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President University Erwin Sitompul Thermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University http://zitompul.wordpress.com 2 0 1 5

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President UniversityErwin SitompulThermal Physics 6/3 The Distribution of Molecular Speeds Chapter 19Kinetic Theory In 1852, Scottish physicist James Clerk Maxwell first solved the problem if finding the speed distribution of gas molecules. The resulting Maxwell’s speed distribution law can be written as: M : molar mass of the gas R: gas constant T: gas temperature v : molecular speed The quantity P(v) is a probability distribution function. For any speed v, the product P(v)dv (an area, dimensionless) is the fraction of molecules with speed in the interval dv centered on speed v.

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Page 1: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/1

Lecture 6Thermal Physics

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com2 0 1 5

Page 2: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/2

The Distribution of Molecular Speeds The root-mean-square speed vrms gives us a general idea of

molecular speeds in a gas at a given temperature. But, we often want to know more about how the possible values of

speed are distributed among the molecules. The speed distribution for oxygen molecules at room temperature

(T = 300 K) is shown below.

Chapter 19 Kinetic Theory

Page 3: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/3

The Distribution of Molecular SpeedsChapter 19 Kinetic Theory

In 1852, Scottish physicist James Clerk Maxwell first solved the problem if finding the speed distribution of gas molecules.

The resulting Maxwell’s speed distribution law can be written as:

2

32 2 2( ) 4

2Mv RTMP v v e

RT

M : molar mass of the gasR : gas constantT : gas temperaturev : molecular speed

The quantity P(v) is a probability distribution function.

For any speed v, the product P(v)dv (an area, dimensionless) is the fraction of molecules with speed in the interval dv centered on speed v.

Page 4: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/4

The Distribution of Molecular SpeedsChapter 19 Kinetic Theory

As shown in the figure, the fraction of molecules with speeds in the interval dv is equal to the area of a strip with height P(v) and width dv.

The total area under the distribution curve corresponds to the fraction of the molecules whose speeds lie between zero and infinity.

Since all molecules fall into this category, the value of this total area is unity:

0

( ) 1P v dv

The fraction (frac) of molecules with speeds in an interval of v1 and v2 is then:

2

1

frac ( )v

v

P v dv

Page 5: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/5

The Distribution of Molecular SpeedsChapter 19 Kinetic Theory

As the speed is a function of temperature, the speed distribution also varies with temperature. The distribution at T = 300 K is compared with the one at T = 80 K in the following figure.

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President University Erwin Sitompul Thermal Physics 6/6

Average, RMS, and Most Probable SpeedsChapter 19 Kinetic Theory

In principle, we can find the average speed vavg of the molecules in a gas by evaluating:

avg0

( )v vP v dv

After substituting P(v) and performing the integral,

Average Speedavg

8RTvM

Similarly, we can find the average of the square of the speeds v2avg with

2 2avg

0

( )v v P v dv

After substituting P(v) and performing the integral,

2avg

3RTvM

Page 7: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/7

Average, RMS, and Most Probable SpeedsChapter 19 Kinetic Theory

Thus,

RMS Speedrms3RTvM

The most probable speed vP is the speed at which P(v) is maximum. To calculate vP, we set dP/dv = 0 and then solve for v. Doing so, we find:

Most Probable Speed2

PRTvM

A molecule is more likely to have speed vP than any other speed, but some molecules will have speeds that are many times vP.

These molecules lie in the high-speed tail of a distribution curve.

Page 8: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/8

ProblemA container is filled with oxygen gas maintained at room temperature (300 K). What fraction of the molecules have speeds in the interval 599 to 601 m/s? The molar mass M of oxygen is 0.0320 kg/mol.

Chapter 19 Kinetic Theory

The interval Δv = 2 m/s is very small compared to the center speed v

= 600 m/s. Thus the integration can be approximated through:2

32 2 2frac 4

2Mv RTM v e v

RT

2

32 (0.0320)(600) 2(8.31)(300)20.03204 (600) (601 599)

2 (8.31)(300)e

2 2.31(1.321 10 )( ) (2)e 32.622 10

0.2622%

Page 9: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/9

The Molar Specific Heats of an Ideal GasChapter 19 Kinetic Theory

In this section, we want to derive an expression for the internal energy Eint of an ideal gas from molecular consideration.

In other words, we want an expression for the energy associated with the random motions of the atoms or molecules in the gas.

We shall then use that expression to derive the molar specific heats of an ideal gas.

Page 10: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/10

The Molar Specific Heats of an Ideal GasChapter 19 Kinetic Theory

Internal Energy Eint Let us firs assume that the ideal gas is a monatomic gas such as

helium, neon, or argon. The internal energy Eint of the ideal gas is simply the sum of the

translational kinetic energies of its atoms. The average translational kinetic energy of a single atom depends

only on the gas temperature and is given as Kavg =3/2·kT. A sample of n moles of such a gas contains nNA atoms. The internal

energy of the sample is then:3

int A avg A 2( ) ( )( )E nN K nN kT

Since k = R/NA, we can rewrite this as:3

int 2E nRT Monatomic Ideal Gas

The internal energy Eint of an ideal gas is a function of the gas temperature only.

