Kinetic-Molecular Theory of an Ideal GasDue: 8:03pm on Thursday, November 17, 2011 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy [Switch to Standard Assignment View] 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 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 with velocity component along . molecules. . Correct Part C What is the rms speed of molecules in air at ? Air is composed mostly of so you may assume that it has molecules of average atomic mass . and other given quantities. . dimensions.1 Equipartition Theorem for three degrees of freedom Hint not displayed Express your answer in terms of ANSWER: . Find . . and other given quantities. the rms speed if the gas is at an absolute temperature Hint B.1 Equipartition for one velocity component Hint not displayed Express your answer in terms of ANSWER: . Correct Part B Now consider the same system--a monatomic gas of particles of mass --except in three . that angular velocity represents another degree of freedom. If a molecule has moment of inertia about an axis and is rotating with angular velocity about that axis with associated . 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 rotational kinetic energy 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. . .mass does not affect the energy (or the pressure) of the gas particles' motion. Now consider a rigid dumbbell with two masses. .Express your answer in meters per second. each of mass Part D Find . . ANSWER: = Correct Express the rms angular speed in terms of ANSWER: = Correct Part E What is the typical rotational frequency )? Assume that for this molecule is for a molecule like at room temperature ( to be . which is consistent with this.85).2.1 Finding of a dumbbell Hint not displayed Express in terms of and . ANSWER: 493 = Correct Not surprisingly. .2 Moment of inertia of a dumbbell What is . the rms angular speed of the dumbbell about a single axis (taken to be the x . assuming that it is in equilibrium at temperature Hint D. Take the atomic mass of . axis). . and other given quantities. to the nearest integer. spaced a distance apart. Commercial jets have a maximum speed of about 85% of the speed of sound (Mach 0. the moment of inertia of the dumbbell? Hint D. this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air. .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. 1 How to approach the problem Hint not displayed Hint A. Average Spacing of Gas Molecules Consider an ideal gas at 27.3 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). . Only nonhomonuclear molecules such as water vapor absorb energy at infrared frequencies.47×1011 = Correct Hz This frequency corresponds to light of wavelength 0. However. 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. the physical volume of a molecule (based on the van der Waals equation constant ) is approximately . with each molecule at the center of a small cube.45×10−9 = Correct For carbon dioxide gas. as compared to the distance between molecules. because the molecules in the air are homonuclear diatomic molecules. which is also the average distance separating one molecule from the next. This implies that the linear dimension of the molecule is only about Comparing this number to the result of your calculation above. ANSWER: 3.4 The edge length of a cube Hint not displayed Express your answer numerically in meters. ANSWER: 9. their symmetry prevents them from interacting strongly with radiation of this frequency.3 Calculate the volume per molecule Hint not displayed Hint A. A small molecular size. Part A What is the length of an edge of each small cube if adjacent cubes touch but don't overlap? Hint A.0 degrees Celsius and 1.2 Calculate the volume per mole Hint not displayed Hint A..00 atmosphere pressure. to three significant figures. is a necessary assumption in . Imagine the molecules to be uniformly spaced. Express numerically in hertz. each molecule has degrees of freedom. Kinetic theory and statistical mechanics provide a way to relate molecular models to thermodynamics. where is the number of moles of gas. When summed over the entire gas. Hint A. Part A Using the equipartition theorem. for each molecular degree of freedom. determine the molar specific heat.1 How to approach the problem Hint not displayed Hint A. 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. is the change in internal energy. Degrees of Freedom Thermodynamics deals with the macroscopic properties of materials. The molar specific heat raise the temperature Mathematically. . where .the kinetic-molecular model of an ideal gas. change in temperature. of a gas in which . where is Boltzmann's . Scientists can make quantitative predictions about these macroscopic properties by thinking on a microscopic scale. this gives is the ideal gas constant.2 Monatomic gas: an example Hint not displayed Express your answer in terms of and . . The equipartition theorem says that each degree of freedom of a molecule has an average kinetic energy equal to constant . of a gas at a constant volume is the quantity of energy required to of one mole of gas by one degree while the volume remains the same. and is the Kinetic theory tells us that the temperature of a gas is directly proportional to the total kinetic energy of the molecules in the gas. 12 3. there are only three translational degrees of freedom.4 20. giving a value of about . 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. Polyatomic molecules also have vibrational degrees of freedom. For example. Part B Given the molar specific heat of a gas at constant volume.50 1.49 2. has molar specific heat .9 25. How many degrees of freedom of cis-2-butene are energetically . 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. Nonlinear molecules have three translational and three rotational degrees of freedom. so .45 2. 