Download JIF 314 Chap 4

JIF 314
Thermodynamics
Chapter 4
Heat and the first law of
thermodynamics
Key concepts
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Work and heat
Adiabatic work
Work done adiabatically is path
independence
Internal energy function, U is a state fuction
of the system.
Wif (adiabatic) = Uf – Ui
Definition: Uf >Ui when work is done on the
system.
Internal energy function
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Interpretation of DU = Uf – Ui
Conservation of energy
Difference in energy function = energy
change in the system
U a function of two thermodynamical
coordinates, e.g. {P,T}, {P,V},{V,T}
The third variable is fixed by the equation of
state.
U is specified by any two
thermodynamical coordinates (out of
three)
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Example: Given the equation of state, PV = RT, there are two
degree of freedom with three variables. Out of three variable,
only two are independent since the equation of state fixes the
third one.
For example, if we choose {V ,T } as the two independent
variables, P is then the dependent variable that is fixed by the
equation of state via
P =RT / V.
Alternatively, we can also choose {P ,T } as the two
independent variables, V is then the dependent variable via the
equation of state,V = RT /P .
As a conclusion, to specify the state of U, we need only any
pair of independent thermodynamical coordinates
U is path-independent
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U is path independent, hence dU is an exact differential
Say U (a, b), U (a+da, b+db), a,b any two thermodynamical
coordinates
dU = U (a+da, b+db) - U (a, b)
U
dU 
a


U
da 
b
b fixed
db
a fixed
{a, b} can be e.g. {T, V } or {T, P } or {V, P }.
In each case, the third variable, c, are P, V and T
respectively.
U
U
a b fixed
a b fixed
U
a
U
,
a
b fixed
are two different functions
c fixed
U
a

U = U (a,b)
U
a
≠
b fixed
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c fixed
U = U (a,c )
Non-adiabatical process
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Diathermal wall
For non-adiabatic process, Wif ≠ Uf – Ui
The difference between Wif and Uf – Ui is called
heat, Q = (Uf – Ui) - Wif
Convention: Q is positive if heat enters a system,
negative when if leave the system
Transit of heat is a result of a temperature difference
Heat is a form of energy
Q = (Uf – Ui) - W
Volume expand from Vi to Vf , hence work is done by the
system. Internal energy increases from Ui to Uf when
positive Q flows in. Work is done by the system.
weight
Volume
expands
Ui
UiUf
Diathermal
wall
permitting
heat flow
Q
Temperature
Tsu > Tsy so
that heat flows
from
surrounding
(su) to the
system (sy)
Q and W have meaning only if a state
undergoes transitional process to a new
one
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Heating and working are transient processes that
causes a system to change from one state to
another.
Heat and work are involved only in the process of
making transition from a state to another.
Once the transition of states ceases and equilibrium
achieved, heat or work exist no more.
Once the transition of state ceases, what endures
finally is the new state, and the final internal energy.
Infinitesimal amount of Q, W are not
exact differentials
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U is a state function of the coordinates of the
system, hence, it is path-independent
The difference in U between two
infinitesimally different states is an exact
differential, dU, and we can write, e.g.
U
U
dU 
dT 
dP
T P
P T
Infinitesimal amount of Q, W are not
exact differentials
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In contrast, Q and W are not state function,
path-dependent.
The difference in Q and W between two
infinitesimally different states are not exact
differential,
that is, e.g., dQ  Q dT  Q dP
T
P
P T
Inexact differential form of Q and W
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Hence, we use dQ to denote an infinitesimal amount of
heat, and not the differential form, dQ.
The same notation goes to W.
Calculation of W and Q are pathdependence
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What all these mean are that: the calculation
involving heat and work is path-dependent,
and normally we have to carry out integration
to determine W and Q. between two states,
that is path-dependent.
Path-independence and path-dependence
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As an example, when we calculate the difference in
internal energy between two states, we only need to
calculate the difference, DU =Uf – Ui. This difference is
always the same since U is a state function. This infers
path-independence.
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However, in calculating the work done, DW
when a system change from one state to
another, we cannot simply calculate DW as
Wf – Wi but we have to perform the
integration which will result in different value
for process carried out via different route (e.g.
adiabatical path result in a final work done
that is different from that is a non-adibatical
one)

path 1
dW   path 2 dW
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Within an adiabatic boundary, the heat lost
(or gained) by system A is equal to the heat
gained (or lost) by system B
Q = - Q’
System A
Q’, heat flow
into system B
Q, heat flom
into system A
System B
Adiabatic wall
diathermal wall
Differential form of the first law
dU  dQ  dW
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Two inexact differentials on the right hand
side (RHS) make one exact differential on the
LHS.
For hydrostatic system (fluid),
dW   PdV
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and the first law reduces to
dU  dQ  PdV
Work done is
path dependent
Heat Capacity
Q
dQ
C  lim

Ti T f T  T
dT
f
i
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Ti
In unit of joules per kelvin (J/K)
It is a extensive quantity (i.e. the larger the
mass the larger is the value of C since a
larger amount of heat is require to heat up
the material for 1 degree.)
Specific heat Capacity
c C/m
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In unit of joules per kelvin per kg (J/kg∙K)
Intensive quantity, i.e. it’s value remains the
same for different amount of mass of the
same material.
Molar heat capacity
c C/n
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n is the amount of material measured in unit of mole.
In unit of joules per kelvin per mole (J/mol∙K)
Intensive quantity, i.e. it’s value remains the same
for different amount of mass of the same material.
Amount of material in mole
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1 mole of material = NA atom
NA = Avogardo number, 6.023  1023
If an atom has a mass of m, N atoms will
have a total mass of M = mN
Given an element with atomic mass m, M kg
of material made up of such element contains
N = M/m atoms.
The ratio of N/NA defines the amount of atom
in mole of that material: n = N/NA
Heat capacity at constant pressure
 dQ 
CP  
  CP  P, T 
 dT  P
Heat capacity at constant volume
 dQ 
CV  
  CV V , P 
 dT V
Equations for a hydrostatic system
dU  dQ  PdV
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If we choose U=U(T,V)
 U
dU  
 T
 U
dQ  PdV  
 T

