Lecture2.ppt first law of thermodynamics

first law of thermodynamics states that the

1. Lecture 2 The First Law of
Thermodynamics (Ch.1)
Outline:
1.Internal Energy, Work, Heating
2.Energy Conservation – the First Law
3.Quasi-static processes
4.Enthalpy
5.Heat Capacity

2. Internal Energy
system
U = kinetic + potential
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 rest and 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 a state function – it 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
V
T
A
B
For an ideal gas (no interactions) : U = U (T) - “pure” kinetic

3. Internal Energy of an Ideal Gas
The internal energy of an ideal gas
with f degrees of freedom:
TNk
f
U
B
2

f  3 (monatomic), 5 (diatomic), 6 (polyatomic)
How does the internal energy of air in this (not-air-tight) room change
with T if the external P = const?
PV
f
Tk
PV
NTkN
f
U
B
roominBroomin
22








(here we consider only trans.+rotat. degrees of freedom, and neglect
the 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 T is compensated by a decrease of their number.

4. Work and Heating (“Heat”)
We are often interested in U , not U. U is due to:
Q - energy flow between a system and its
environment due to T across a boundary and a finite
thermal conductivity of the boundary
– heating (Q > 0) /cooling (Q < 0)
(there is no such physical quantity as “heat”; to
emphasize this fact, it is better to use the term
“heating” rather than “heat”)
W - any other kind of energy transfer across
boundary - work
Heating/cooling processes:
conduction: the energy transfer by molecular contact – fast-moving
molecules transfer energy to slow-moving molecules by collisions;
convection: by macroscopic motion of gas or liquid
radiation: by emission/absorption of electromagnetic radiation.
HEATING
WORK
Work and Heating are both defined to describe energy transfer
across a system boundary.

5. The First Law
For a cyclic process (U
i = U
f)  Q = - W.
If, in addition, Q = 0 then W = 0
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 conservation of energy.
P
V
T
An equivalent formulation:
Perpetual motion machines of the first type do not exist.
Sign convention: we consider Q and W to be positive if energy
flows into the system.

6. Quasi-Static Processes
Quasi-static (quasi-equilibrium) processes – sufficiently
slow processes, any intermediate state can be considered
as an equilibrium state (the macroparamers are well-
defined for all intermediate states).
Examples of quasi-
equilibrium processes:
isochoric: V = const
isobaric: P = const
isothermal: T = const
adiabatic: Q = 0
For quasi-equilibrium processes, P, V, T are
well-defined – the “path” between two states is
a continuous lines in the P, V, T space.
P
V
T
1
2
Advantage: the state of a system that participates in a quasi-equilibrium
process can be described with the same (small) 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-
equilibrium processes (e.g. turbulent flow of gas), we need a huge number
of macro parameters.

7. Work
The sign: if the volume is decreased, W is positive (by
compressing gas, we increase its internal energy); if the
volume is increased, W is negative (the gas decreases
its internal energy by doing some work on the
environment).



2
1
),(
21
V
V
dVVTPW
The work done by an external force on 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 changes – e.g.,
in the Joule’s experiments on determining the “mechanical equivalent of
heat”, the system (water) was heated by stirring.
dU = Q – PdV
A – the
piston
area
force

8. W and Q are not State Functions
P
V
P
2
P
1
V
1
V
2
A B
CD
- the work is negative for the “clockwise” cycle; if
the cyclic process were carried out in the reverse
order (counterclockwise), the net work done on
the gas would be positive.
  
