Thermal Physics Lecture Note 3

8/10/2019 Thermal Physics Lecture Note 3

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3. The first law of thermodynamics F

3.1 Work done

Mechanical work = force applied distance (displacement)

dW = F cos ds (applied to a point centre of gravity)

3.2 Work done to a surface:

P = pressure of gas ; Pe= external pressure applied on the system

P >Pe # the gas expands

P


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Sign convention :

dV = +ve (expansion)

dW = +ve # Work donebythe system

dV = -ve (compression)

dW = -ve # Work done to the system

If the compression is performed quasistaticallyso the process is reversible,

then the externally applied pressure Pe=P, the pressure of the gas

then dW = P dV

In a finite reversible process, to change the volume from Vato Vb,

total work done required:b

a

V

V

ba PdVW

Wcan be obtained by expressingP in terms of V using the equation of state.


3/20

Example 1 : Ideal gas

PV = nRT # PdVnRW

(a) If the process is isothermal,

then Wa

b is represented by the area under

theP V curve, hence#

$

%

&

'

a

b

V

V

baV

VnRTdV

VnRTW

b

a

ln1

(b) If the process is isobaric, then

)(ab

V

V

ba VVPdVPW

b

a


4/20

Example 2 : Paramagnetic substance

Consider cylindrical paramagnet being magnetised

by current flowing through coil wound around it(solenoid)

L : length ; A: cross-sectional area

N : number of turns of the coil

I : current flowing through the coil

The magnetic field intensity produced byIflowing through the coil is

L

NIH

Magnetic flux density, HA

Bo

if centre of coil is free space

B = !o(H +!) if centre of coil is filled with paramagnetic

material

dB = !o(dH + d!) ; ! is the magnetic moment per unit

volume


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IfB is varying with time, an electromagnetic force (emf) will be induced

emf,dt

dBNA

The power input :#

=$

I

dt

dBHV

dt

dB

L

VHL

dt

dBANI

A

NI

#

$

%

&

'

Work done on the system : dW = # dt = $ I dt dBHV = % !oVH dH % !oVH d!

The term !oVH dH is the work done required for free space

!oVH d! is the work done required to magnitise the paramagnet by d!

Unit: H - ampere per m ( A m-1

)

! A m-1

!o= 4$ 10-7henry per meter ( H m-1 )

!oVH d! henry ampere2( H A

2) % joules ( J )

Work of magnitization, dWm=H dM, M = !oV d!


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Example 3 : Surface tension

Take a "U" shape wire frame with another

wire placed on top to form a closed rectangle

Dip the wire frame into soap water so that

a thin film of soap is formed inside the frame

& : surface tension ( N/m )

The wire will experience a force due to this

surface tension given by

F = 2& L ( 2 because the film has 2 surfaces)

If we pull the wire so as to increasexby dx, the work done required is

dW = % 2& L dx = % & dA

[Note: We use%

vesign here since work is done to the system]


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3.3 Work done to change the state of a system depends on path

The total work done to change the state of a system from a to b:b

a

dWW

For a P-V-T system, take isothermal process, the path taken from a to bcan be

represented by

VfP=

and the work done to perform the process isb

a

V

V

PdVW which is the area under the

curve VfP= on the P-V surface

The state of the system can change from a to b through different path, each on

represented by a different function. We can see that if 2 paths are different, the areas

under the 2 curves will be different. Hence,

b

a

V

V

II dVVfW ' b

a

V

V

IIII dVVfW # WI% WII ' 0


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This means that if the state of the system is changed from ato b through path I and then

return to athrough path II, there will be a net work done by the system [ W= +ve ]

Net work PdVdWW = +ve

Hence, dW is an inexact differential

WI( WII

II

Va Vb


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The work done in the expansion (by the system) and compression (to the system) of a gas

is an example of the configuration work

Note that in this case, P is an intensive variableV is an extensive variable

In general, if a system has a set of intensive variables Y1, Y2, Y3 ...........

and a set of extensive variables X1,X2,X3 ...........

then, the configuration work is dW = ) (Y dX)

If a gas expand into vacuum , no work is required # free expansion

If there is energy loss during a process , the work done to bring about the process is said

to be dissipative work

All dissipative work is always irreversible

Example of dissipative work: Electric current flowing through resistor

Dissipative work : W = * I2Rdt


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3.4 Internal energy

For adiabaticprocess ( no heat flow in or out of the system ), the work done (by the

system or to the system) to change the state of the system will go into the internal energyof the system.

Since work done bythe system is +ve

work done to the system is -ve

We write internal energy, dU= - dW

because we expect the internal energy to increase when work is done to the system.

