Kinetic-Molecular Theory of an Ideal GasDue: 8:03pm on Thursday, November 17, 2011

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Equipartition Theorem and Microscopic Motion

Learning Goal: To understand the Equipartition Theorem and its implications for the mechanical motion of small objects.

In statistical mechanics, heat is the random motion of the microscopic world. The average kinetic or potential energy of each degree of freedom of the microscopic world therefore depends on the temperature. If heat is added, molecules increase their translational and rotational speeds, and the atoms constituting the molecules vibrate with larger amplitude about their equilibrium positions. It is a fact of nature that the energy of each degree of freedom is determined solely by the temperature. The Equipartition Theorem states this quantitatively:

The average energy associated with each degree of freedom in a system at absolute

temperature is , where is Boltzmann's constant.

The average energy of the ith degree of freedom is , where the angle brackets represent "average" or "mean" values of the enclosed variable. A "degree of freedom" corresponds to any dynamical variable that appears quadratically in the energy. For instance,

is the kinetic energy of a gas particle of mass with velocity component along the x axis.

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically accessible quantum state of a system has equal probability of being populated, which in turn leads to the Boltzmann distribution for a system in thermal equilibrium. From the standpoint of an introductory physics course, equipartition is best regarded as a principle that is justified by observation.

In this problem we first investigate the particle model of an ideal gas. An ideal gas has no interactions among its particles, and so its internal energy is entirely "random" kinetic energy. If we consider the gas as a system, its internal energy is analogous to the energy stored in a spring. If one end of the gas container is fitted with a sliding piston, the pressure of the gas on the piston can do useful work. In fact, the empirically discovered ideal gas law,

, enables us to calculate this pressure. This rule of nature is remarkable in that the

value of the mass does not affect the energy (or the pressure) of the gas particles' motion, only the temperature. It provides strong evidence for the validity of the Equipartition Theorem as applied to a particle gas:

or

for a particle constrained by a spring whose spring constant is . If a molecule has moment

of inertia about an axis and is rotating with angular velocity about that axis with

associated rotational kinetic energy , that angular velocity represents another degree of freedom.

Part A

Consider a monatomic gas of particles each with mass . What is , the root mean square (rms) of the x component of velocity of the gas particles if the gas is at an

absolute temperature ?

Hint A.1

Equipartition for one velocity component

Hint not displayed

Express your answer in terms of , , , and other given quantities.

ANSWER:

Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three

dimensions. Find , the rms speed if the gas is at an absolute temperature .

Hint B.1

Equipartition Theorem for three degrees of freedom

Hint not displayed

Express your answer in terms of , , , and other given quantities.

ANSWER:

Correct

Part C

What is the rms speed of molecules in air at ? Air is composed mostly of molecules, so you may assume that it has molecules of average atomic mass

.Express your answer in meters per second, to the nearest integer.

ANSWER:    =

493Correct  

Not surprisingly, this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air. Commercial jets have a maximum speed of about 85% of the speed of sound (Mach 0.85), which is consistent with this.

Now consider a rigid dumbbell with two masses, each of mass , spaced a distance apart.

Part D

Find , the rms angular speed of the dumbbell about a single axis (taken to be the x

axis), assuming that it is in equilibrium at temperature .

Hint D.1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem?Express your answer in terms of the x component of the angular velocity and the

moment of inertia about this axis

ANSWER:    = Correct

Hint D.2

Moment of inertia of a dumbbell

What is , the moment of inertia of the dumbbell?

Hint D.2.1 Finding of a dumbbell

Hint not displayed

Express in terms of and .

ANSWER:

   = Correct

Express the rms angular speed in terms of , , , , and other given quantities.

ANSWER:    =

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature (

)? Assume that for this molecule is . Take the atomic mass of to be

.

Express numerically in hertz, to three significant figures.

ANSWER:    =

9.47×1011

Correct Hz

This frequency corresponds to light of wavelength 0.3 mm and is in the far-infrared region of the electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced with microwave techniques). However, because the molecules in the air are homonuclear diatomic molecules, their symmetry prevents them from interacting strongly with radiation of this frequency. Only nonhomonuclear molecules such as water vapor absorb energy at infrared frequencies.

 

Average Spacing of Gas Molecules

Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with each molecule at the center of a small cube.

Part A

What is the length of an edge of each small cube if adjacent cubes touch but don't overlap?

Hint A.1

How to approach the problem

Hint not displayed

Hint A.2

Calculate the volume per mole

Hint not displayed

Hint A.3

Calculate the volume per molecule

Hint not displayed

Hint A.4

The edge length of a cube

Hint not displayed

Express your answer numerically in meters.

