1. MODULE 6
IDEAL OR PERFECT GAS
1. CHARACTERISTIC EQUATION
C
T
P
o r
C
T
PV
RT
P
P
RT
RT
P
mRT
PV
=

=
=

=

=

=
where:
P - absolute pressure in KPa
V - volume in m3
m -mass in kg
R -Gas constant in KJ/kg-K
T - absolute temperature in K
 - specific volume in m3/kg
 - density in kg/m3
2. GAS CONSTANT
K
-
kg
KJ
M
8.3143
M
R
R =
=
R = 8.3143 KJ/kgm-K
R - universal gas constant, KJ/kgm-K
M - molecular weight, kg/kg mol
moles
of
no.
-
n
kg
,
mass
m
kg
kg
n
m
M
mol
−
=
3. BOYLE`S LAW (T = C) Robert Boyle (1627-1691)
If the temperature of a certain quantity of gas is held constant, the volume V is inversely proportional to the absolute
pressure P, during a quasi-static change of state.
C
V
P
V
P
or
C
PV
P
1
C
V
or
P
1
V
2
2
1
1 =
=
=
=

4. CHARLE`S LAW (P = C and V = C) Jacques Charles (1746-1823)
and Joseph Louis Gay-Lussac (1778-1850)
A) At constant pressure (P=C), the volume V of a certain quantity of gas is directly proportional to the absolute temperature
T, during a quasistatic change of state.
V  T or V = CT
2
2
1
1
T
V
T
V
or
C
T
V
=
=
B) At constant volume (V = C), the pressure P of a certain quantity of gas is directly proportional to the absolute temperature
T, during a quasi-static change of state.
P  T or P = CT

2. 1
2
2
1
1
T
P
T
P
o r
C
T
P
=
=
5. AVOGADRO`S LAW: Amedeo Avogadro (1776-1856)
All gases at the same temperature and pressure, under the action of a given value of g, have the same number of
molecules per unit of volume. From which it follows that the
1
2
2
1
2
1
2
2
1
1
R
R
M
M
or
V
n
V
n
gases
all
for
pressure
and
re
temperatu
same
t the
A
=
=


=
6. SPECIFIC HEATS
A. SPECIFIC HEAT AT CONSTANT PRESSURE (Cp)
1
2
p
h
h
h
h
m
)
T
(
mC
Q
−
=


=

=
From: dh = dU + PdV + VdP
but dU + VdP = dQ ; therefore
dh = dQ + VdP
but at P = C ; dP = O; therefore
dh = dQ and by integration
Q = h considering m,
h = m(h2 - h1)
Q = h = m (h2 - h1)
From the definition of specific heat, C = dQ/T
Cp = dQ /dt
Cp = dh/dT, then
dQ = CpdT
and by considering m,
dQ = mCpdT
then by integration
Q = m Cp T
but T = (T2 - T1)
Q = m Cp (T2 - T1)
B. SPECIFIC HEAT AT CONSTANT VOLUME (Cv)
1
2
V
U
U
U
U
m
T
mC
Q
−
=


=

=
At V = C, dV = O, and from dQ = dU + PdV
dV = 0, therefore
dQ = dU
then by integration
Q = U
then the specific heat at constant volume Cv'
Cv = dQ/dT = dU/dT
dQ = CvdT
and by considering m,
dQ = mCvdT
and by integration
Q = mU
Q = mCvT

3. 2
Q = m(U2 - U1)
Q = m Cv(T2 - T1)
From: h = U + P and P = RT
h = U + RT
and by differentiation,
dh = dU + Rdt
but dh =CpdT and dU = CvdT, therefore
CpdT = CvdT + RdT
and by dividing both sides of the equation by dT,
Cp = Cv + R
7. RATIO OF SPECIFIC HEATS
1
-
k
Rk
C
an d
1
-
k
R
C
th en
R
C
C
Fro m
k
C
C
k C
C
U
h
C
C
k
p
v
v
p
p
v
v
p
V
P
=
=
+
=
=
=


