thermal physics lecture note 3
TRANSCRIPT
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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.
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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
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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
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Copper atP= 1 atm
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Mercury at T = 0 C
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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