ANSWER    = 3.45×10−9  

: Correct

For carbon dioxide gas, the physical volume of a molecule (based on the van der Waals

equation constant ) is approximately . This implies that the

linear dimension of the molecule is only about .

Comparing this number to the result of your calculation above, you can see that the size of the molecule is less than one eighth of the length of the cube edge that surrounds the molecule, which is also the average distance separating one molecule from the next. A small molecular size, as compared to the distance between molecules, is a necessary assumption in the kinetic-molecular model of an ideal gas.

 

Degrees of Freedom

Thermodynamics deals with the macroscopic properties of materials. Scientists can make quantitative predictions about these macroscopic properties by thinking on a microscopic scale. Kinetic theory and statistical mechanics provide a way to relate molecular models to thermodynamics. Predicting the heat capacities of gases at a constant volume from the number of degrees of freedom of a gas molecule is one example of the predictive power of molecular models.

The molar specific heat of a gas at a constant volume is the quantity of energy required to

raise the temperature of one mole of gas by one degree while the volume remains the same. Mathematically,

,

where is the number of moles of gas, is the change in internal energy, and is the change in temperature.

Kinetic theory tells us that the temperature of a gas is directly proportional to the total kinetic energy of the molecules in the gas. The equipartition theorem says that each degree of

freedom of a molecule has an average kinetic energy equal to , where is Boltzmann's

constant . When summed over the entire gas, this gives , where

is the ideal gas constant, for each molecular degree of freedom.

Part A

Using the equipartition theorem, determine the molar specific heat, , of a gas in which each molecule has degrees of freedom.

Hint A.1

How to approach the problem

Hint not displayed

Hint A.2

Monatomic gas: an example

Hint not displayed

Express your answer in terms of and .

ANSWER:    =

Correct  

Experimentally, kinetic theory and the equipartition theorem do a good job of predicting the specific heat of many gases at room temperature as shown in the chart below.

Molecule Degrees of Freedom

Argon (Ar) 12.5 1.50 3Helium (He) 12.5 1.50 3

Carbon Monoxide (CO) 20.7 2.49 5Hydrogen (H2) 20.4 2.45 5

Nitric Oxide (NO) 20.9 2.51 5Hydrogen Sulfide (H2S) 25.92 3.12 6

Water Vapor (H2O) 25.24 3.03 6

For a monatomic gas, there are only three translational degrees of freedom, so is about

. Diatomic molecules and linear molecules have three translational degrees of freedom and two rotational components to the motion (rotation about the axis of the molecule does

not contribute much except at high temperatures), giving a value of about . Nonlinear molecules have three translational and three rotational degrees of freedom, which gives

.

Polyatomic molecules also have vibrational degrees of freedom, where the bonds between atoms can vibrate back and forth like a spring being compressed and released as well as possible side-to-side swinging motion from the bonds bending. The vibrational motion does not normally contribute to the degrees of freedom until a high temperature of 400 degrees Celsius or more is reached.

Part B

Given the molar specific heat of a gas at constant volume, you can determine the number of degrees of freedom that are energetically accessible.

For example, at room temperature cis-2-butene, , has molar specific heat

. How many degrees of freedom of cis-2-butene are energetically accessible?

Hint B.1

How to approach the problem

Hint not displayed

Express your answer numerically to the nearest integer.

ANSWER:    =

17Correct

 

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 .

Note that 1.0 of diatomic hydrogen at 50 has a total translational kinetic energy of

4000 .

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-

square speed for diatomic oxygen at 50 is:

Hint A.1

Definition of root-mean-square speed

Hint not displayed

Choose the correct value of .

ANSWER:

none of the above

Correct

Part B

The total translational kinetic energy of 1.0 mole of diatomic oxygen at 50 is:

Hint B.1

Definition of average and total kinetic energy

Hint not displayed

Choose the correct total translational kinetic energy.

ANSWER:

none of the above

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 . The root-

mean-square speed for diatomic hydrogen at 100 is:Choose the correct .

ANSWER:

none of the above

Correct

 

Particle Gas Review

A particle gas consists of monatomic particles each of mass all contained in a volume

at temperature . Your answers should be written in terms of the Boltzmann constant and

Avagadro's number rather than .

Part A

Find , the average speed squared for each particle.

Hint A.1

how to approach the problem

Hint not displayed

Hint A.2 Find for each particle

Hint not displayed

Hint A.3

Relating the , , and velocities

Hint not displayed

Express the average speed squared in terms of the gas temperature and any other given quantities.

ANSWER:

   = Correct

Part B

Find , the internal energy of the gas.

Hint B.1

How to approach the problem

Hint not displayed

Hint B.2

Kinetic energy of a single gas particle

Hint not displayed

Express the internal energy in terms of the gas temperature and any other given quantities.

ANSWER:    =

Correct

Part C

Find , the molar heat capacity (heat capacity per mole) of the gas at constant volume.