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). at room temperature cis-2-butene. which gives .5 20.24 1.51 3. .7 20. The vibrational motion does not normally contribute to the degrees of freedom until a high temperature of 400 degrees Celsius or more is reached.5 12. you can determine the number of degrees of freedom that are energetically accessible.50 2.92 25.ANSWER: = Correct Experimentally.03 Degrees of Freedom 3 3 5 5 5 6 6 is about For a monatomic gas. Molecule Argon (Ar) Helium (He) Carbon Monoxide (CO) Hydrogen (H2) Nitric Oxide (NO) Hydrogen Sulfide (H2S) Water Vapor (H2O) 12. 1 How to approach the problem Hint not displayed Express your answer numerically to the nearest integer.0 mole of diatomic oxygen at 50 Hint B. has a total translational kinetic energy of 4000 none of the above Correct Part B The total translational kinetic energy of 1.accessible? Hint B. of diatomic hydrogen at 50 is 2000 . Part A Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. ANSWER: 17 = Correct Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 Note that 1.0 . The root-meansquare speed for diatomic oxygen at 50 is: Hint A. ANSWER: is: .1 Definition of root-mean-square speed Hint not displayed Choose the correct value of ANSWER: .1 Definition of average and total kinetic energy Hint not displayed Choose the correct total translational kinetic energy. The root-mean- none of the above Correct Particle Gas Review A particle gas consists of at temperature Avagadro's number Part A monatomic particles each of mass . the average speed squared for each particle.2 Find for each particle . Hint A. Your answers should be written in terms of the Boltzmann constant rather than Find .1 how to approach the problem Hint not displayed Hint A. all contained in a volume and .none of the above Correct Part C The temperature of the diatomic hydrogen gas sample is increased to 100 square speed for diatomic hydrogen at 100 Choose the correct . ANSWER: is: . . .1 How to approach the problem Hint not displayed .1 How to approach the problem Hint not displayed Hint B. Since is the total heat capacity of the gas. ANSWER: = Correct Part B Find . and any other given .1 Definition of heat capacity Hint not displayed Express the molar heat capacity in terms of ANSWER: = Correct You can use the molar heat capacity to find the total heat capacity of the gas at constant volume . and any other given quantities. Hence. Hint B.2 Kinetic energy of a single gas particle Hint not displayed Express the internal energy in terms of the gas temperature ANSWER: = Correct Part C Find . the internal energy of the gas.3 Relating the . it is just the heat capacity per mole multiplied by the number of moles of gas particles Part D Express the pressure of the gas in terms of its energy density Hint D. . the molar heat capacity (heat capacity per mole) of the gas at constant volume. Hint C. and velocities Hint not displayed Express the average speed squared in terms of the gas temperature quantities. and .Hint not displayed Hint A. 1 How to approach the problem Hint not displayed ANSWER: 1 = Correct to the old molar heat capacity ? In fact. such as a gas of photons. .1 Definition of rms speed ANSWER: to the old rms speed ? to the old molar mass ? Hint not displayed 0. when the mass of the particles increases. ANSWER: 0. For example. All other information given in the problem introduction remains the same. neon.577 = Correct So if the internal energy (temperature) remains the same. note that for an ultra-relativistic particle gas. and argon all have almost the .667 = Correct in your expression for to at least three As an interesting aside. the velocity correspondingly decreases. helium. Now imagine that the mass of each gas particle is increased by a factor of 3. the molar heat capacity is (almost) the same for all monoatomic gases. Part G What will be the ratio of the new molar heat capacity Hint G.Hint D. Part E What will be the ratio of the new molar mass ANSWER: 3 = Correct Part F What will be the ratio of the new rms speed Hint F. as the general expression derived earlier shows.2 Find an expression for pressure Hint not displayed Enter the numerical factor that multiplies significant figures. Observe that the pressure and so the average force that the particles exert on the walls also remains the same. . For this problem you will need to know Boltzmann's constant: . to the nearest integer. where is defined ).1 Find the rms speed Hint not displayed State your answer numerically. as do hydrogen and neon. The tiny differences in the values tell us that the ideal gas equation describes gases pretty well. Compare a gas of helium atoms to a gas of 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. ANSWER: 300 = Correct 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.2 Total translational kinetic energy Hint not displayed Express the temperature numerically in kelvins. in kelvins.same molar heat capacity. 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. Velocity and Energy Scaling Helium atoms have a mass of as an atomic mass unit ( oxygen molecules. helium still behaves approximately like an ideal gas. At this temperature. Part A and oxygen molecules have a mass of . but it is not perfect. ANSWER: 38 = Correct K Note that 38 K is is a very low temperature. 3 Use the Equipartition Theorem Hint not displayed Express the mean translational kinetic energy numerically. 