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PHY2009S Thermodynamics and statistical models in physicsPart 3

Gas LawsMatter & Interactions

Chapter 12

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Gases: molecules mostly beyond range of intermolecular forces.

Solids: molecules close and cannot move far.

Liquids: intermediate case … molecules always close, but can

move.

Gases, solids and liquids

molecules in a gas

molecules in a liquid

M&I12.1

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1-d gas leak through hole

Consider N

molecules travelling to the right with same speed v inside a tube which has volume V

and cross sectional area A.

Number density

v

end viewNnV

v

v t

Distance a single molecule travels in time ∆t

is v∆t.

M&I12.2

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1-d gas leak through tube …2

Number N

of molecules leaving the tube in ∆t is

v t

A V A v t

N nV nA v t

Flux through A is nvA.

Real gases move at different speeds, therefore take weighted average:

1 1 2 2 3 3 ...n v n v n vvn

Then flux is . nvA

(one dimensional case, molecules have various speeds)

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3-d gas leaks through hole

In 3-d maybe 6 directions (x, y, z and both ways) gives factor 1/6. However, since we must average over angle to wall, etc. …a proper calculation (integration) gives rather a factor 1/4:

Consider a 1 mm diameter hole in 1 litre balloon in vacuum at STP.

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nAv

2

3

1 8 3Flux 4 22.4 10 4 3

AN d kTnAvm

= 6.61×

1021 atoms s-1

V

= 1 ×

10-3

m3

N

= NA ( V / 22.4 ×

10-3 )NnV

Then flux is .

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Leak causes gas to cool

The faster molecules are more likely to escape …… thus the fraction of slower molecules remaining increases, and the gas in the balloon is colder.

If the gas is a real gas, then escaping gas expands, and molecules must do work against attractive intermolecular forces.

This cools the escaping gas.

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Mean free path

2

1nAd

A r RNnV

M&I12.3

Let d be the average distance travelled by “special”

molecule (dashed

circle) in an imaginary cylinder of radius R

+ r

before it has a collision with a gas molecule (dark).

R+r

R

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Mean free path in nitrogen at STP

Consider 1 mole of nitrogen at STP25 -3

3 3 2.7 10 molecules m22.4 10 m

ANNnV

210 -19 22 2 10 m 5 10 mA

-8

25 -3 -19 2

1 1 7 10 m2.7 10 molecules m 5 10 m

dnA

Compare average distance between molecules:

1

13 25 -3 -831 2.7 10 molecules m 0.3 10 mLn

Radius of molecule: -102 10 mr

So d L r

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Pressure and temperature

Can we understand the ideal gas law

Pressure caused by gas molecules bombarding walls.

Temperature related to average kinetic energy of molecules.

for moles of gas, in terms of kinetic theory?PV RT

M&I12.4

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Elastic collisions of helium atom and wall

Consider an elastic collision of a helium atom with a wall.

21 x xp pFPA A t A t

Energy is conserved,but momentum changes.

Hence

2x xp p For the helium atom:

2x xp p For the wall:

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Pressure from collisions: one speed

If nright

helium atoms per unit volume move with +vx

then the number of such atoms that hit area A

to their right in time Δt is

right

1

x

tn Av

Then right

right

1 1 21/

x xx x

x

p pFP n v pA A t A n Av

right2 xx

pn pm

2

right2 xpnm

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Pressure from collisions: all speeds (in one dimension)

Average over all speeds

right left 2nn n

2 2x xp p

If n1

have px,1

, and n2

have px,2

, …

then2 2

1 ,1 2 ,22 ...x xx

n p n pp

n

Then the pressure 2 2,1 ,2

1 2 ...x xp pP n n

m m

2xpP n

m

On average:

or

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Pressure from collisions: all directions

Momentum is a vector, . Magnitude is p.p

Random motion, so average 0p

Thus, also 0x y zp p p

But and2 2 2 2x y zp p p p 2 2 2

x y zp p p

Therefore .2

2

3xpp

NnV

Pressure thus is where21

3pP nm

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Reasonable at room temperatureEsurroundings

<< Enuclear/electronic

Boltzmann distribution also applies to gases.We will focus on ideal gases …

molecules don’t interact.

For real gases we assume very dilute, i.e. low density, few interactions.