=
=
8. ENTROPY CHANGE (S)
Entropy is that property of a substance that determines the amount of randomness and disorder of a substance. If during a process,
an amount of heat is taken and is by divided by the absolute temperature at which it is taken, the result is called the
ENTROPY CHANGE.
dS = dQ/T
and by integration
S = ∫dQ/T
and from eq. 39
dQ = TdS
1
2
1
2
v
υ
υ
Rln
T
T
ln
C
ΔS +
=


+
=


+
=
=

=

+
=
+
=

+
=
n
in teg ratio
By
d
R
T
d T
C
d S
T
b y
d iv id in g
d
RT
d T
C
Td S
Td S
d Q
an d
RT
P
Pd
d U
d Q
RT
U
h
P
U
h
:
Fro m
v
v

4. 3
1
2
1
2
p
P
P
Rln
T
T
ln
C
ΔS −
=
−
=
−
=
−
=

+
=
n
in teg ratio
b y
P
d P
R
T
d T
C
d S
T
b y
d iv id in g
P
d P
RT
d T
C
Td S
d P
P
RT
d h
d Q
d P
d Q
d h
Fro m
p
p
Actual – Gas equation of State
In actual gases the molecular collision is inelastic; at high densities in particular there are intermolecular forces that the simplified
equation of the state does not account for. There are many gas equations of state that attempt to correct for the non-ideal behavior
of gases. The disadvantage of all methods is that the equations are more complex and require the use of experimental coefficients.
a. Van der Waals Equation
( )
mol
3
2
kg
m
in
gas.
the
of
behavior
nonideal
the
for
compensate
that
ts
coefficien
a
b
and
a
:
where
T
R
b
a
P

=
−










+
b. Beattie-Bridgeman Equation
3
0
0
2
2
T
c
b
1
B
B
a
1
A
A
:
where
T
R
)
B
)(
1
(
A
P

=








−
=







−
=
=
+


−










+
Compressibility Factor
Z
RT
P
gases
of
behavior
nonideal
For
1
RT
P
=

=

Where: Z – compressibility factor

5. 4
system)
on the
done
is
(Work
KW
24.67
W
KW
67
.
24
8.2)
0.1785(-13
W
KW
In
kg
KJ
2
.
138
26
.
1
101
86
.
4
546
)
82
(
)
24
(
W
P
P
-
)
U
-
(U
-
Q
W
0
-
0
-
)
P
-
(P
-
)
U
-
(U
-
Q
W
PE
-
KE
-
)
(P
-
U
-
Q
W
W
PE
KE
)
(P
U
Q
kg/sec
0.1785
kg/min
71
.
10
)
5
.
8
(
26
.
1
V
m
V
m
1
1
2
2
1
2
1
1
2
2
1
2
=
−
=
=
−
=






−
−
−
−
=









−

=


=





=
+

+

+


+

=
=
=
=

=
=

Sample Problems (IDEAL GAS)
1. An air compressor handles 8.5 m3
/min of air with a density of 1.26 kg/m3
and a pressure of 101 KPaa and discharges
at 546 KPaa with a density of 4.86 kg/m3
. The changes in specific internal energy across the compressor is 82 KJ/kg
and the heat loss by cooling is 24 KJ/kg. Neglecting changes in kinetic and potential energies, find the work in KW.
2. Calculate the change of entropy per kg of air when heated from 300K to 600K while the pressure drops from 400
Kpa to 300 KPa. (S = 0.78 KJ/kg-K)
Given;
R = 0.287 KJ/kg-K
k = 1.4
T1 = 300K ; T2 = 600K
P1 = 400 KPa ; P2 = 300 KPal
1
2
1
2
p
P
P
ln
R
T
T
ln
C
S −
=

