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Lecture 2 The First Law of Thermodynamics (Ch.1)

Lecture 1- we introduced macroscopic parameters that describethe state of a thermodynamic system (including temperature), the

equation of state f (P,V,T) = 0, and linked the internal energy of

the ideal gas to its temperature.

Outl ine:

1. Internal Energy, Work, Heating

2. Energy Conservationthe First Law

3. Enthalpy

4. Heat Capacity

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Internal Energy

systemU = kinet ic + po tential

system

boundary

environment

The internal energy of a system of particles, U, is the sum of the kinetic

energy in the reference frame in which the center of mass is at restand the

potential energy arising from the forces of the particles on each other.

Difference between the total energy and the internal energy?

The internal energy is astate functionit depends only on

the values of macroparameters (the state of a system), not

on the method of preparation of this state (the path in the

macroparameter space is irrelevant).

U = U (V, T)In equilibrium [ f (P,V,T)=0 ] :

U depends on the kinetic energy of particles in a system and an average

inter-particle distance (~ V-1/3)interactions.

P

VT A

B

For an ideal gas (no interactions) : U = U(T

) - pure kinetic

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Internal Energy of an Ideal Gas

The internal energy of an ideal gas

with fdegrees of freedom: TNkf

U B2

f 3 (monatomic), 5 (diatomic), 6 (polyatomic)

How does the internal energy of air in this (not-air-tight) room change

with Tif the external P = const?

PVf

Tk

PVNTkN

fU

B

roominBroomin22

(here we consider only trans.+rotat. degrees of freedom, and neglectthe vibrational ones that can be excited at very high temperatures)

- does not change at all, an increase of the kinetic energy of individual

molecules with Tis compensated by a decrease of their number.

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Work and Heating (Heat)

We are often interested in U, not U. U isdue to:

Q - energy flow between a system and itsenvironment due to Tacross a boundary and a finite

thermal conductivity of the boundary

heat ing (Q > 0)/c oo l ing (Q < 0)

(there is no such physical quantity as heat; to

emphasize this fact, it is better to use the term

heatingrather than heat)

W - any other kind of energy transfer across

boundary - work

Heating/cool in g pro cesses:

conduct ion: the energy transfer by molecular contact fast-moving

molecules transfer energy to slow-moving molecules by collisions;

convect ion: by macroscopic motion of gas or liquid

radiat ion: by emission/absorption of electromagnetic radiation.

HEATING

WORK

Work and Heating are both defined to describe energy transfer

across a system boundary.

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The First Law

For a cyclic process (Ui= Uf) Q= - W.

If, in addition, Q = 0 then W = 0

Perpetual motion machines come in two types: type 1 violates

the 1stLaw (energy would be created from nothing), type 2 violates

the 2ndLaw (the energy is extracted from a reservoir in a way that

causes the net entropy of the machine+reservoir to decrease).

The first law of thermodynamics:the internal energy of a system can be

changed by doing work on it or by heating/cooling it.

U = Q + W

From the microscopic point of view, this statement is equivalent to

a statement of conservation of energy.

P

VT

An equivalent formulation:

Perpetual motion machines of the first type do not exist.

Sign convent ion:we consider Qand Wto be posi t iveif energy flows

in tothe system.

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Quasi-Static Processes

Quasi-stat ic (quasi-equi l ibr ium ) processessufficiently

slow processes, anyintermediate state can be consideredas an equilibrium state (the macroparamers are well-

defined for all intermediate states).

Examples ofquasi-

equi l ibr ium processes:

isochoric: V= const

isobaric: P= const

isothermal: T= const

adiabatic: Q= 0

For quasi-equilibrium processes, P, V, T are

well-def ined the path between two statesis a continuouslinesin the P, V, Tspace.

P

V

T

1

2

Advantage: the state of a system that participates in a quasi-equilibrium

process can be described with the same (smal l ) number of macro

parameters as for a system in equilibrium (e.g., for an ideal gas in quasi-equilibrium processes, this could be T and P). By contrast, for non-

equi l ibr ium processes(e.g. turbulent flow of gas), we need a huge number

of macro parameters.

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Work

The sign:if the volume is decreased, Wis posi t ive(by

compressing gas, we increase its internal energy); if the

volume is increased, W is negat ive(the gas decreases

its internal energy by doing some work on the

environment).

