### Lecture 10: Ideal Gases

```EGR 334 Thermodynamics
Chapter 3: Section 12-14
Lecture 10:
Ideal Gas Law
Quiz Today?
Today’s main concepts:
•
•
•
•
Be
Be
Be
Be
able to explain the Ideal Gas Law
able to explain when it is appropriate to use the Ideal Gas Law
able to use the Ideal Gas Law to determine State Properties
able to apply the Ideal Gas Law to the solution of
1st Law problems.
• Read Chap 3: Sections 15
Homework Assignment:
From Chap 3: 102, 107,115, 125
3
Ideal Gas Law
When the general compressibility factor, Z  1,
then the following relationship between pressure, temperature, and
volume of a gas applies.
p  ab so lu te p ressu re
pv  RT
which can also be written
pV  m RT
or
pV  nRT
m = mass
n = number of moles
T  ab so lu te tem p eratu re
v  sp ecific vo lu m e
 8 .3 1 4 k J/k m o l  K

o
R   1 .9 8 6 B tu /lb m o l  R
o
1 5 4 5 ft  lb /lb

R
f
m ol

R may be found on Table 3.1
Sec 3.13 : Ideal gases and u, h, cv, cp
4
If a gas behaves as an ideal gas, then its specific
internal energy, u, depends only on temperature.
u  u (T )
Enthalpy, h, was defined as:
h  u  pv
For an ideal gas, since
pv  RT
then
h  u  RT
Since
and
u  u (T )
h  h (T )  u (T )  R T
Therefore:
If a gas can be treated as an ideal gas, its intensive properties of specific
energy and enthalpy are entirely functions of temperature.
Sec 3.13 : Ideal gases and u, h, cv, cp
5
If u and h are only functions of temperature, then the specific
heats may be used to determine relations between
temperature change and energy levels.
cV
 u 



T

V
cP
 h 



T

P
For an ideal gas, the expressions for u and h can be simplified
since u = f(T) and h = f(T)
du 

T2
T1
cV dT
dh 

T2
T1
c P dT
For many cases, the specific heats will be treated as constant
values over a limited temperature range and these integrals will
be approximated as:
u 2  u 1  cV (T 2  T1 )
h 2  h1  c p (T 2  T1 )
Sec 3.13 : Ideal gases and u, h, cv, cp
6
Another important ideal gas equation may be written as
h T
  u T  
RT
dh

dT

du
R
dT
c P  T   cV  T   R

When the specific heat ratio
may also be written as
cP T

kR
k 1
k 
cP
cV
and
For monotonic gases (Ar, Ne, He) with k = 1.4
is used, this equation
cV  T  
R
k 1
c P _ m o n o g a ses 
5
2
R
Sec 3.13 : Ideal gases and u, h, cv, cp
7
Temperature Dependence:
Specific heats cv and cp are
functions of temperature.
du 

T2

T2
T1
dh 
T1
cV ( T ) dT
c P ( T ) dT
If possible, you should look up
their values from tables which
give the specific heat at the
temperature indicated.
(see Tables A-20 and A-20E)
k 
cP
cV
An alternative method is to use a formula to represent cp based on
cp
R
   T  T
2
T
3
 T
4
where Table A-21 has values of , , , ,  for different gases
Sec 3.13 : Ideal gases and u, h, cv, cp
8
It best to use an actual function of the specific heats to evaluate u and h,
by integration
du 

T2
T1
dh 
cV d T

T2
T1
cP dT
But, often this is simplified, evaluating cV and cP at an average T
either
cV a ve 
c v (T1 )  c v (T 2 )
2
and
c p a ve 
c p (T1 )  c p (T 2 )
2
or the specific heat at the average temperature may be used.
