igeal gas

Introduction to Chemical Engineering Calculations

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños

Lecture 5.

Ideal Gas Calculations

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Ideal Gas Calculations5

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE

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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 gasV = total volume occupied by the gasn = number of moles of the gasR = ideal gas constants in appropriate unitsT = absolute temperature of the gas



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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)



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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 GasIndustry

333.15 K 14.696 psia 379.4 ft3 1 lbmol



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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 nor P RTt t t t

P V = n RT



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Solving for volumetric flowrate:

n RTV =P

Obtaining the molar flowrate from mass flowrate:

1100 kg / h 19.0kmol / h58kg / kmol

mn =MW

Using absolute temperatures and pressure:

T = 633 K and P = 3 atm



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Using the following value of R:

L atm 1000gmol L atmR 0.08206 82.06gmol K 1kmol kmol K

The volumetric flowrate is

3 3

n RT (19.0kmol / h)(82.06L atm / kmol K)(633K)VP 3atm

L 1m mV 328,978.5 329h 1000L h

=

=



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Method B. By comparison to standard conditions

S S S S

PV nTP V n T

Using a basis of 1 hr, then n = 19 kmol

The following standard conditions will be used.

PS = 1 atmVS = 22.41 m3

nS = 1 kmolTS = 273 K



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Solving for V:

SS

S S

3

3

Pn TV Vn T P

19.0kmol 633K 1atmV 22.415m1kmol 273K 3atm

V 329m

In terms of volumetric flowrate

3mV 329h

=


Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE10

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.

1 1 1 1 1

2 2 2 2 2

P V n T TP V n T T

Solving for V2:

03 31 2

2 1 02 1

P T 1.00atm 1070 RV V 10.0ft 8.08ftP T 2.50atm 530 R



Example 5-3. Calculation of Ideal Gas Density

What is the density of N2 at 270C and 100 kPa in SI units?

S S S S

PV nTP V n T

Solving for (n/V) and obtaining the density from this :

S S

S S

3 3

n Tn PMW MWV V P T

1kmol 100 kPa 273K kg kg28 1.123101.3kPa 300K kmol22.41m m



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.



Ideal Gas Mixtures and Partial Pressures

For the gas mixture:

PtotalVtotal = ntotalRTtotal

Dividing the two equations,

i T i T ii T i T

T T T T T

P V n RT nor P P y PP V n RT n

According to Dalton,

PA + PB + PC + . . . . = Ptotal



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

igeal gas

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