Gas Mixture 1. An ideal gas mixture has the following stoichiometric analysis Component % by mass N2 40 CO2 10 CO 13 O2 17 H2 08 CH4 12 Find the analysis on a volume basis. (B) What is the volume of 1...

1. An ideal gas mixture has the following stoichiometric analysis
Component % by mass
N2 40
CO2 10
CO 13
O2 17
H2 08
CH4 12
(A) Find the analysis on a volume basis.
Answer
Compund Molecular weight Mass Percentage Mass/mole No of moles Volume
Percentage
N2 29 40 11.6 6 38.36%
CO2 44 10 4.4 1.0 6.39%
CO 28 13 3.64 2.0 12.78%
O2 32 17 5.44 2.30 14.70%
H2 2 08 0.16 1.08 6.90%
CH4 16 12 1.92 3.26 20.84%

Total mass = 27.64kg
(B) What is the volume of 1 kg of this gas when the P = 1.5 MPa and T = 30oC?
Rm = Ru/Mm
Rm = 8.314/27.64
Rm = 0.3
Hence Volume can be calculated as,
Vm = RmTmmm/Pm
Vm = 60.6 x 10-6 liters (Answer)
(C) Find the specific heats at 300 K. Using Table A.2 for CO2 and N2
Cp = 0.488*1.039 + 0.512*0.846
Cp = 0.940 kJ/kg.K (Answer)
CV = CP – R
CV = 0.940 -0.3
CV = 0.64kJ/kgK (Answer)
(D) This mixture undergoes an expansion isentropic process from 2.0 bar, 130oC to 1 MPa. Find T2.
Ratio of specific heat capcities will be(k) = CP/CV
k = 1.468
Hence T2 can be evaluated as
T2 = T1(P2/P1)(k-1)/k
T2 = 300 (0.2/0.1)(1.468-1)/1.468
T2 = 362.2K (Answer)
(F) Find ∆sm per kg of mixture when the mixture is compressed isothermally from 0.1 MPa to 0.2 MPa.
But, the compression process is isothermal, T2 = T1. The partial pressures are given by
The entropy change becomes
For this problem the components are already mixed before the compression process. So,
Then,
P y Pi i m=
y yi i, ,2 1=
∆ ∆s mf s
kg
kg
kJ
kg K
kg
kg
kJ
kg K
kJ
kg K
m i i
i
N
m N
CO
m CO
m
=
= −
⋅
+ −
⋅
= −
⋅
=
∑
1
2
0 488 0 206 0 512 0131
0167
2
2
2
2
( . )( . ) ( . )( . )
.
(G) All the gases are supplied in separate lines at 0.2 MPa and 300 K to a mixing chamber and are mixed adiabatically.
The resulting mixture has the composition as given in Part (A). Determine the entropy change due to the mixing process
per unit mass of mixture.
Take the time to apply the steady-flow conservation of energy and mass to show that the temperature of the mixture at
state 3 is 300 K.
But the mixing process is isothermal, T3 = T2 = T1. The partial pressures are given by
The entropy change becomes
But here the components are not mixed initially. So,
P y Pi i m=
y
y
N
CO
2
2
1
2
1
1
,
,
=
=
y
y
N
CO
2
2
3
3
0 6
0 4
,
,
.
.
=
=

Then,
2. An ideal gas mixture has the following volumetric analysis: N2 = 55%, CO2 = 25%, O2 = 20% is compressed adiabatically
in a Piston-Cylinder assembly from 1 bar 20 0C to 5 bar.
(A) Find the analysis on a mass basis.
Compund Molecular weight Mass Percentage Mass per mole % amount
N2 29 .55 15.95 47.82%
CO2 44 .25 11 32.98
O2 32 .20 6.4 19.19
Total Mass = 33.15 g/mole
(B) Find the specific heats at 350 K. Using Table A.2
Find the specific heats at 350 K. Using Table A.2 for CO2 and N2
Cp = 0.488*1.041 + 0.512*0.895
Cp = 0.966 kJ/kg.K (Answer)
CV = CP – R
CV = 0.966 -0.28
CV = 0.686kJ/kgK (Answer)
(C) Determine the work required for compression process?
ϒ can be calculated as = Cp/Cv = 1.408
Since process is isothermal then work can be calculated as,
W = -RTlnP2/P1
∆ ∆s mf s
kg
kg
kJ
kg K
kg
kg
kJ
kg K
kJ
kg K
m i i
i
N
m N
CO
m...