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The Molar Specific Heats of an Ideal GasChapter 19 Kinetic Theory

Molar Specific Heat at Constant Volume The figure on the right shows n moles of an

ideal gas at pressure p and temperature T, confined to a cylinder of fixed volume V.

This initial state is denoted as i. Suppose a small amount of energy to the gas

as heat Q is added to the gas. The gas temperature rises a small amount

T+ΔT, and its pressure rises to p+Δp. This final state is denoted as f.

We would find that the heat Q is related to the temperature change ΔT by:

VQ nC T ConstantVolume

CV is a constant called the molar specific heat at constant volume [J/mol·K].

Page 12: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/12

The Molar Specific Heats of an Ideal GasChapter 19 Kinetic Theory

The first law of thermodynamics can now be written as:

Monatomic Gas

int VE nC T W

With the volume held constant, the gas cannot do any work, W = 0. This yields:

intV

EC

n T

32 nR Tn T

Thus,

32VC R 12.5 J mol K

We can now generalize the equation for the internal energy as:int VE nC T Any Ideal Gas

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President University Erwin Sitompul Thermal Physics 6/13

The Molar Specific Heats of an Ideal GasChapter 19 Kinetic Theory

When an ideal gas that is confined to a container undergoes a temperature change ΔT, then we can write the resulting change in its internal energy as:

int VE nC T Any Ideal Gas, Any Process

The change in the internal energy Eint of a confined ideal gas depends on the change in the gas temperature only. It does not depend on what type of process produces the change in the temperature.

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The Molar Specific Heats of an Ideal GasChapter 19 Kinetic Theory

Molar Specific Heat at Constant Pressure We now assume that the temperature of our

ideal gas is increased by the same small amount ΔT as previously but now the necessary energy (heat Q) is added with the gas under constant pressure.

From such experiments we find that the heat Q is related to the temperature change ΔT by:

PQ nC T ConstantPressure

CP is a constant called the molar specific heat at constant pressure [J/mol·K].

CP > CV, because energy must now be supplied not only to raise the temperature of the gas but also for the gas to do work.

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The Molar Specific Heats of an Ideal GasChapter 19 Kinetic Theory

To relate the molar specific heats CP and CV, we start with the first law of thermodynamics:

intE Q W

V PnC T nC T nR T

We will find that:V PC C R

P VC C R

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President University Erwin Sitompul Thermal Physics 6/16

The Molar Specific Heats of an Ideal GasChapter 19 Kinetic Theory

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CheckpointThe figure here shows five paths traversed by a gas on a p-V diagram. Rank the paths according to the change in internal energy of the gas, greatest first.

5, then tie of 1, 2, 3, and 4.

Chapter 19 Kinetic Theory

int VE nC T

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President University Erwin Sitompul Thermal Physics 6/18

Problem

PQ nC T 52(5)( )(20)R 2077.5 J

A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase ΔT of 20.0°C at constant pressure. As a result, the bubble expands. The helium is monatomic an ideal.(a) How much energy is added to the helium as heat during the

increase and expansion?

Chapter 19 Kinetic Theory

(b) What is the change ΔEint in the internal energy of the helium during the temperature increase?

int VE nC T 32(5)( )(20)R 1246.5 kJ

(c) How much work W is done by the helium as it expands against the pressure of the surrounding water during the temperature increase?

intW nR T Q E 2077.5 1246.5 831kJ

Page 19: President UniversityErwin SitompulThermal Physics 6/1 Lecture 6 Thermal Physics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul Thermal Physics 6/19

Class Group Assignments1. The rms velocity of oxygen molecules at 27°C is 318 m/s. Thus, the rms velocity of

hydrogen molecules at 127 °C is:(a) 1470 m/s (b) 1603 m/s (c) 1869 m/s (d) 2240 m/s (e) 3211 m/s

2. An ideal gas of N monoatomic molecules is in thermal equilibrium with an ideal gas of the same number of diatomic molecules and equilibrium is maintained as the temperature is increased. The ratio of the changes in the internal energies ΔEdia/ΔEmon is:(a) 1/2 (b) 3/5 (c) 1/1 (d) 5/3 (e) 2/1

3. An ideal gas has molar specific heat CP at constant pressure. When the temperature of n moles is increased by ΔT the increase in the internal energy is:(a) nCPΔT (d) n(CP+R)ΔT (b) n(CP–R)ΔT (e) n(2CP+R)ΔT (c) n(2CP–R)ΔT

Chapter 19 Kinetic Theory

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No Homework This WeekChapter 19 Kinetic Theory

Prepare well for the midterm exam.