  0
1212
211122


VVPP
VVPVVPWWW
CDABnet



2
1
),(
21
V
V
dVVTPW
- 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, Q is not a state function either.
U = Q + W
Since the work done on a system depends not
only on the initial and final states, but also on the
intermediate states, it is not a state function.
PV diagram
P
V
T
1
2

9. the difference between the values of some (state) function
z(x,y) at these points:
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(x
1
,y
1
)
z(x
2
,y
2
)
dy
y
z
dx
x
z
zd
xy



















 , ,
x ydz A x y dx A x y dy  - it is an exact differential if it is
 
x
yxA
y
yxA yx




 ,,
 , ,dz z x dx y dy z x y   
A necessary and sufficient condition for this:
If this condition
holds:




y
yxz
yxA
x
yxz
yxA
yx






,
,
,
,
e.g., for an ideal gas: 





 dV
V
T
dT
f
NkPdVdUQ
B
2

- cross derivatives
are not equal
dVPSdTUd 
U
V
S- an exact differential
In math terms, Q and W are not exact differentials of some functions
of macroparameters. To emphasize that W and Q are NOT the state
functions, we will use sometimes the curled symbols  (instead of d)
for their increments (Q and W).

10. Problem
Imagine that an ideal monatomic gas is taken from its initial state A to state
B by an isothermal process, from B to C by an isobaric process, and from
C back to its initial state A by an isochoric process. Fill in the signs of Q,
W, and U for each step.
V, m
3
P,
10
5
Pa
A
B
C
Step Q W U
A  B
B  C
C  A
2
1
1 2
+ -- 0
-- + --
+ 0 +
T=const
TNk
f
U
B
2
 B
PV Nk T

11. Quasistatic Processes in an Ideal Gas
isochoric ( V = const )
isobaric ( P = const )
0
21
W
   TCTTNkQ
VB 
 0
2
3
1221
 0),(
12
2
1
21  VVPdVTVPW
   TCTTNkQ
PB


0
2
5
1221
21
QdU
2121  QWdU
V
P
V
1,2
PV= Nk
B
T
1
PV= Nk
B
T
2
1
2
V
P
V
1
PV= Nk
B
T
1
PV= Nk
B
T
2
1
2
V
2
(see the last slide)

12. Isothermal Process in an Ideal Gas
1
2
21 ln),(
2
1
2
1
V
V
TNk
V
dV
TNkdVTVPW
B
V
V
B
V
V
 
f
i
Bfi
V
V
TNkW ln

W
i-f
> 0 if V
i
>V
f
(compression)
W
i-f
< 0 if V
i
<V
f
(expansion)
isothermal ( T = const ) :
V
P
PV= Nk
B
T
V
1
V
2
W
2121 
WQ
0dU

13. Adiabatic Process in an Ideal Gas
adiabatic (thermally isolated system)
PdVdTNk
f
dUTNk
f
U
BB 
22
( f – the # of “unfrozen” degrees of freedom )
dTNkVdPPdVTNkPV
BB  PVPdV
f
VdPPdV 
2
fP
dP
fV
dV 2
1,0
2
1 








 

constVPPV
P
P
V
V

















11
1
1
lnln
The amount of work needed to change the state of a thermally isolated system
depends only on the initial and final states and not on the intermediate states.
0
21


Q
21
WdU
to calculate W
1-2 , 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
V
1
PV= Nk
B
T
1
PV= Nk
B
T
2
1
2
V
2
Adiabatic
exponent

14. Adiabatic Process in an Ideal Gas (cont.)
V
P
V
1
PV= Nk
B
T
1
PV= Nk
B
T
2
1
2
2
2 2
1 1 1
11 1
1 2 1 1
1 1 1 1
2 1
1
( , )
1
1 1 1
1
VV V
V V V
PV
W P V T dV dV PV V
V
PV
V V

 


 


 

 
  
 
 
   

 
 
 1+2/31.67 (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 “steeper” than 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.
V
2
Prove 
1 2
2 2
B
f f
W PV Nk T U
    

15. Summary of quasi-static processes of ideal gas
Quasi-Static
process
U Q W
Ideal gas
law
isobaric
(P=0)
isochoric
(V=0)
0
isothermal
(T=0)
0
adiabatic
(Q=0)
0
fi
i f
VV
T T

2 2
B
f f
U Nk T P V    
f i
U U U  
2
2
f
P V

 P V 

2 2
B
f f
U Nk T P V     
2
f
P V
fi
i f
PP
T T

i i f fPV PVln
f
B
i
V
Nk T
V
W
i i f fPV PV
 


2 2
B
f f
U Nk T PV     U

16. Problem
Imagine that we rapidly compress a sample of air whose initial pressure is
10
5
Pa and temperature is 22
0
C (= 295 K) to a volume that is a quarter of
its original volume (e.g., pumping bike’s tire). What is its final temperature?