3. A certain mass of sulfur dioxide (SO2) is contained in a vessel of 0.142 m3
capacity, at a pressure and temperature of
2310 KPa and 18C, respectively. A valve is open momentarily and the pressure falls immediately to 690 KPa.
Sometime later the temperature is again 18C and the pressure is observed to be 910 KPa. Estimate the value of specific
heat ratio. (k = 1.29)
K
9 2
.
8 6
T
P
P
T
T
P
T
P
C
V
A t
k g
6 7
.
8
)
2 9 1
(
1 3
.
0
)
1 4 2
.
0
(
2 3 1 0
m
RT
m
V
P
1 3
.
0
6 4
3 1 4 3
.
8
R
1
1
2
2
2
2
1
1
1
1
1
1
1

=












=
=
=
=
=
=
=
=

6. 5
4. Two unequal vessel A and B are connected by a pipe with a valve. Vessel A contains 150 L of air at 2760 KPa and
95C. Vessel B contains an unknown volume of air at 70 KPa and 5C. The valve is opened and when the properties
have been determined, it was found out that the pressure of the mixture is 1380 KPa and the temperature is 45C. What
is the volume of vessel B.(0.166 m3)
Given:
VA = 0.150 m3
; PA = 2760 KPa ; TA = 95 + 273 = 368 K
PB = 70 KPa ; TB = 5 + 273 = 278 K
P = 1380 KPa ; T = 45 + 273 = 318 K
3
B
B
B
B
B
B
B
B
B
A
A
A
A
B
A
B
A
m
1 1 6
.
0
)
2 5 2
.
0
3 4
.
4
(
)
6 5
.
0
1 2 5
.
1
(
V
V
2 5 2
.
0
1 2 5
.
1
V
3 4
.
4
6 5
.
0
)
2 7 8
(
R
)
V
(
7 0
)
3 6 8
(
R
)
1 5 0
.
0
(
2 7 6 0
)
3 1 8
(
R
)
V
1 5 0
.
0
(
1 3 8 0
RT
V
P
m
;
RT
V
P
m
;
RT
PV
m
m
m
m
V
V
V
=
−
−
=
+
=
+
+
=
+
=
=
=
+
=
+
=
5. A vessel of volume 0.2 m3
contains nitrogen at 101.3 KPa and 15C. If 0.2 kg of nitrogen is now pumped into the
vessel, calculate the new pressure when the vessel has returned to its initial temperature. For nitrogen: M = 28; k =
1.399. (187 KPa) (Sample Prob. June 18, 2014)
K Pa
8 2
.
1 8 6
)
2
.
0
(
2 8
2 7 3 )
4 3 )(1 5
0 .4 3 7 (8 .3 1
P
V
V
V
mRT
P
K
2 8 8
2 7 3
1 5
T
mass
fin al
k g
4 3 7
.
0
0 .2
0 .2 3 7
m
k g
2 3 7
.
0
)
2 7 3
1 5
(
3 1 4 3
.
8
2 8
)
2
.
0
(
3
.
1 0 1
m
K
-
k g
K J
2 8
3 1 4 3
.
8
R
RT
PV
m
mRT
PV
ad d ed
N
o f
(mass
k g
0 .2
-
m
K
2 8 8
2 7 3
1 5
T
K Pa;
1 0 1 .3
P
;
m
2
.
0
V
:
G iv en
2
1
2
2
2
1
2
a
1
1
3
1
=
+
=
=
=

=
+
=
→
=
+
=
=
+
=
=
=
=

=
+
=
=
=

7. 6
C
8
.
1 1 1
t
K
8
.
3 8 4
V
P
V
P
T
T
T
V
P
T
V
P
C
T
PV
k g
k g
1 6
R
3 1 4 3
.
8
M
M
3 1 4 3
.
8
R
K
-
k g
KJ
5 2
.
0
R
mT
V
P
R
mRT
PV
2
1
1
2
2
1
2
2
2
2
1
1
1
mol
1
1
1