Energy of molecule in an ideal gas in the gravitational field near the Earth’s surface:

molecule trans vib rot cmE K E E Mgy

We ignore:rest energynuclear energyelectronic energy

Energy of a diatomic gas

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Average translational kinetic energy in a gas

Energy depends on a quadratic quantity ω2, e.g.21

trans cm2K Mv 212s sU k x 21

rot 2E I

Average value of ω2

is denoted 2

2

2

22 0

0

1......2

kT

kT

e dkT

e d

If , average of a quadratic energy term is0kT 1

2kT

The number of quadratic terms in the expression for the energy is often called the “degrees of freedom.”For an ideal monatomic and diatomic gas:

2 2 21trans 2

1 332 2x y zK M v v v kT kT

(integrate by parts)

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Ideal gas law

In terms of , pressure is2 2

trans1 2 23 3 2 3

p pP n n nKm m

NnV

Previously found that trans32

K kT

Therefore P nkT

For moles, and and

putting NnV

AN N AR kN

PV RT mole version of ideal gas law

molecular version of ideal gas law

23 -1 23 -1

-1 -1

1.38 10 J K 6.023 10 mole

8.31 J K moleAR kN

Gas constant

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Temperature from entropy or ideal gas law

… in turn based on statistical mechanics:

… based on Boltzmann distribution applied to a low density gas.1 ET S

… to get …

Previously showed that trans32

K kT

Now inserted into trans32

K kT21

3pP nm

PV RT

Thus PVTR

Gas thermometer

… same T

is in 1 ET S

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Real gases

PV RT “ideal”

gas law works well for low density gases

For “real”

high density gases …

two problems …

Real gas molecules have volume …

hence V

needs to be smallerand short range electric forces …

“van der

Waals forces”

…

… hence P

needs to be smaller

Replace with van der Waals equation:2

2 ( )aP V b RTV

where a

: attraction factor

b: volume factor

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Weight of gas in a box

Fy

= −Mg

P

slightly smaller

P

slightly greater

gas

gas

vacuum

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Weight of a bouncing molecule

Consider a single molecule that bounces up and down on a scale without losing significant energy.

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P mv mgt v g

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Energy transfer between a gas and its surroundings

cylinder filled with gas

piston

We need a device that lets us control the flow of energy in and out of a gas, in the form of work done W or energy transfer Q

…

… use a system consisting of a cylindrical container of gas that is enclosed by a piston that can move in and out of the cylinder with little friction but which fits tightly enough to keep the gas from leaking out.

M&I12.5

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gas

sand

Energy transfer between a gas and its surroundings …2

Consider a cylinder of gas with a vertical running piston, on which we can load varying amounts of sand …

in order to control the

pressure in the contained gas.

PA

mg

MgPair

A

airMg mgP P

A

Then the pressure on the gas in the cylinder is

What will happen if we suddenly dump a whole lot of sand onto the piston?

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Slow compression: a quasistatic process

… maybe isothermal.… maybe adiabatic.… maybe neither.

gas

add or remove one grain at a time

Instead of dumping a lot of sand, we add (or remove) sand very slowly, one grain at a time …

a new equilibrium is

established almost immediately (“quasistatic compression” or “quasistatic expansion”

of the gas).

Consider quasistatic compression …

PV NkT holds …

… but is T constant?

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Isothermal compression

P1

V1

= P2

V2

Compressgas

water at temperature T

Energy transfer due to temperature difference

add sand

Consider a metal cylinder (good thermal conductor) immersed in a large bath of water.

The big thermal reservoir keeps the temperature of the gas constant as it is compressed.

“Isothermal”

or constant temperature compression.

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Isothermal compression: work and heat

… and there is thermal transfer of energy out of gas (into the water).

Fon

gas = PA

Piston does work on gas.

gasE W Q

Since T

= constant, gas 0E

0W Q

Therefore if W > 0 , then Q

< 0 , and

2

1

V

V

W PdV … work done by the piston of the gas.

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Isothermal compression: integration

2 2

1 1

V V

V V

NkTW PdV dVV

2

1

2

1

1

2

[ln ]

ln

V

V

VV

dVNkTV

NkT V

VNkTV

2 1V V

2 1

1 2

ln lnV VV V

1by piston into water

2

ln VW Q NkTV

Since the process was a compression

and since

initial volume V1

final volume V2

more sand

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Isothermal compression: work done on the gas

P

V1 V2

V

1by piston

2

ln VW NkTV

Isotherm …

PV

= constant

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Isothermal compression: first law of thermodynamics

ΔEint

= W + Q = 0

First law of thermodynamics (energy conservation)

ΔEsys

= W + Q

In this case of isothermal compression of an ideal gas:

… where Eint

is the “internal”

energy of the gas (the sum of the translational, rotational, vibrational, and other energy terms of all the molecules).