2211
222
111
VPVP
TNkVP
TNkVP
B
B



1
2
1
2
1
2
1
11
21
2
11
2
11
2
T
T
V
V
T
T
VP
TNk
V
VP
V
VP
P
B 
















constVTVT 
 1
22
1
11

KKK
V
V
TT 51474.12954295
4.0
1
2
1
12 










For adiabatic processes:
Rapid compression – approx. 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

17. Non-equilibrium Adiabatic Processes
- applies only to quasi-equilibrium processes !!! constTV
1
2.On the other hand, U = Q + W = 0
U ~ T  T – unchanged
(agrees with experimental finding)
Contradiction – because approach
#1 cannot be justified – violent
expansion of gas is not a quasi-
static process. T must remain the
same.
constTV
1
1. V – increases
 T – decreases (cooling)
Free expansion

18. The Enthalpy
Isobaric processes (P = const):
dU = Q - PV = Q -(PV)  Q =  U + (PV)
The enthalpy is a state function, 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 enthalpy of an ideal gas:
(depends on T only)
TNk
f
TNkTNk
f
PVUH
BBB 





 1
22
Q =  U

19. Heat Capacity
T
Q
C



The heat capacity of a system - the amount of energy
transfer due to heating required to produce a unit
temperature rise in that system
C is NOT a state function (since Q is not a
state function) – it depends on the path
between two states of a system 
T
V
T
1
T
1+dT
i
f
1
f
2
f
3
The specific heat capacity
m
C
c
( isothermic – C = , adiabatic – C = 0 )

20. C
V
and C
P
dT
PdVdU
dT
Q
C



V = const
P
=
c
o
n
s
t
V
V
T
U
C 







 the heat capacity at
constant volume
the heat capacity at
constant pressure
P
P
T
H
C 








To find C
P and C
V, we need f (P,V,T) = 0 and U = U (V,T)
nR
f
Nk
f
C
BV
22

For an ideal gas TNk
f
U
B
2

# of moles
1
2
B
f
H Nk T
 
 
 
 
1
2
P
f
C nR
 
 
 
 
For one mole of a
monatomic ideal gas:
RCRC
PV
2
5
2
3


21. Another Problem
During the ascent of a meteorological helium-gas filled balloon,
its volume increases from V
i
= 1 m
3
to V
f
= 1.8 m
3
, and the
pressure inside the balloon decreases from 1 bar (=10
5
N/m
2
) to
0.5 bar. Assume that the pressure changes linearly with volume
between V
i
and V
f
.
(a) If the initial T is 300K, what is the final T?
(b) How much work is done by the gas in the balloon?
(c) How much “heat” does the gas absorb, if any?
P
V
P
f
P
i
V
i
V
f
K270
1mbar1
1.8mbar5.0
K300
3
3




ii
ff
if
B
B
VP
VP
TT
Nk
PV
TTNkPV
(a)
(b)


f
i
V
V
ON dVVPW )(
 bar625.1bar/m625.0
3
 VVP
(c)
ON
WQU 
  Jmbarmbarmbar
333 4
1066.04.05.08.05.0)( 
f
i
V
V
BY dVVPW
- work done on a system


f
i
V
V
BY
dVVPW )( - work done by a system
BYON WW  
   JJJ
445
105.41061.0105.11
2
3
2
3










BY
i
f
iiONifBON W
T
T
VPWTTNkWUQ 