Hint C.1

Definition of heat capacity

Hint not displayed

Express the molar heat capacity in terms of and .

ANSWER:    =

Correct

You can use the molar heat capacity to find the total heat capacity of the gas at constant volume . Since is the total heat capacity of the gas, it is just the heat capacity per mole

multiplied by the number of moles of gas particles . Hence, .

Part D

Express the pressure of the gas in terms of its energy density .

Hint D.1

How to approach the problem

Hint not displayed

Hint D.2

Find an expression for pressure

Hint not displayed

Enter the numerical factor that multiplies in your expression for to at least three significant figures.

ANSWER:    =

0.667Correct

 

As an interesting aside, note that for an ultra-relativistic particle gas, such as a gas of photons,

.

Now imagine that the mass of each gas particle is increased by a factor of 3. All other information given in the problem introduction remains the same.

Part E

What will be the ratio of the new molar mass to the old molar mass ?

ANSWER:

   =3Correct

Part F

What will be the ratio of the new rms speed to the old rms speed ?

Hint F.1 Definition of rms speed

Hint not displayed

ANSWER:

   =

0.577Correct

So if the internal energy (temperature) remains the same, when the mass of the particles increases, the velocity correspondingly decreases. Observe that the pressure and so the average force that the particles exert on the walls also remains the same.

Part G

What will be the ratio of the new molar heat capacity to the old molar heat capacity ?

Hint G.1 How to approach the problem

Hint not displayed

ANSWER:

   =

1Correct

In fact, the molar heat capacity is (almost) the same for all monoatomic gases, as the general expression derived earlier shows. For example, helium, neon, and argon all have almost the same molar heat capacity. The tiny differences in the values tell us that the ideal gas equation describes gases pretty well, but it is not perfect.

 

Velocity and Energy Scaling

Helium atoms have a mass of and oxygen molecules have a mass of , where is

defined as an atomic mass unit ( ). Compare a gas of helium atoms to a gas of oxygen molecules.

Part A

At what gas temperature would the average translational kinetic energy of a helium atom be equal to that of an oxygen molecule in a gas of temperature 300 K?

Hint A.1

Find the energy associated with one degree of freedom

Hint not displayed

Hint A.2

Total translational kinetic energy

Hint not displayed

Express the temperature numerically in kelvins.

ANSWER:    =

300Correct

 K

Part B

At what gas temperature would the root-mean-square (rms) speed of a helium atom be equal to that of an oxygen molecule in a gas at 300 K?

Hint B.1

Find the rms speed

Hint not displayed

State your answer numerically, in kelvins, to the nearest integer.

ANSWER:

   =

38Correct

 K

Note that 38 K is is a very low temperature. At this temperature, helium still behaves approximately like an ideal gas, as do hydrogen and neon, but all other gasses would be frozen solid!

 

± Dust Equipartitions

Small dust particles suspended in air seem to dance randomly about, a phenomenon called Brownian motion.

For this problem you will need to know Boltzmann's constant: .

Part A

What would you expect the mean translational kinetic energy of such particles to be if they are in air at a temperature of 290 K?

Hint A.1

Kinetic energy in terms of average velocity

Hint not displayed

Hint A.2

Equipartition Theorem

Hint not displayed

Hint A.3

Use the Equipartition Theorem

Hint not displayed

Express the mean translational kinetic energy numerically, in joules, to two significant

figures. Note that has been factored out already to make your answer simpler.

ANSWER:    =

6.0Correct

 

Part B

Find an expression for the rms (root-mean-square) speed of these particles, assuming

them to be spheres of diameter and density

Hint B.1

Find the rms speed in terms of temperature

Hint not displayed

Hint B.2

Find the mass of the particles

Hint not displayed

Express the rms speed in terms of , , , , and .

ANSWER:    =

Correct

Part C

Now calculate the rms (root-mean-square) speed of these particles, assuming them to be

spheres of diameter and density . The mass of such a dust

particle is .Express your answer in millimeters per second to one decimal place only.

ANSWER:    =

0.3Correct

 

This speed is several orders of magnitude smaller than the typical velocities of gas molecules at this temperature (which are of the order of hundreds of meters per second). This is simply because the mass of these particles is much larger than the mass of typical gas molecules. For particles larger than the ones described here, the weight can no longer be ignored. Such particles tend to settle quite quickly on account of their weight. Then such a calculation is no longer valid.

 

± Gas Scaling

When doing numerical calculations involving temperature, you need to pay particular attention to the temperature scale you are using. In general, you should use the Kelvin scale

(for which represents absolute zero) in such calculations. This is because the standard thermodynamic equations (i.e., the ideal gas law and the formula for energy of a gas in terms

of temperature) assume that zero degrees represents absolute zero.