cV a ve  c v
 T T 
1
2



2

c p a ve  c p
 T T 
1
2



2

where if cV and cP are treated as constants, then
u  T 2   u  T1   cV  T 2  T1 
h  T 2   h  T1   c P  T 2  T1 
Sec 3.13 : Ideal gases and u, h, cv, cp
9
Summary:
1) Decide if a substance can be treated as an ideal gas…if yes, then
pv  RT
pV  m RT
pV  nRT
or
2) To evaluate changes in internal energy, u, and enthalpy, h:
i) Integrate with cv and cp as function of T (see Table A-21)
T2
T2
dh   cP dT
du 
cV d T

T1
T1
or
ii) Use value of cv or cp at an average temperature (see Table A-20)
u 2  u 1  cV _ a ve  T 2  T1 
or
iii) Use k and R to define
cv 
u 2  u 1  cV  T 2  T1 
or
h 2  h1  c P _ a ve  T 2  T1 
R
k 1
and
cP 
kR
k 1
then
h 2  h1  c P  T 2  T1 
iv) Look up temperature dependent values of u and h on property tables.
 u  u (T 2 )  u (T1 )
 h  h (T 2 )  h (T1 )
Table A-22 has property values for Air
Table A-23 has property values for CO2, C0, H20, O2, and N2.
Sec 3.13 : Ideal gases and u, h, cv, cp
Example: (3.111) A piston cylinder assembly contains air at 2 bar,
300K, and a volume of 2 cubic meters. the air undergoes a process to
a state where the pressure is 1 bar, during which the pressure-volume
relationship is pV = constant. Assuming ideal gas behavior, determine
a) the mass of the air, b) the work, and c) the heat transfer.
10
Sec 3.13 : Ideal gases and u, h, cv, cp
11
Example: (3.111) A piston cylinder assembly contains air at 2 bar,
300K, and a volume of 2 cubic meters. the air undergoes a process to
a state where the pressure is 1 bar, during which the pressure-volume
relationship is pV = constant. Assuming ideal gas behavior, determine
a) the mass of the air, b) the work, and c) the heat transfer.
State 1: p1 = 2 bar
T1 = 300 K
V1 = 2 m3
pV  m RT
For Ideal Gas:
m 
p1V1
R T1
State 2: p2 = 1 bar
T2 = ?
V2 = ?
3

5
( 2 b a r )( 2 m )
10 N / m
(0 .2 8 7 0 kJ / kg  K )(3 0 0 K )
1b a r
For constant pV:
p1V1  p 2V 2

2
1kJ
J
1000 J N  m
 4 .6 5 kg
 2bar 
3
3
V2 
V1  
2
m

4
m

p2
1
b
a
r


p1
Sec 3.13 : Ideal gases and u, h, cv, cp
12
Example: (3.111) continued...
State 2: p2 = 1 bar
T2 = ?
V2 = 4 m3
State 1: p1 = 2 bar
T1 = 300 K
V1 = 2 m3
find T2:
p 2V 2
T2
 R 
p1V1
T1
T2 

p 2V 2
p1V1
T1 
1b a r 4 m
3
2bar 2 m
2
(3 0 0 K )  3 0 0 K
1st Law of Thermodynamics:
 U  c v (T 2  T 1 )  0
U  Q  W
W 

pdV 
C
 V dV
 V2 
 V2 
W  C ln 
  p1V1 ln 

V
V
 1 
5 1 
3
p1V1  p 2V 2  C
 4m  10 N / m 2
J
1kJ
 ( 2 b a r )( 2 m ) ln 
 2 7 7 .3 kJ
3 
2
m
1
b
a
r
N

m
1
0
0
0
J


3
Q   U  W  0  2 7 7 .3 kJ
Sec 3.13 : Ideal gases and u, h, cv, cp
Example: (3.124) Two kilograms (2 kg) of air, initially at 5 bar, 350 K
and 4 kg of CO initially at 2 bar, 450 K are confined to opposite sides
of a rigid, well-insulated container by a partition. The partition is free
to move and allows conduction from one gas to the other without
energy storage in the partition itself. The air and CO each behave as
ideal gases with constant specific heat ratio, k = 1.395. Determine at
equilibrium (a) the temperature in K, (b) the pressure, in bar, and (c)
the volume occupied by each gas, in m3.
13
Sec 3.13 : Ideal gases and u, h, cv, cp
Example: (3.124)
k = 1.395.