6.3 = Correct 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). Express your answer in millimeters per second to one decimal place only. assuming them to be .1 Kinetic energy in terms of average velocity Hint not displayed Hint A. The mass of such a dust particle .2 Find the mass of the particles Hint not displayed Express the rms speed in terms of ANSWER: = Correct Part C Now calculate the rms (root-mean-square) speed spheres of diameter and density of these particles. This is simply because the mass of these particles is much larger than the mass of typical gas molecules.Part A What would you expect the mean translational kinetic energy they are in air at a temperature of 290 K? Hint A. Note that ANSWER: Part B Find an expression for the rms (root-mean-square) speed them to be spheres of diameter and density Hint B.1 Find the rms speed in terms of temperature Hint not displayed Hint B. in joules. of these particles.2 Equipartition Theorem Hint not displayed Hint A. . assuming has been factored out already to make your answer simpler. ANSWER: 0. and . to two significant figures. .0 = Correct of such particles to be if is . . . ANSWER: 293 = Correct Part B The molecules in an ideal gas at have a root-mean-square (rms) speed . (You may then want to convert back into the initial temperature unit to give your answer.e. ANSWER: 859 = Correct . ? Hint A. convert them to kelvins before plugging them into these equations.2 Convert from Celsius to Kelvin scale Hint not displayed Express the temperature to the nearest integer. If you are given temperatures measured in units other than kelvins. Such particles tend to settle quite quickly on account of their weight. Then such a calculation is no longer valid. you should use the Kelvin scale (for which represents absolute zero) in such calculations.. the ideal gas law and the formula for energy of a gas in terms of temperature) assume that zero degrees represents absolute zero.) Part A The average kinetic energy of the molecules of an ideal gas at what temperature twice this value. the weight can no longer be ignored.1 What is rms speed? Hint not displayed Hint B. At (in degrees Celsius) will the average kinetic energy of the same gas be temperature (in degrees Celsius) will the molecules have twice the rms speed. At what ? has the value .2 Find the change in translational kinetic energy Hint not displayed Express the temperature to the nearest integer. you need to pay particular attention to the temperature scale you are using.For particles larger than the ones described here. Hint B. In general. This is because the standard thermodynamic equations (i. ± Gas Scaling When doing numerical calculations involving temperature.1 Formula for energy in terms of temperature Hint not displayed Hint A. this measure does not average to zero over the entire gas. There are several different ways to describe statistically the average velocity of a molecule in a gas. A third measure is the rootmean-square (rms) speed. since the molecules in a gas are moving in random directions. equal to the square root of . 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. One of the most important microscopic properties of gas molecules is velocity. The gas molecules exert no long-range forces on each other. The rms speed is a good approximation of the the typical speed of the molecules in a gas. The gas molecules are in continual random motion with collisions being perfectly elastic. the average squared velocity. the average velocity is approximately zero. you'll . The most obvious measure is the average velocity . Another measure of velocity is .1 How to use the histogram Hint not displayed shows a theoretical ) gas. This histogram distribution of speeds of molecules in a sample of nitrogen ( use the histogram to compute properties of the gas. Since the square of velocity is always positive. Part A What is the average speed of the molecules in the gas? Hint A.± 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. . In this problem. However. In contrast. .2 Find the mean square velocity Hint not displayed Express your answer numerically to three significant digits. Part C What is the temperature of the sample of gas described in the histogram? Hint C. ANSWER: 474 = Correct 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.2 More on computing the average Hint not displayed Express your answer numerically to three significant digits. a and the speed of sound in air is Boeing 747 jet airliner has a maximum air speed of 270 only about 330 .2 Find the molar mass of N2 Hint not displayed Express your answer in degrees Celsius to three significant figues. What is the rms speed of the molecules in the nitrogen gas? Hint B.1 How to approach the problem Hint not displayed Hint C.Hint A. or its square root . ANSWER: 29.4 = Correct Part D .1 How to approach the problem Hint not displayed Hint B. 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. ANSWER: 519 = Correct A speed of 519 is comparable to that of a bullet shot from a handgun. The quantity is more common than because it has the dimensions of velocity instead of the less-familiar velocity-squared. The peak moves to the left. while the distribution becomes less spread out. Correct At the higher temperatures. ANSWER: The peak moves to the right. indicating a higher average velocity. the majority of the molecules have a speed near the average speed.The histogram used in this problem is obviously only an approximation of the true distribution of velocities in a gas. The peak moves to the left. while the distribution becomes more spread out. while the distribution becomes less spread out. 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. To good approximation. indicating that a greater percentage of molecules are traveling at a higher velocity than in the lowtemperature case. [ Print ] . the peak of the curve shifts to the right. In reality. This applet allows you to see the curves for the Maxwell-Boltzmann distribution at many different temperatures. with a few molecules traveling very fast or very slow.7%. You received 91. while the distribution becomes more spread out. The peaks of the higher temperature curves are also broader.52 out of a possible total of 90 points. the molecules span a continuous range of velocities. the speeds of molecules in a gas follow what is known as the MaxwellBoltzmann distribution. Score Summary: Your score on this assignment is 101. 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. For a given temperature. The peak moves to the right.