Dr. Belal Al Zaitone Zuriqat

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King Abdulaziz University

Faculty of Engineering

Department of Chemical and Materials Engineering

Dr. Belal Al Zaitone Zuriqat

eMail: [email protected] http://balzaitone.kau.edu.sa Office: Building No. 40, Room No. 24G58

ChE 201

Introduction to Chemical Engineering

Chapter 7

IDEAL & REAL GASES

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Chapter 7: Ideal & Real Gases

1 Ideal Gases

3 Real Gases: Compressibility Charts 4 Real Gas Mixtures

2 Real Gases: Equations of State

1 The Ideal Gas Law

2 Ideal Gas Mixtures and Partial Pressure 3 Material Balances Involving Ideal Gases

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Chapter 7: Ideal & Real Gases

30 kg I 25 kg

30 mole I A

10 mole B

30 mole I A

60 mole B

30 m³ I A

20 m³ A

No !!! Mass conservation

Yes, chemical reaction

Yes, process parameters 3A  B

A  2B

P₁,T₁ P₂,T₂

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Chapter 7: Ideal & Real Gases

Ideal Gas Law

The ideal gas law relates the variables of pressure, volume, temperature, and number of moles of gas within a closed system.

The ideal gas law: PV = nRT

P = Pressure of the confined gas in atmospheres (atm) V = Volume of the confined gas, in liters (L) n = Number of moles of gas in moles (mol)

T = Temperature in Kelvin (K)

R = Ideal Gas Constant equal= 0.082057 L·atm/mol·K R = 0.082057 L·atm/mol·K

R= 62.364 L Torr mol^-1 K^-1 R= 8.3145 m³ Pa mol^-1 K^-1 R= 8.3145 J mol^-1 K^-1

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Chapter 7: Ideal & Real Gases

Conditions for a gas to behave as predicted by the ideal gas law:

1. The molecules do not occupy any space; they are infinitesimally small.

2. No attractive forces exist between the molecules so the molecules move completely independently of each other.

3. The gas molecules move in random, straight-line motion.

4. The collisions between the molecules, and between the molecules and the walls of the container, are perfectly elastic.

These condition are met at: high temperature and low pressure.

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Chapter 7: Ideal & Real Gases

Example 7.1 : Calculate the volume, in cubic meters, occupied by 40 kg of CO₂ at 1 atm and 0 °C?

Assume CO₂ acts as an ideal gas.

_CO2= 1.977 kg/m3 (gas at 1 atm and 0 °C) MW_CO2= 44 g/mol

Solution:

1 m³

1.977 kg 40 kg

= 20.37 m³at standard conditions

V=

PV = nRT

Calculate CO₂ volume at 2 atm and 50 °C!??

We need to know density of CO₂ gas at 2 atm and 50 °C!!!!

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Example 1: Recalculate the volume using ideal gas law the volume, occupied by 40 kg of CO₂ at 1 atm and 0 °C?

Assume the gas acts as an ideal gas.

MW_CO2= 44 g/mol Solution:

𝑀𝑊 = 𝑚 𝑛

𝑉 = ^{40∗1000 𝑔}

44 𝑔/𝑚𝑜𝑙

0.082057 (L·atm/mol·K) ^{0+273.15 𝐾}

1 𝑎𝑡𝑚

V=20376 L = 20.376 m³

𝑉 = 𝑛 ^𝑅𝑇_𝑃 𝑉 = 𝑚

𝑀𝑊

𝑅𝑇 𝑃𝑉 = 𝑛𝑅𝑇 𝑃

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Chapter 7: Ideal & Real Gases

• Calculate the volume 40 kg CO₂ at 2 atm and 50 °C!??

𝑉 = ^{40∗1000 𝑔}

0.082057 (L·atm/mol·K) 0+273.15 𝐾

1 𝑎𝑡𝑚 =20367 L =20.376 m³

𝑉 = ^{40∗1000 𝑔}

0.082057 (L·atm/mol·K) 50+273.15 𝐾

2𝑎𝑡𝑚 =12053 L =12.053 m³

• Calculate the volume 40 kg CO₂ at 0.5 atm and 100 °C!??

• Calculate the volume 40 kg CO₂ at 101325 Pa and 273.15 K ??