If you are given temperatures measured in units other than kelvins, convert them to kelvins before plugging them into these equations. (You may then want to convert back into the initial temperature unit to give your answer.)

Part A

The average kinetic energy of the molecules of an ideal gas at has the value . At

what temperature (in degrees Celsius) will the average kinetic energy of the same gas be

twice this value, ?

Hint A.1

Formula for energy in terms of temperature

Hint not displayed

Hint A.2

Convert from Celsius to Kelvin scale

Hint not displayed

Express the temperature to the nearest integer.

ANSWER:

   =

293Correct  

Part B

The molecules in an ideal gas at have a root-mean-square (rms) speed . At what

temperature (in degrees Celsius) will the molecules have twice the rms speed, ?

Hint B.1

What is rms speed?

Hint not displayed

Hint B.2

Find the change in translational kinetic energy

Hint not displayed

Express the temperature to the nearest integer.

ANSWER:

   =

859Correct  

 

± The Speed of Nitrogen Molecules

The kinetic theory of gases states that the kinetic energy of a gas is directly proportional to the temperature of the gas. A relationship between the microscopic properties of the gas molecules and the macroscopic properties of the gas can be derived using the following

assumptions:

The gas is composed of pointlike particles separated by comparatively large distances.

The gas molecules are in continual random motion with collisions being perfectly elastic.

The gas molecules exert no long-range forces on each other.

One of the most important microscopic properties of gas molecules is velocity. There are several different ways to describe statistically the average velocity of a molecule in a gas.

The most obvious measure is the average velocity . However, since the molecules in a gas are moving in random directions, the average velocity is approximately zero. Another

measure of velocity is , the average squared velocity. Since the square of velocity is always positive, this measure does not average to zero over the entire gas. A third measure is

the root-mean-square (rms) speed, , equal to the square root of . The rms speed is a good approximation of the the typical speed of the molecules in a gas.

This histogram shows a theoretical

distribution of speeds of molecules in a sample of nitrogen ( ) gas. In this problem, you'll use the histogram to compute properties of the gas.

Part A

What is the average speed of the molecules in the gas?

Hint A.1

How to use the histogram

Hint not displayed

Hint A.2

More on computing the average

Hint not displayed

Express your answer numerically to three significant digits.

ANSWER:    =

474Correct  

Part B

Because the kinetic energy of a single molecule is related to its velocity squared, the best measure of the kinetic energy of the entire gas is obtained by computing the mean squared

velocity, , or its square root . The quantity is more common than because it has the dimensions of velocity instead of the less-familiar velocity-squared.

What is the rms speed of the molecules in the nitrogen gas?

Hint B.1

How to approach the problem

Hint not displayed

Hint B.2

Find the mean square velocity

Hint not displayed

Express your answer numerically to three significant digits.

ANSWER:    =

519Correct  

A speed of 519 is comparable to that of a bullet shot from a handgun. In contrast, a

Boeing 747 jet airliner has a maximum air speed of 270 and the speed of sound in air is

only about 330 . The speed of sound in air must be slower than the average speed of the molecules because it is the movement of the molecules that transmits sound.

Part C

What is the temperature of the sample of gas described in the histogram?

Hint C.1

How to approach the problem

Hint not displayed

Hint C.2

Find the molar mass of N2

Hint not displayed

Express your answer in degrees Celsius to three significant figues.

ANSWER:    =

29.4Correct  

Part D

The histogram used in this problem is obviously only an approximation of the true distribution of velocities in a gas. In reality, the molecules span a continuous range of velocities. For a given temperature, the majority of the molecules have a speed near the average speed, with a few molecules traveling very fast or very slow. To good approximation, the speeds of molecules in a gas follow what is known as the Maxwell-Boltzmann distribution. This applet allows you to see the curves for the Maxwell-Boltzmann distribution at many different temperatures. It also lets you move a small interval around on the histogram to highlight all of the molecules within the speed range of that part of the histogram.

Which of the following describes the qualitative behavior of the Maxwell-Boltzmann distribution as temperature increases? You will have to press the "reset" button on the applet before you can change the temperature using the thermometer on the right side.

ANSWER:

The peak moves to the right, while the distribution becomes more spread out.

The peak moves to the right, while the distribution becomes less spread out.

The peak moves to the left, while the distribution becomes more spread out.

The peak moves to the left, while the distribution becomes less spread out.

Correct

At the higher temperatures, the peak of the curve shifts to the right, indicating a higher average velocity. The peaks of the higher temperature curves are also broader, indicating that a greater percentage of molecules are traveling at a higher velocity than in the low-temperature case.

Score Summary:Your score on this assignment is 101.7%.You received 91.52 out of a possible total of 90 points.

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