In theP-V-Tsystem, dU = - PdV

For process from state a and state b,ba

b

a

ab

U

U

WdWUUdU

b

a

#

Thus when a gas expand adiabatically, its internal energy will decrease

Whereas if the gas is compressed adiabatically, its internal energy will increase


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3.5 Heat flow

For nonadiabaticprocess, there is heat flow into the system or out of the system.

Write dQ to be the amount of heat flow during a process, and define

dQ = +ve # heat flow intothe system

dQ = - ve # heat flow out ofthe system

Heat flow from region of high temperature to region of low temperature

Hence, during a nonadiabatic process, the total work done W is given by

W = Wadiabatic+ Q or Wadiabatic= W Q = Ua Ub

That means Ub Ua = Q W

Differential form (for small Q and W), dU = dQ dW

For aP-V-T system, dU = dQ PdV

In general, dU = dQ + YdX

Since dUdoes not depend on path, while dW depends on path, so dQ also depends on path


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3.6 First Law of Thermodynamics

The expression dU = dQ dW

is the differential form of the First Law of Thermodynamics, which states that

In any process in which there is no change in the kinetic and potential

energy of the system, the increase in internal energy of the system equals

the net heat flow into the system minus the total work done by the system.

In a more general form, the First Law of Thermodynamics states that

The total work is the same in all adiabatic processes between any two

equilibrium states having the same kinetic and potential energy.

Generalised by including the kinetic and potential energy, we can write

dE = dQ dW, where dE = dU + dEK + dEP


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3.7 Unit of heat

1 calorie % heat required to flow into 1 gram of water to raise its temperature by 1C

[ 1 15-degree calorie # temperature raised from 14.5 to 15.5 C ]

1 BTU % heat required to flow into 1 lb of water to raise its temperature by 1 F

Joule's experiment (1940 1978) : Mechanical work , Heat

#

1 calorie of heat%

4.19 joules of work done

or 1 BTU of heat % 778.28 ft-lb


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3.8 Heat capacity

For processes without change of phase, the temperature change of a system -T is

proportional to the heat flow into or out of the system,

i.e. CT

Q

, where C is a constant called Heat Capacity of the system

For infinitely small change in temperature,dT

dQ

T

QC

T

0

lim ( J K-1

)

where Q is given as a function of -T ( Note: NOT a function of T)

The heat capacity of aP-V-Tsystem is different for isobaric and isochoric processes.

For isobaric process (constantP),P

PdT

dQC

#

$

%

&

For isochoric process ( constant V),

VV

dT

dQC

#

$

%

&

Specific heat capacity,m

dT

dQ

c

#

$

%

&

orn

dT

dQ

c

#

$

%

&

Heat flow :2

1

T

T

CdTQ


15/20

Copper atP= 1 atm


16/20

Mercury at T = 0 C


17/20

3.9 Heats of transformation; enthalpy

During an isobaric process, the volume is changed from v1to v2.

If a phase change has occured, the heat flow (into or out of the system) is called

the heat of transformation

Using the first law, du= dq dw

Write dw = P( v2 v1 )

dq= l heat of transformation

du = u2 u1

then u2 u1= l P (v2 v1) # l = ( u2+ P v2 ) (u1+ P v1)

The quantity ( u + P v) is called the enthalpy h,

Hence l = h2 h1

h = u + P v


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3.10 Energy equation of steady flow

As an example for the application of the first law of thermodynamics, consider the

following system :


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Processes: piston 1 pushed to the right with velocity v1 and byx1

piston 2 pulled to the right with velocity v2and byx2

heat Q flow into the system

Result from these processes # Work done output through the shaft : Wsh

Work done by the pistons : P2V2 P1V1

Water flows from positionz1toz2 # change in potential energy, -Ep= mg (z2 z2)

Piston movement # change in kinetic energy, -Ek= !m(v22 v1

2)

Total work done of the system : W = Wsh+ (P2V2 P1V1)

First Law of Thermodynamics : -U + -Ek+ -Ep= Q W, -U= m( u2 u1)

.

)()()()( 11221221222112 VPVPWQzzmgmuum sh

In terms of specific quantities

: )()()()( 112212

21

222

112 vPvPwqzzguu sh

Or, shwqgzvPugzvPu )()( 1212

11112

222

1222


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This is the energy equation of steady flow

For a Bernoulli flow, wsh= 0 , assuming adiabatic flow

0)()( 1212

11112

222

1222 gzvPugzvPu

gzPvu 2

2

1

constant

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Thermal Physics Lecture Note 3