CO State 1:
Air: State 1:
14
TC O _ 1  4 5 0 K
T a ir _ 1  3 5 0 K
pCO _1  2b a r
p a ir _ 1  5 b a r
m C O  4 kg
m a ir  2 kg
State 2:
TC O _ 2  T a ir _ 2  T 2
p C O _ 1  p air _ 2  p 2
m C O  m air  m total
1st Law of Thermo:
 U system  Q system  W system
 U C O   U air  0
or
 U air   U C O
since system is isolated
Sec 3.13 : Ideal gases and u, h, cv, cp
15
CO State 1:
Example: (3.124) continued…
 U  m  u 2  u1 
 m cV  T 2  T1 
For CO:
Air: State 1:
TC O _ 1  4 5 0 K
T a ir _ 1  3 5 0 K
pCO _1  2b a r
p a ir _ 1  5 b a r
mCO  4 kg
m a ir  2 k g
 U C O  m C O cV C O  T 2  T C O _ 1 
cV , C O 
RC O
k 1

 0 .2 9 6 8 kJ
/ kg  K
1 .3 9 5  1

 0 .7 5 1 4
kJ
kg  K
 U C O  (4 kg )(0.7514 kJ / kg  K )  T 2  450 K
For Air:
 U air  m air cV air  T 2  T air _ 1 
cV
_ a ir

R a ir
k 1

 0 .2 8 7 0 kJ
/ kg  K
1 .3 9 5  1

 0 .7 2 6 6
kJ
kg  K
 U air  (2 kg )(0.7266 kJ / kg  K )  T 2  350 K


Sec 3.13 : Ideal gases and u, h, cv, cp
Example: (3.124) continued…
 U C O   U air  0
16
CO State 1:
Air: State 1:
TC O _ 1  4 5 0 K
T a ir _ 1  3 5 0 K
pCO _1  2b a r
p a ir _ 1  5 b a r
mCO  4 kg
m a ir  2 k g
(4 kg )(0.7514 kJ / kg  K )  T 2  450 K   (2 kg )(0.7266 kJ / kg  K )  T 2  350 K   0
(3.006 kJ / K )  T 2  450 K   (1.453 kJ / K )  T 2  350 K   0
(3.006 kJ / K )T 2  1352.7 kJ  (1.453 kJ / K ) T 2  508.55 kJ  0
(4.459 kJ / K )T 2  1858.3 kJ
T 2  416.8 K
Sec 3.13 : Ideal gases and u, h, cv, cp
17
Example: (3.124) continued
Find Vtotal
V T  V C O  V A ir
 mRT 
 mRT 





 P  C O ,i
 P  A ir,i

  2   0.2870   350  
 4   0.2968   450  



 
 
5
5




 2  10 
 5  10 
C O ,i
A ir,i

 2.67  0.40 = 3.07 m
Then find pfinal


 ( kg )( kJ / kg  K )( K ) 1000 N  m
2

N /m
kJ

3
pf 
mRT

 m R T CO , f
V to ta l
  mRT
 A ir , f
V to ta l
  4   0.2968   416.8      2   0.2870   416.8   ( kg )( kJ / kg  K )( K ) 1000 J N  m
 3.07 
m
3
kJ
J
1bar
5
10 N / m
 2.39 b a r
Sec 3.13 : Ideal gases and u, h, cv, cp
18
Example: (3.124) continued…
The final volumes are then,
 mRT 


 P  C O ,f
VC O
VC O
  4   0.2968   416.8  

 
5

 2.39  10  
V A ir
V A ir
( kg )( kJ / kg  K )( K ) 1000 N  m
N /m
2
 2.079 m
3
kJ
C O ,f
 mRT 
 

 P  A ir,f
  2   0.2870   416.8  

 
5

 2.39  10  
( kg )( kJ / kg  K )( K ) 1000 N  m
N /m
A ir,f
2
kJ
 1.001 m
3
19
Solution using IT:
Note: The results are slightly different
as the cv and cp values that IT pulled
out slightly different.
20
End of Slides for Lecture 10
```