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Introduction to Chemical Engineering Calculations
Lecture 5.
Ideal Gas Calculations
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
5
Ideal Gas Calculations
What is an ideal gas?
An ideal gas is an imaginary gas that obeys exactly the
following relationship:
PV = nRT
where P = absolute pressure of the gas
V = total volume occupied by the gas
n = number of moles of the gas
R = ideal gas constants in appropriate units
T = absolute temperature of the gas
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Ideal Gas Calculations
The Ideal Gas Constant, R
R = 1.987 cal/(gmol)(K)
= 1.987 Btu/(lbmol)(0R)
= 10.73 (psia)(ft3)/(lbmol)(0R)
= 8.314 (kPa)(m3)/(kmol)(K)
= 8.314 J/(gmol)(K)
= 82.06 (atm)(cm3)/(gmol)(K)
= 0.08206 (atm)(L)/(gmol)(K)
= 21.9 (in. Hg)(ft3)/(lbmol)(0R)
= 0.7302 (atm)(ft3)/(lbmol)(0R)
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Ideal Gas Calculations
Standard Conditions for the Ideal Gas
Several arbitrarily specified standard states of temperature
and pressure have been selected by custom.
System
TS
PS
VS
nS
SI
273.15 K
101.325 kPa
22.415 m3
1 kmol
Am. Eng.
4920R
1 atm
359.05 ft3
1 lbmol
Natural Gas
Industry
333.15 K
14.696 psia
379.4 ft3
1 lbmol
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Ideal Gas Calculations
Example 5-1. Ideal Gas Calculation
Butane (C4H10) at 3600C and 3.00 atm absolute flows into a
reactor at a rate of 1100 kg/h. Calculate the volumetric flow
rate of this stream.
Method A. Computation using a known value of R.
The ideal gas equation in terms of flowrate:
PV nRT
V n


or P      RT
t
t
 t  t


P V = n RT
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
SLIDE
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5
Ideal Gas Calculations
Example 5-1. Ideal Gas Calculation
Solving for volumetric flowrate:


n RT
V=
P
Obtaining the molar flowrate from mass flowrate:


m
1100 kg / h
n=

 19.0 kmol / h
MW 58 kg / kmol
Using absolute temperatures and pressure:
T = 633 K and P = 3 atm
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Ideal Gas Calculations
Example 5-1. Ideal Gas Calculation
Using the following value of R:
L  atm  1000gmol 
L  atm
R  0.08206
 82.06


gmol  K  1kmol 
kmol  K
The volumetric flowrate is


V=
n RT (19.0kmol / h)(82.06 L  atm / kmol  K)(633K)

P
3atm
L  1m3 
m3
V =328,978.5 
  329
h  1000 L 
h

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
SLIDE
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Ideal Gas Calculations
Example 5-1. Ideal Gas Calculation
Method B. By comparison to standard conditions
PV
nT

PS VS n STS
Using a basis of 1 hr, then n = 19 kmol
The following standard conditions will be used.
PS = 1 atm
VS = 22.41 m3
nS = 1 kmol
TS = 273 K
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
SLIDE
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5
Ideal Gas Calculations
Example 5-1. Ideal Gas Calculation
Solving for V:
 n  T   P 
V      S  VS
 n S  TS   P 
 19.0kmol  633K  1atm 
3
V
22.415m
 273K  3atm 
1kmol




V  329m 3
In terms of volumetric flowrate
m3
V = 329
h

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Ideal Gas Calculations
Example 5-2. Ideal Gas at Two Different Conditions
Ten cubic feet of air at 700F and 1 atm is heated to 6100F
and compressed to 2.5 atm. What volume does the gas
occupy in its final state?
Let 1 denote the initial state of the gas and 2 the final state.
P1V1 n1T1 T1


P2 V2 n 2T2 T2
Solving for V2:
0

 P1  T2 

1.00
atm

1070
R
3
3
V2  V1     10.0 ft 

8.08
ft
  530 0 R 
P
T
2.50
atm


 2  1 

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Ideal Gas Calculations
Example 5-3. Calculation of Ideal Gas Density
What is the density of N2 at 270C and 100 kPa in SI units?
PV
nT

PSVS n STS
Solving for (n/V) and obtaining the density from this :
 n S   P   TS  
n
   MW          MW
V
 VS   PS   T  
 1kmol   100 kPa   273K  
kg 
kg

 28
  1.123 3



3 
m
 22.41m   101.3kPa   300 K   kmol 
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Ideal Gas Calculations
Ideal Gas Mixtures and Partial Pressures
In a mixture of ideal gases, the partial pressure of a gas
component is the pressure that would be exerted by a that
component if it existed by itself in the same volume as
occupied by the mixture and the same temperature of the
mixture.
PiVtotal = niRTtotal
where Pi and ni are the partial pressure and number of
moles of component i.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
SLIDE
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Ideal Gas Calculations
Ideal Gas Mixtures and Partial Pressures
For the gas mixture:
PtotalVtotal = ntotalRTtotal
Dividing the two equations,
Pi VT n i RTT

PT VT n T RTT
or
Pi 
ni
PT  yi PT
nT
According to Dalton,
PA + PB + PC + . . . . = Ptotal
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Ideal Gas Calculations
Example 5-4. Ideal Gas Mixtures and Partial Pressures
A flue gas analyzes 14.0% CO2, 6.0% O2, and 80.0% N2. The mixture
is at 4000F and 765 mmHg pressure. Calculate the partial pressure
of each component.
Component
y
Pi (mmHg)
CO2
0.140
0.140(765) = 107.1
O2
0.060
0.060(765) = 45.9
N2
0.800
0.800(765) = 612.0
Total
1.000
765 mmHg
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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