• Temperature ~ Average KE of each particle • Particles have different speeds • Gas Particles are in constant RANDOM motion • Average KE of each particle is: 3/2 kT • Pressure is due to momentum transfer

Speed ‘Distribution’ at CONSTANT Temperature is given by the Maxwell Boltzmann Speed Distribution

Pressure and Kinetic Energy • Assume a container is a cube with

edges d. • Look at the motion of the molecule in

terms of its velocity components and momentum and the average force

• Pressure is proportional to the number of molecules per unit volume (N/V) and to the average translational kinetic energy of the molecules.

• This equation also relates the macroscopic quantity of pressure with a microscopic quantity of the average value of the square of the molecular speed

• One way to increase the pressure is to increase the number of molecules per unit volume

• The pressure can also be increased by increasing the speed (kinetic energy) of the molecules

___22 1

3 2 oNP m vV

=

Molecular Interpretation of Temperature

• We can take the pressure as it relates to the kinetic energy and compare it to the pressure from the equation of state for an ideal gas

• Temperature is a direct measure of the average molecular kinetic energy

= = =

___2

B2 13 2

NP mv nRT Nk TV

Molecular Interpretation of Temperature

• Simplifying the equation relating temperature and kinetic energy gives

• This can be applied to each direction, – with similar expressions for vy and vz

___2

B1 32 2om v k T=

___2

B1 12 2xmv k T=

Total Kinetic Energy

• The total kinetic energy is just N times the kinetic energy of each molecule

• If we have a gas with only translational energy, this is the internal energy of the gas

• This tells us that the internal energy of an ideal gas depends only on the temperature

___2

tot trans B1 3 32 2 2

K N mv Nk T nRT

= = =

Kinetic Theory Problem

A 5.00-L vessel contains nitrogen gas at 27.0°C and 3.00 atm. Find (a) the total translational kinetic energy of the gas molecules and (b) the average kinetic energy per molecule.

Hot Question Suppose you apply a flame to 1 liter of water for a certain time and its temperature rises by 10 degrees C. If you apply the same flame for the same time to 2 liters of water, by how much will its temperature rise? a) 1 degree b) 5 degrees c) 10 degrees d) zero degrees

Ludwig Boltzmann or Dean Gooch?

• 1844 – 1906 • Austrian physicist • Contributed to

– Kinetic Theory of Gases – Electromagnetism – Thermodynamics

• Pioneer in statistical mechanics

Distribution of Molecular Speeds • The observed speed distribution of gas

molecules in thermal equilibrium is shown at right

• NV is called the Maxwell-Boltzmann speed distribution function

• mo is the mass of a gas molecule, kB is Boltzmann’s constant and T is the absolute temperature

2

3 / 2/ 22

B

42

Bmv k ToV

mN N v ek T

ππ

− =

Molecular Speeds and Collisions

Speed Summary • Root mean square speed

• The average speed is somewhat lower than

the rms speed

• The most probable speed, vmp is the speed at which the distribution curve reaches a peak

• vrms > vavg > vmp

B Brms

3 1.73o o

k T k Tvm m

= =

B Bmp

2 1.41k T k Tvm m

= =

π= =B B

avg8 1.60

o o

k T k Tvm m

Some Example vrms Values

At a given temperature, lighter molecules move faster, on the average, than heavier molecules

Speed Distribution • The peak shifts to the right

as T increases – This shows that the average

speed increases with increasing temperature

• The asymmetric shape occurs because the lowest possible speed is 0 and the highest is infinity

23 / 2 1/ 2 rmskT KE mv= =

2 3rms

kTv vm

= =Root-mean-square speed:

The Kelvin Temperature of an ideal gas is a measure of the average translational kinetic energy per particle:

k =1.38 x 10-23 J/K Boltzmann’s Constant

Kinetic Theory Problem Calculate the RMS speed of an oxygen molecule in the air if the temperature is 5.00 °C. The mass of an oxygen molecule is 32.00 u (k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)

3rms

kTvm

= What is m? m is the mass of one oxygen molecule in kg.

What is u? How do we get the mass in kg?

Kinetic Theory Problem Calculate the RMS speed of an oxygen molecule in the air if the temperature is 5.00 °C. The mass of an oxygen molecule is 32.00 u (k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)

3rms

kTvm

=

23

27

3(1.38 10 / )278(32 )(1.66 10 / )

x J K Ku x kg u

−

−=

466 /m s=

What is m? m is the mass of one oxygen molecule.

Is this fast? YES! Speed of sound: 343m/s!

A cylinder contains a mixture of helium and argon gas in equilibrium at 150°C. (a) What is the average kinetic energy for each type of gas molecule? (b) What is the root-mean-square speed of each type of molecule?

More Kinetic Theory Problems A gas molecule with a molecular mass of 32.0 u has a speed of 325 m/s. What is the temperature of the gas molecule? A) 72.0 K B) 136 K C) 305 K D) 459 K

E) A temperature cannot be assigned to a single

molecule.

Temperature ~ Average KE of all particles

Equipartition of Energy • Each translational degree of freedom contributes an

equal amount to the energy of the gas – In general, a degree of freedom refers to an

independent means by which a molecule can possess energy

• Each degree of freedom contributes ½kBT to the energy of a system, where possible degrees of freedom are those associated with translation, rotation and vibration of molecules

• With complex molecules, other contributions to internal energy must be taken into account

• One possible energy is the translational motion of the center of mass

• Rotational motion about the various axes also contributes

• There is kinetic energy and potential energy associated with the vibrations

Monatomic and Diatomic Gases The thermal energy of a monatomic gas of N atoms is

A diatomic gas has more thermal energy than a monatomic gas at the same temperature because the molecules have rotational as well as translational kinetic energy.