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Heat capacity at constant volume CV

Lock the piston in place ... volumeof the gas is constant …

hence no

work is done (W = 0).Water hotter than gas, so thermalenergy transfer Q

into the gas. Temperature

rise ∆T

water at hightemperature T

Lock pistonV

= constant

QDefine the specific heat capacity at constant volume CV

…

thermal VE Q NC T

for a monatomic gas (He, etc.)for other gases (N2

, etc.)32VC k32VC k

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Heat capacity at constant pressure CP

Now piston is movable …

pressure in the gas is constant.Water hotter than gas, so thermal energy transfer Q into the gas and the gas does work on piston. Temperature

rise ∆T

water at hightemperature T

Piston risesP

= constant

QP VQ NC T NC T W

2 2

1 1

V V

V V

W PdV P dV

2 1

2 1

PV PVNkT NkT Nk T

P VQ NC T NC T Nk T

P VC C k

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Adiabatic (No Q) compression

initial volume V1

final volume V2

more sand

If the cylinder and piston are made of insulating material, then no thermal energy transfer Q

can occur between the

ideal gas and its surroundings …

… called and adiabatic compression. 2

1

V

VV

W PdV NC T

Note that we use CV

for an ideal gas (although V

is not constant) since the

total energy of an ideal gas is entirely determined by the temperature.

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Adiabatic compression: integration

2

1

V

VV

W PdV NC T VPdV NC dT

VdVNkT NC dTV

Then 0VC dT dVk T V

ln ln constantVC T Vk

ln constantVCkT V

or constantVCkT V

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Adiabatic compression

constantVCkT V

Using …PV NkT

… get whereconstantPV P

V

CC

P

V Isotherm (PV

= constant)

Adiabat

(PVγ

= constant)

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A random walkM&I12.5

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Fundamental limits on efficiency

We have sources of energy in the form of heat ……. fossil fuel, hydro, wind, solar, . . . , nuclear.

… industrial societies want machines (engines) to do work.

… need to convert thermal energy transfer to work.

There is a fundamental limitation: entropy and the second law of thermodynamics.

M&I12.7

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An engine converts heat into work

Input to the engine is net thermal energy transfer Q

Output is net work W

Can |W| = Q ??

Net thermal energy

transfer Q

Engine

Net work W

… always irreversible processes, e.g. friction.

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Thermal energy transfer and entropy increase

Two blocks connected by a metal bar of length L, cross sectional area A

and thermal

conductivity σ.

Block at TH

… the “source”Block at TL

… the “sink”

H LT TdQ Q Adt L

0HT

0CT

QST

H

H

Q tST

LL

Q tST

0 0 0H LH L

Q t Q tS ST T

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Reversible and irreversible processes

Consider entropy of the Universe, SU

.

SU

increases in irreversible process,

SU

constant in reversible process,

For a reversible process, implies Q = 0 or TH

= TL

.

Thus and the process is infinitely slow.

0US

0US

0H LQ T T

0US

Therefore we can carry out (nearly) reversible processes, hardly changing the entropy of the Universe, as long as we transfer energy very slowly …

so as to be of no practical

use as a (mechanical) engine.

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A maximally efficient process

Assume no friction.

Assume a reversible process (slow changes).

Since the process is reversible, we can run engine backwards (i.e. a refrigerator).

Ideas due to Sadi

Carnot (1824) … before the principle of energy conservation was established!

M&I12.8

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A cyclic process of a reversible engine

… and a hot and cold reservoir.

Four parts to the cycle:… in contact with hot reservoir: isothermal expansion… adiabatic expansion… in contact with cold reservoir: isothermal compression… adiabatic compression

Details of engine do not matter.

Overall principle …

use cylinder of ideal gas, with a piston and some sand on the piston to adjust the pressure on the gas. gas

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Isothermal expansion at TH

Engine in contact with the hot reservoir

gas HT T

Lift the piston by sliding some sand off sideways slowly.

Isothermal expansion … engine gets QH

from source.

Source: H H HS Q T

Gas: gas gasHS Q T but gasHT T

Therefore 0US

No change in gas energy.All energy went into work on gas HW Q

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Adiabatic expansion cools gas to TL

Allow reversible adiabatic expansion …

by slowly moving

more weight from the piston.

Q = 0 so ∆S = 0.

Gas does work.