2

1

),(21V

V

dVVTPW

The workdon e by an external forceon a gas

enclosed within a cylinder fitted with a piston:

W = (PA ) dx = P (Adx)= - PdV

x

P

W = - PdV - applies to any

shape of system boundary

The work is not necessarily associated with the volume changese.g.,

in the Joulesexperiments on determining the mechanicalequivalent of

heat,the system (water) was heated by stirring.

dU = Q PdV

A the

pistonarea

force

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Wand Qare no tState Functions

P

V

P2

P1

V1 V2

A B

CD

- the work is negative for the clockwisecycle; if

the cyclic process were carried out in the reverse

order (counterclockwise), the net work done on

the gas would be positive.

01212

211122

VVPP

VVPVVPWWW CDABnet

2

1

),(21V

VdVVTPW

- we can bring the system from state 1 to

state 2 along infinite # of paths, and for each

path P(T,V)will be different.

U is a state function, W - is not

thus, Qisnot a state function either.U = Q + W

Since the work done on a system depends not

only on the in i t ia land f inalstates, but also on the

intermediatestates, itis not a state function.

PV diagram

P

V

T1

2

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Comment on State Functions

U, P, T, and V are the state functions, Q and W are not. Specifying an initial and final

states of a system does not fix the values of Q and W, we need to know the whole

process (the intermediate states). Analogy: in classical mechanics, if a force is not

conservative (e.g., friction), the initial and final positions do not determine the work, the

entire path must be specified.

x

y z(x1,y1)

z(x2,y2)

dyy

zdx

x

zzd

xy

dyyxAdxyxAad yx ,, - it is an exact differential if it is

x

yxA

y

yxA yx

,,

yxzdyydxxzad ,, the difference between the values of some (state) function

z(x,y)at these points:

A necessary and sufficient condition for this:

If this condition

holds:

y

yxzyxA

x

yxzyxA yx

,,

,,

e.g., for an ideal gas:

dV

V

TdT

fNkPdVdUQ

B 2 - cross derivatives

are not equal

dVPSdTUd

U

VS- an exact differential

In math terms, Qand Ware not exact differentials of some functions

of macroparameters. To emphasize that Wand Qare NOT the state

functions, we will use sometimes the curled symbols (instead of d)

for their increments (Qand W).

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Problem

Imagine that an ideal monatomic gas is taken from its initial stateA

to stateBby an isothermalprocess, from Bto Cby an isobaricprocess, and from

Cback to its initial state A by an isochoricprocess. Fill in the signs of Q,

W, and U for each step.

V, m3

P,

105PaA

B

C

Step Q W U

AB

B C

C A

2

1

1 2

+ -- 0

-- + --

+ 0 +

T=const

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Another Problem

During the ascent of a meteorological helium-gas filled balloon,

its volume increases from Vi = 1 m3 to Vf = 1.8 m

3, and the

pressure inside the balloon decreases from 1 bar (=105N/m2) to

0.5 bar. Assume that the pressure changes linearly with volume

between Viand Vf.

(a) If the initial Tis 300K, what is the final T?

(b) How much work is done bythe gas in the balloon?

(c) How much heatdoes the gas absorb, if any?

P

V

Pf

Pi

Vi Vf

K2701mbar1

1.8mbar5.0K300

3

3

ii

ff

if

B

BVP

VPTT

Nk

PVTTNkPV

(a)

(b) f

i

V

V

ON dVVPW )(

bar625.1bar/m625.0 3 VVP

(c) ONWQU

Jmbarmbarmbar 333 41066.04.05.08.05.0)( f

i

V

V

BY dVVPW

- work done ona system f

i

V

V

BY dVVPW )( - work done bya system

BYON WW

JJJ 445 105.41061.0105.112

3

2

3

BY

i

f

iiONifBON WT

TVPWTTNkWUQ

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Quasistatic Processes in an Ideal Gas

isochoric ( V= const )

isobaric ( P= const )

021 W

TCTTNkQ VB 02

31221

0),( 122

1

21 VVPdVTVPW

TCTTNkQ PB 02

51221

21QdU

2121 QWdU

V

P

V1,2

PV= NkBT1

PV= NkBT21

2

V

P

V1

PV= NkBT1

PV= NkBT21

2

V2

(see the last slide)

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Isothermal Process in an Ideal Gas

1

221 ln),(

2

1

2

1V

VTNk

V

dVTNkdVTVPW B

V

V

B

V

V

f

iBfi

V

VTNkW ln

Wi-f > 0 if Vi>Vf (compression)

Wi-f < 0 if Vi

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Adiabatic Process in an Ideal Gas

adiabatic(thermallyisolatedsystem)

PdVdTNkf

dUTNkf

U BB 22

( f the # of unfrozen degrees of freedom )

dTNkVdPPdVTNkPV BB PVPdVfVdPPdV 2

fP

dP

fV

dV 21,0

21

constVPPVP

P

V

V

111

1lnln

The amount of work needed to change the state of a thermally isolated systemdepends onlyon the initialand finalstates and not on the intermediatestates.