=

=
=
=
=
=
=
=
=
=
=
6. A certain perfect gas of mass 0.1 kg occupies a volume of 0.03 m3
at a pressure of 700 KPa and a temperature of
131C. The gas is allowed to expand until the pressure is 100 KPa and the final volume is 0.2 m3
. Calculate;
a) the molecular weight of the gas (16)
b) the final temperature
Given;
m = 0.1 kg ; V1 = 0.03 m3
; P1 = 700 KPa ; T1 = 131 +273 = 404 K
P2 = 100 KPa ; V2 = 0.2 m3
7. An ideal gas with R = 2.077 KJ/kg-K and a constant k= 1.659 undergoes a constant pressure process during which
527.5 KJ are added to 2.27 kg of the gas. The initial temperature is 38C. Find the S in KJ/K.
Given:
R = 2.077 KJ/kg-K; k = 1.659
Q = 527.5 KJ; m = 2.27 kg
T1 = 38 + 273 = 311 K
Process: P = C
Q = mCp(T2 – T1) ;
K
kg
/
KJ
72
.
5
1
k
RK
Cp −
=
−
=
K
352
T
mCp
Q
T 1
2 
=
+
=
K
/
KJ
6
.
1
T
T
ln
mCp
S
1
2
=
=

8. A certain perfect gas of mass 0.1 kg occupies a volume of 0.03 m3
at a pressure of 700 KPa and a temperature of
131C. The gas is allowed to expand until the pressure is 100 KPa and the final volume is 0.02 m3
. Calculate;
a) the molecular weight of the gas (16)
b) the final temperature (11.5C)
C
7
.
15
t
16
)
273
2
t
)(
3143
.
8
(
1
.
0
)
15
.
0
(
100
mRT
V
P
16
M
M
)
273
131
)(
3143
.
8
(
1
.
0
)
03
.
0
(
700
2
2
2
2

=
+
=
=
+
=
=

8. 7
Continuation of Module 6
IDEAL GAS MIXTURE
• TOTAL MASS OF A MIXTURE

= i
m
m
• MASS FRACTION OF A COMPONENT
m
m
x i
i =
• TOTAL MOLES OF A MIXTURE

= i
n
n
• MOLE FRACTION OF A COMPONENT
n
n
y i
i =
• EQUATION OF STATE
Mass Basis
a. For the Mixture
b. Fort the Components
Mole Basis
c. For the Mixture
d. Fort the Components
AMAGAT’S LAW
The total volume of a mixture of gases is equal to the sum of the volume occupied by each component at the mixture pressure P
and temperature T.
P = P1 = P2 = P3
T = T1 = T2 = T3
T
R
PV
n
;
T
R
PV
n
;
T
R
PV
; n
T
R
PV
n
n
n
n
n
3
3
2
2
1
1
3
2
1
=
=
=
=
+
+
=
1
V1
2
V2
3
V3
P, T
mR
T
PV=
i
i
i
i
i T
R
m
V
P =
T
R
n
PV=
i
i
i
i T
R
n
V
P =

9. 8
V
V
n
n
y
V
V
V
V
V
V
P
T
R
T
R
PV
T
R
PV
T
R
PV
T
R
PV
T
R
PV
T
R
PV
T
R
PV
T
R
PV
i
i
i
i
3
2
1
3
2
1
3
2
1
=
=
=
+
+
=














+
+
=
+
+
=

DALTON’S LAW
The total pressure of a mixture of gases P is equal to the sum of the partial pressure that each gas would exert at the mixture volume
V and temperature T.
V = V1 = V2 = V3
T = T1 = T2 = T3
T
R
V
P
n
;
T
R
V
P
n
;
T
R
V
P
; n
T
R
PV
n
n
n
n
n
3
3
2
2
1
1
3
2
1
=
=
=
=
+
+
=
P
P
n
n
y
P
P
P
P
P
P
V
T
R
T
R
V
P
T
R
V
P
T
R
V
P
T
R
PV
T
R
V
P
T
R
V
P
T
R
V
P
T
R
PV
i
i
i
i
3
2
1
3
2
1
3
2
1
=
=
=
+
+
=