Work done on gas

Gas cools to .

No thermal contact with hot reservoir.

on gas 0W

gas LT T

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Isothermal compression at TL

Engine in contact with the cold reservoir

gas LT T

Lower the piston by sliding some sand on slowly.

Isothermal compression… engine gives QL

to sink.

Sink: L L LS Q T Gas: gas gasHS Q T

but gasLT TTherefore 0US

No change in energy of the gas.All energy went into work on gas LW Q

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Adiabatic compression heats gas to TH

Allow reversible adiabatic compression by sliding some more sand on slowly.

Q = 0 so ∆S = 0.

Work done on gas

Gas warms to

Was there net work done over the whole cycle?

No thermal contact with cold reservoir.

on gas 0W

gas HT T

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Carnot Cycle

The four steps of a Carnot cycle:… isothermal expansion at TH

while absorbing heat… adiabatic expansion to TC… isothermal compression at TC

while expelling heat… adiabatic compression back to TH

.

P

V

Isotherm (TH

)

adiabatadiabat

Isotherm (TC

)

Net work done on gas

Work done by gas equals enclosed area.

on gasW PdV

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Entropy change in one cycle of a reversible engine

Engine is reversible, so ∆SU

= 0

After one cycle, gas returns to initial state.

Entropy S is a state function, ∆Sgas

= 0.

gasU H LS S S S

0 0H L

H L

Q QT T

H L

H L

Q QT T

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Energy change in one cycle of a reversible engine

Heat into gas H LQ Q

Work done by surroundings W

Apply conservation of energy

Engine does work

engine H LE Q Q W

HH L L

L

TQ Q QT

engine 0E

on gas 0L HW Q Q

on gas H LW W Q Q

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Efficiency of a reversible engine

Efficiency does not depend on details, only on input and output temperatures. Less than 100 % efficient since we have to reject heat.

This wastes heat .

Best to use a big TH .

Define efficiency H L

H H

W Q QQ Q

1 1L L

H H

Q TQ T

L LQ T

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No other engine can be more efficient

Then

… is in violation of second law of thermodynamics.

Thus no engine of greater efficiency than ε

can exist.

The maximum efficiency is

Proof by contradiction.

Assume an engine has efficiency more than 1 L

H

TT

Run between TH

and TL

.

Then for same QH

, QL

is less.

max 1 L

H

TT

0H LU

H L

Q QST T

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Real engines are less efficient

Real processes are irreversible, so

0H LU

H L

Q QST T

0US

L L

H H

Q TQ T

irH L

H H

W Q QQ Q

max1 1L L

H H

Q TQ T

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Running engine backwards: refrigerator

Disconnect from hot reservoir at TH

.

Adiabatic expansion, cooling to TL

.

Connect to cold reservoir at TL

.

Isothermal expansion, QL

goes in.

Disconnect from cold reservoir at TL

.

Adiabatic compression, warming to TH

.

Connect to hot reservoir at TH

.

Isothermal compression, QH

goes out.

We do work W on the gas.

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The refrigerator

Energy conservation

If reversible, so

Heat removal “efficiency”

We simply moved the energy.

0 L HW Q Q

0US

H L

H L

Q QT T

1

1 1L H

L

Q TW T

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The heat pump

We simply moved the energy.

Same as the refrigerator, but wewant to be warm!

Heat arrival “efficiency”1

1 1H L

H

Q TW T

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Why don’t we attain theoretical efficiency?

Friction? No. Minimized by good design, lubrication.

To reduce entropy

increase, need small temperaturedifference between parts of the machine.

… length L, cross sectional area A

and thermal conductivity σ.

H LT TdQ Q Adt L

Machine is slow …… in fact, infinitely slow for a perfectly reversible cycle.

Recall:

0 as 0H LQ T T

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Efficiency of a non-reversible engine

If hot reservoir is connected, engine is at T1

.If cold reservoir is connected,engine is at T2

.

H LT TdQ Q Adt L

Write 1H HQ b T T

2L LQ b T T

Energy conservation gives

H LQ W Q

Choose b to maximize power .W

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There must be maximum power output

(i) Reversible

(i) (ii)

1 2 and H LT T T T Very slow 0W

(ii) 1 2T T H LQ Q thenand again 0W

Somewhere between these two extreme designs …

can find

the maximum output power.

Can show that the efficiency of a maximum power engine

1 L

H H

TWQ T

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Refrigerator with non-zero LQ

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Heat pump with non-zero HQ