, a second valve that can release air from the bell into the sea, a third valve that regulates the entry of seawater for ballast, a pump that removes the ballast to increase buoyancy, and an electric heater that maintains a constant temperature of \(20.0^{\circ} \mathrm{C}\). The total mass of the bell and all of its apparatuses is \(4350 \mathrm{~kg}\). The density of seawater is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). (a) An \(80.0 \mathrm{~kg}\) adventurer enters the bell. How many liters of seawater should be moved into the bell so that it is neutrally buoyant? (b) By carefully regulating ballast, the bell is made to descend into the sea at a rate of \(1.0 \mathrm{~m} / \mathrm{s}\). Compressed air is released from the tank to raise the pressure in the bell to match the pressure of the seawater outside the bell. As the bell descends, at what rate should air be released through the first valve? (Hint: Derive an expression for the number of moles of air in the bell \(n\) as a function of depth \(y ;\) then differentiate this to obtain \(d n / d t\) as a function of \(d y / d t .)\) (c) If the compressed air tank is a fully loaded, specially designed, \(600 \mathrm{ft}^{3}\) tank, which means it contains that volume of air at standard temperature and pressure ( \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) ), how deep can the bell descend?

Estimate the number of atoms in the body of a \(50 \mathrm{~kg}\) physics student. Note that the human body is mostly water, which has molar mass \(18.0 \mathrm{~g} / \mathrm{mol}\), and that each water molecule contains three atoms.

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