+
+
=
+
+
=

• MOLECULAR WEIGHT OF A MIXTURE (M)
K
-
k g
K J
R
3 1 4 3
.
8
R
R
M
M
y
M i
i
=
=
= 
• GAS CONSTANT (R)
K
-
k g
K J
M
3 1 4 3
.
8
M
R
R
R
x
R i
i
=
=
= 
1
P1
2
P2
3
P3 P
Mixture
Components
V, T

10. 9
• SPECIFIC HEAT OF A MIXTURE
At Constant Volume

= vi
i
v C
x
C
At Constant Pressure
R
C
C
C
x
C
v
p
Pi
i
P
+
=
= 
• RATIO OF SPECIFIC HEAT
1
k
R
C
1
k
Rk
C
C
C
k
V
P
V
P
−
=
−
=
=
GRAVIMETRIC AND VOLUMETRIC ANALYSIS
Gravimetric analysis gives the mass fractions of the components in the mixture.
Volumetric analysis gives the volumetric or molal fractions of the components in the mixture.
CONVERSION


=
=
=
i
i
i
i
i
i
i
i
i
i
i
M
x
M
x
y i
M
M
y
M
y
M
y
x
SAMPLE PROBLEMS (Gas Mixture)
1. A 6 m3
tank contains helium at 400K and is evacuated from atmospheric pressure to 740 mm Hg vacuum. Determine
a) the mass of helium remaining in the tank
b) the mass of helium pumped out
c) the temperature of the remaining helium falls to 10C , what is the pressure in KPa
For Helium:
R = 2.077 KJ/kg-K
Cp = 5.1954 KJ/kg-K
Cv = 3.1189 KJ/kg-K
Given:
P1 = 101.325 KPa
P2 = 760 – 740 =20 mm Hg = 2.67 KPa
V1 = V2 = 6 m3
T = 400K
KPa
9 6
.
1
6
)(28 3)
0.02 (2.0 77
P
K
28 3
to
falls
re
temperatu
the
If
ou t)
pu mped
heliu m
o f
(mass
k g
71 2
.
0
0.02
-
0.73 2
m
tank )
in the
remain ing
heliu m
o f
(mass
k g
0 2
.
0
2.07 7(40 0)
2.67 (6)
m
tank )
in the
heliu m
o f
mass
(original
k g
73 2
.
0
)
40 0
(
07 7
.
2
)
6
(
32 5
.
10 1
m
mRT
PV
p
2
1
=
=

=
=
=
=
=
=
=

11. 10
KPa
3 0
0 .30 (1 0 0)
P
KPa
4 0
0 .40 (1 0 0)
P
KPa
3 0
)
1 0 0
(
3 0
.
0
P
P
Pi
y i
Law
s
Dalto n '
From
K
-
k g
KJ
2 3 4
.
0
R
6
.
3 5
3 1 43
.
8
M
3 1 43
.
8
R
2
2
2
N
CO
O
=
=
=
=
=
=
=