021 Q 21WdU

to calculate W1-2, we need to know P(V,T)

for an adiabatic process

2

1

),(21

V

V

dVTVPW

0

11

P

P

V

V P

dP

V

dV

V

P

V1

PV= NkBT1

PV= NkBT21

2

V2

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Adiabatic Process in an Ideal Gas (cont.)

V

P

V1

PV= NkBT1

PV= NkBT21

2

1

1

1

2

11

1121

11

1

1

),(2

1

2

1

VVVP

dVV

VPdVTVPW

V

V

V

V

1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)(again, neglecting the vibrational degrees of freedom)

constVPPV

11

An adiabata is steeperthan an isotherma:

in an adiabatic process, the work flowing

out of the gas comes at the expense of its

thermal energy its temperature will

decrease.V2

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Problem

Imagine that we rapidly compress a sample of air whose initial pressure is

105Pa and temperature is 220C (= 295 K) to a volume that is a quarter of

its original volume (e.g., pumping bikestire). What is its final temperature?

2211

222

111

VPVP

TNkVP

TNkVP

B

B

1

2

1

2

12

1

1121

2

11

2

112T

T

V

VTT

VPTNkV

VP

V

VPP B

constVTVT 1

22

1

11

KKKV

VTT 51474.12954295 4.0

1

2

112

For adiabatic processes:

Rapid compressionapprox. adiabatic, no time for the energy

exchange with the environment due to thermal conductivity

constTP

/1

also

- poor approx. for a bike pump, works better for diesel engines

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Non-equilibrium Adiabatic Processes

- applies only to quasi-equi l ibr iumprocesses !!!constTV 1

2. On the other hand, U = Q + W = 0

U ~ T Tunchanged

(agrees with experimental finding)

Contradiction because approach

#1 cannot be justified violent

expansion of gas i s not a quasi-

static process. T must remain the

same.

(Joules Free-Expansion Experiment)

constTV 11. Vincreases Tdecreases (cooling)

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The Enthalpy

Isobaric processes (P = const):

dU = Q - PV = Q - (PV) Q = U + (PV)

The enthalpy is a state fun ct ion, because U, P,

and V are state functions. In isobaric processes,the energy received by a system by heating equals

to the change in enthalpy.

Q= H

isochoric:

isobaric:

in both cases, Q

does not

depend on the

path from 1 to 2.

Consequence: the energy released (absorbed) in chemical reactions at constant

volume (pressure) depends only on the initial and final states of a system.

H U + PV - the enthalpy

The enth alpy of an id eal gas:

(depends onT only)

TNkf

TNkTNkf

PVUH BBB

1

22

Q= U

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Heat Capacity

T

QC

The heat capacity o f a system- the amount of energy

transfer due to heating required to produce a unit

temperature rise in that system

Cis NOT a state function(since Qis not a

state function)it depends on the path

between two states of a system

T

V

T1

T1+dT

i

f1 f2 f3

The specif ic heat capacitym

Cc

( isothermic C =, adiabatic C = 0 )

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CVand CP

dT

PdVdU

dT

QC

VV

T

U

C

the heat capacity atconstant vo lume

the heat capacity at

con stant pressureP

PT

HC

To find CPand CV, we need f (P,V,T) = 0and U = U (V,T)

dT

dVP

V

U

T

U

dVV

UdT

T

UdU

dT

PdVdUC

TV

TV

PT

VPdT

dVP

V

UCC

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CVand CPforan Ideal Gas

nRf

Nkf

C BV22

CV = dU/dT

3/2NkB

5/2NkB

7/2NkB

10 100 1000 T, K

Translation

Rotation

Vibration

CV

of one mole of H2

R

NkP

Nk

PdT

dV

PV

U

CC BB

PTVP

( for one mole )0 nR

f

Nkf

NkCC BBVP

12

1

2

For an ideal gas

TNkf

U B2

# of moles

For one mole of a monatomic ideal gas: RCRCPV 2

5

2

3