 U 
dT



 dV
V
 V T

 U 
dT



 dV
V
 V T
 U 

 U 
dQ  
dT


P

 dV


 T V
 V T

 dT
 dV
dQ  U   U 

  
  P
dT  T V  V T
 dT
Special case, dV = 0 (for the case of CV)

Raising the temperature by heating
substance but without changing the volume:
 dV
dQ  U   U 
dQ
 U 


 CV  
  
  P

dT  T V  V T
dT V
 T V
 dT

Specific heat at constant volume of a
substances CV can be calculated from theory
if the internal energy function of that
substance is known.
Special case, dP = 0 (for the case of CP)

Raising the temperature by heating
substance but without changing the pressure:
 dV
dQ  U   U 


  
  P
dT  T V  V T
 dT
dQ
dT
P
  V 
 U   U 
 CP  
  
  P 


T

V

T

V 
T
P

 U 

CP  CV  

P
V 

 V T

CP  CV  PV 
 U 

 
V
 V T

Specific heat at constant pressure of a substances CP can
be calculated from theory if the internal energy function
and  of that substance is known.
Heat reservior
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A body of such a large mass that it may
absorbed or reject an unlimited quantity of
heat without experiencing an appreciable
change in temperature or in any other
thermodynamic coordinate.
Calculating quasi-static isobaric heat
transfer process via a temperature
difference
Tf
QP   CP dT
Ti
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If CP is constant in temperature in the range
of Ti – Tf,
QP  CP T f  Ti 
Calculating quasi-static isochoric heat
transfer process via a temperature
difference
Tf
QV   CV dT
Ti
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If CV is constant in temperature in the range
of Ti – Tf,
QV  CV T f  Ti 
Three mechanism of heat conduction
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Conduction
Convection
Radiation
Heat conduction
Heat flow from high
temperature to low
temperature
Thermal conductivity
dQ
dT
  KA
dt
dx
Cross section
perpendicular to
direction of heat flow
Temperature gradient
Heat convection
Convection coefficient
dQ
 hADT
dt
Temperature difference
Thermal radiation
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Emission of heat as electromagnetic radiation
Absorbitivity
Radiant exitance, R
Emissivity, e
Black body
Kirchhoff’s law
Radiated heat
dQ
 Ae T   Rbb TW   Rbb T 
dt
Stefan-Boltzmann law
Rbb T    T
4
Stefan-Boltzmann constant, = 5.67051  10-8 W/m2∙K4

dQ
4
4
 Ae TW  T
dt

Experimental determination of 
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Nonequilibrium method
Equilibrium method
Question 4.1
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Regarding the internal energy of a hydrostatic
system to be a function of T and P, derive the
following equations:
a)
b)
c)
Solution for 4(a)
U =U (T, P )
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First law of Thermodynamics =>
Combining both
Eq. (1)
Solution for 4(a) (cont.)
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For a hydrostatic system or PVT system,
we can write V as a function of T and P.
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By substituting the expression of dV into
equation Eq(1), we get

Eq. (2)
Solution for 4(b)
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At constant pressure, dP=0. Setting dP=0, and
dividing Eq. (2) by dT, we get
dQ
 CP
dT P

Since

Therefore,
Solution for 4(c)
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At constant volume, dV=0. Setting dV=0, and dividing
Eq. (1) by dT, we get
dQ
 U   U  dP
 U   U   P 
 CV  

 

 
 
 ;
dT V
 T  P  P T dT V  T  P  P T  T V
dP
 P 


dT V  T V
Eq. (3)

1  V 
1  T 
 T 

 V  
 
  V
V  T  P
  V  P  V  P
Eq. (4)
Solution for 4(c)

Combining Eq. (3), (4), and
 U   
CV   CP   PV   
 
 P T  
 CV  CP   PV  U 



 

 P T
 
 
Question 4.3

One mole of a gas obeys the van der Waals
equation of state:
and its molar internal energy is given by
where a,b,c and R are constants. Calculate
the molar heat capacities cv and cP.
Solutions

We write U = U(T,V)
 u 
 u 
du  
dT


  dv
 T v
 v T
dq  du  Pdv
Eq. (1)
Eq. (2)
Eq. (1) combined with Eq. (2)
 u 

 u 
dq  
 dT     P  dv
 T v
 v T

dT
 dv
dq  u   u 



P

  
dT  T v  v T
 dT
Eq. (5)
 u 
cV  

 T v
Solutions

At constant volume, Eq. (5) becomes
dq
 u 
 cV  
  c,since u  cT  a / v
dT v
 T v
Solutions

At constant volume, Eq. (5) becomes
 dv
 u   u 
 cP  
     P 
 T V  v T
P
 dT P
 u 
  v 
a
  v 
cP  cV     P  

c


P
V


 since u  cT  a / v
2


v
  T  P
 v T
  T  P
dq
dT

From
 v 

 
 T  P
R
a 2a  v  b 
P 2 
v
v3


a
P


2
 a
  V 
V
CP  CV   2  P  
R
 
2
a
v

b
 
a
V
  T  P 
P 2 


v
v3