=
=
=
2. Two unequal vessel A and B are connected by a pipe with a valve. Vessel A contains 150 L of air at 2760 KPa and 95C.
Vessel B contains an unknown volume of air at 70 KPa and 5C. The valve is opened and when the properties have been
determined, it was found out that the pressure of the mixture is 1380 KPa and the temperature is 45C. What is the volume
of vessel B.(0.166 m3)
Given:
VA = 0.150 m3 ; PA = 2760 KPa ; TA = 95 + 273 = 368 K
PB = 70 KPa ; TB = 5 + 273 = 278 K
P = 1380 KPa ; T = 45 + 273 = 318 K
3
B
B
B
B
B
B
B
B
B
A
A
A
A
B
A
B
A
m
1 1 6
.
0
)
2 5 2
.
0
3 4
.
4
(
)
6 5
.
0
1 2 5
.
1
(
V
V
2 5 2
.
0
1 2 5
.
1
V
3 4
.
4
6 5
.
0
)
2 7 8
(
R
)
V
(
7 0
)
3 6 8
(
R
)
1 5 0
.
0
(
2 7 6 0
)
3 1 8
(
R
)
V
1 5 0
.
0
(
1 3 8 0
RT
V
P
m
;
RT
V
P
m
;
RT
PV
m
m
m
m
V
V
V
=
−
−
=
+
=
+
+
=
+
=
=
=
+
=
+
=
3. A gaseous mixture has the following volumetric analysis: O2 = 30%; CO2 = 40% ; N2 = 30%. Determine
a. the gravimetric analysis
b. the partial pressure of each component if the total pressure is 100 KPa and the temperature is 32C
c. the molecular weight and gas constant of the mixture
For
O2: M = 32 ; k = 1.395
CO2: M = 44 ; k = 1.288
N2: M = 28 ; k = 1.399
Given
Volumetric analysis
O2 = 30% ; CO2 = 40%; N2 = 30% ; P = 100 KPa
%
24
24
.
0
6
.
35
4
.
8
x
%
49
49
.
0
6
.
35
6
.
17
x
%
27
27
.
0
6
.
35
6
.
9
x
kg/kg
6
.
35
4
.
8
6
.
17
6
.
9
M
)
28
(
30
.
0
)
44
(
40
.
0
)
32
(
30
.
0
M
yiMi
M
M
yiMi
xi
2
2
N
CO
2
O
mol
=
=
=
=
=
=
=
=
=
=
+
+
=
+
+
=
=
=

A
B

12. 11
kg
3
.
2
3.6
-
5.9
m
kg
9
.
5
m
m
6
.
3
61
.
0
m
m
m
m
m
x
39
.
0
36
14
x
61
.
0
36
22
x
kg /kg
36
14
22
M
)
28
(
50
.
0
)
44
(
50
.
0
M
M
yiMi
xi
2
2
2
2
2
2
2
N
N
CO
CO
CO
N
CO
mol
=
=
=
=
+
=
=
=
=
=
=
=
=
+
=
+
=
=
KPa
138
138
276
P
P
P
P
KPa
276
P
P
138
50
.
0
P
Pi
yi
K
-
kg
KJ
231
.
0
36
8.3143
R
36
M
2
2
2
N
N
CO
=
−
=
+
=
=
=
=

=
=
=
4. A. How many kilograms of N2 must be mixed with 3.6 kg of CO2 in order to produce a gaseous mixture that is 50% by
volume of ach constituents.
B. For the resulting mixture, determine M and R, and the partial pressure of the N2 if that of the CO2 is
138 KPa.
Given
mCO2 = 3.6 kg ; yCO2 = 0.50
mN2 = ? yN2 = 0.50
5. Assume 2 kg of O2 are mixed with 3 kg of an unknown gas. The resulting mixture occupies a volume of 1.2 m3
at 276
KPa and 65C. Determine
a) R and M of the unknown gas constituent
b) the volumetric analysis
c) the partial pressures
For O2: M = 32 ;k = 1.395
Given; mO2 = 2 kg; mx = 3 kg
V = 1.2 m3 ; P = 276 KPa; T = 338 K
a)
m = 5 kg
xO2 = 0.40 ; xx = 0.60
R = 0.1361 KJ/kg-K
R = .40(0.26) + 0.60(Rx)
Rx = 0.535 KJ/kg-K
Mx = 15.54 kg/kgm
b)
yO2 = 0.245 ; yx = 0.755
c)
PO2 = .245(276) = 67.62 KPa ; Px = 0.755(276) = 208.38 KPa
END OF MODULE 6