Molar Specific Heat • We define specific heats for two processes

that frequently occur: – Changes with constant pressure – Changes with constant volume

• Using the number of moles, n, we can define molar specific heats for these processes

• Molar specific heats: – Q = nCV DT for constant-volume processes – Q = nCP DT for constant-pressure processes

Ideal Monatomic Gas

• Therefore, ∆Eint = 3/2 nRT − ∆E is a function of T only

• In general, the internal energy of an ideal gas is a function of T only – The exact relationship depends on the type of

gas • At constant volume, Q = ∆Eint = nCV ∆T

– This applies to all ideal gases, not just monatomic ones

Ratio of Molar Specific Heats

• We can also define the ratio of molar specific heats

• Theoretical values of CV , CP , and γ are in excellent agreement for monatomic gases

• But they are in serious disagreement with the values for more complex molecules – Not surprising since the analysis was for monatomic gases

5 / 2 1.673 / 2

P

V

C RC R

γ = = =

Agreement with Experiment • Molar specific heat is a function of

temperature • At low temperatures, a diatomic gas

acts like a monatomic gas CV = 3/2 R • At about room temperature, the value

increases to CV = 5/2 R – This is consistent with adding

rotational energy but not vibrational energy

• At high temperatures, the value increases to CV = 7/2 R – This includes vibrational energy

as well as rotational and translational

Sample Values of Molar Specific Heats

In a constant-volume process, 209 J of energy is transferred by heat to 1.00 mol of an ideal monatomic gas initially at 300 K. Find (a) the increase in internal energy of the gas, (b) the work done on it, and (c) its final temperature

Molar Specific Heats of Other Materials

• The internal energy of more complex gases must include contributions from the rotational and vibrational motions of the molecules

• In the cases of solids and liquids heated at constant pressure, very little work is done, since the thermal expansion is small, and CP and CV are approximately equal

Adiabatic Processes for an Ideal Gas

• An adiabatic process is one in which no energy is transferred by heat between a system and its surroundings (think styrofoam cup)

• Assume an ideal gas is in an equilibrium state and so PV = nRT is valid

• The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV γ = constant

γ = CP / CV is assumed to be constant All three variables in the ideal gas law (P, V, T ) can change during an adiabatic process

Adiabatic Process

• The PV diagram shows an adiabatic expansion of an ideal gas

• The temperature of the gas decreases – Tf < Ti in this process

• For this process Pi Vi

γ = Pf Vfγ and

Ti Viγ-1 = Tf Vf

γ-1

A 2.00-mol sample of a diatomic ideal gas expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and final temperatures? (c) Find Q, W, and ∆Eint.

Cyclic Processes

• A cyclic process is one that starts and ends in the same state – On a PV diagram, a cyclic process appears as a

closed curve • If ∆Eint = 0, Q = -W • In a cyclic process, the net work done on the

system per cycle equals the area enclosed by the path representing the process on a PV diagram

intE Q W∆ = +

Isothermal Process • At right is a PV diagram of an isothermal

expansion • The curve is a hyperbola • The curve is called an isotherm

• The curve of the PV diagram indicates PV = constant – The equation of a hyperbola

• Because it is an ideal gas and the process is quasi-static,

PV = nRT and f f f

i i i

V V V

V V V

nRT dVW P dV dV nRTV V

= − = − = −∫ ∫ ∫

ln i

f

VW nRTV

=

Isothermal Process • An isothermal process is one that occurs at

a constant temperature • Since there is no change in temperature,

∆Eint = 0 • Therefore, Q = - W • Any energy that enters the system by heat

must leave the system by work

intE Q W∆ = + f

i

V

VW P dV= −∫PV nRT=

Isobaric Processes • An isobaric process is one that occurs at a

constant pressure • The values of the heat and the work are

generally both nonzero • The work done is W = P (Vf – Vi) where P

is the constant pressure

intE Q W∆ = + f

i

V

VW P dV= −∫PV nRT=

Isovolumetric Processes • An isovolumetric process is one in which there is

no change in the volume • Since the volume does not change, W = 0 • From the first law, ∆Eint = Q • If energy is added by heat to a system kept at

constant volume, all of the transferred energy remains in the system as an increase in its internal energy

intE Q W∆ = + f

i

V

VW P dV= −∫PV nRT=

Special Case: Adiabatic Free Expansion

• This is an example of adiabatic free expansion

• The process is adiabatic because it takes place in an insulated container

• Because the gas expands into a vacuum, it does not apply a force on a piston and W = 0

• Since Q = 0 and W = 0, ∆Eint = 0 and the initial and final states are the same and no change in temperature is expected. – No change in temperature is expected

Thermo Processes • Adiabatic

– No heat exchanged – Q = 0 and ∆Eint = W

• Isobaric – Constant pressure – W = P (Vf – Vi) and

∆Eint = Q + W • Isovolumetric

– Constant Volume – W = 0 and ∆Eint = Q

• Isothermal – Constant temperature ∆Eint = 0 and Q = -W

intE Q W∆ = +

ln i

f

VW nRTV

=

A gas is taken through the cyclic process as shown.

(a) Find the net energy transferred to the system by heat during one complete cycle. (b) What If? If the cycle is reversed—that is, the process follows the path ACBA—what is the net energy input per cycle by heat?

intE Q W∆ = +f

i

V

VW P dV= −∫

A sample of an ideal gas goes through the process as shown. From A to B, the process is adiabatic; from B to C, it is isobaric with 100 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 150 kJ of energy leaving the system by heat. Determine the difference in internal energy E(B) – E(A).

intE Q W∆ = + f

i

V

VW P dV= −∫PV nRT=

HW Problem 28

Important Concepts