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Specific gravity of gases, SG:

• Defined as the ratio of the density of the gas to the density of air of at Standard Conditions, SC:

𝑆𝐺 = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑔𝑎𝑠 𝐴 @ 𝑆𝐶 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓𝑎𝑖𝑟 @ 𝑆𝐶

Standard Conditions, S.C. of Ideal gas are: 0 °C and 1 atm

𝑃𝑉 = 𝑛𝑅𝑇 𝑃𝑉 =

^𝑚

𝑀𝑊

𝑅𝑇

Density of ideal gases

𝑃 = 𝜌

^𝑅𝑇

𝑀𝑊

𝜌 = 𝑃

^𝑀𝑊

𝑅𝑇

𝑃 =

^𝑚

𝑉

𝑅𝑇 𝑀𝑊

Density of ideal gases

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Chapter 7: Ideal & Real Gases

Example 2: Calculation of Gas Density

What is the density of N2 at 27°C and 100 kPa in SI Units, Assume N2 acts as an ideal gas.

Mw_N2=28 g/mol Solution:

𝜌 = 𝑃

^𝑀𝑊_𝑅𝑇

Density of N2 R = 0.082057 L·atm/mol·K R= 62.364 L Torr mol^-1 K^-1 R= 8.3145 m³ Pa mol^-1 K^-1 R= 8.3145 J mol^-1 K^-1

R=10.73159 ft³psi R⁻¹lb-mol⁻¹ R=0.7302413 ft³atm R⁻¹ lb-mol⁻¹

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Chapter 7: Ideal & Real Gases

I II

P₁,V₁, T₁, n₁

Gas A @ initial state Gas A @ final state

P₂,V₂, T₂, n₂ P₁V₁= n₁RT₁ P₂V₂= n₂RT₂

P₁V₁

P₂V₂ = n₁RT₁ n₂RT₂ P₁V₁

P₂V₂ = n₁T₁ n₂T₂ P₁

P₂

V₁

V₂ = n₁ n₂

T₁ T₂

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Example 7.2: Calculation of Gas Volume

If the volume of CO₂ at 0 °C and 1 atm is 20.376 m³ Calculate the CO₂ volume at 27°C and 0.5 atm?

Solution:

P₁ P₂

V₁

V₂ = n₁ n₂

T₁ T₂ 1 𝑎𝑡𝑚

0.5 𝑎𝑡𝑚

20.365 𝑚³

𝑉₂ = 0 + 273.15 27 + 273.15

V₂= 44.76 m³

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Chapter 7: Ideal & Real Gases

Example 3: Calculation of Gas Density

If the density of N2 at T₁=27°C and P₁= 100 kPa is ₁=1.122 kg/m³

Calculate the density, ₂ of N2 at 48°C and 150 kPa Assume N2 acts as an ideal gas.

Solution:

𝑃𝑉 = 𝑛𝑅𝑇 𝜌

₁

= 𝑃

₁ ^𝑀𝑊

𝑅𝑇₁

𝜌

₂

= 𝑃

₂ ^𝑀𝑊

𝑅𝑇₂ ρ₁

ρ₂ = P₁ P₂

T₂ T₁ 1.122

ρ₂ = 100 150

(48 + 273.15) (27 + 273.15)

₂= 1.712 kg/m³

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Chapter 7: Ideal & Real Gases

Gas A, P_A Gas B, P_B Gas C, P_C

Mixture of all three gases P_tot=P_A+P_B+P_C

Ideal Gas Mixtures and Partial Pressure

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Chapter 7: Ideal & Real Gases

Partial pressure, p_i

The partial pressure of Dalton, namely the pressure that would be exerted by a single component in a gaseous mixture if it existed alone in the same volume as that occupied by the mixture and at the same temperature of the mixture, and is defined by the ideal gas law:

p_iV_total=n_iRT_total

where p_i is the partial pressure of component i in the mixture.

P_iV_tot

P_totV_tot = n_iRT_tot

n_totRT_tot P_i = n_i

n_tot P_tot P_i = y_i ∙ P_tot

p_iV_total=n_iRT_total p_totV_total=n_totRT_total

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Chapter 7: Ideal & Real Gases

Example 7.3: Calculation of the Partial Pressures of the Components in a Gas.

In one process the off-flue gas analyzes 14.0% CO2, 6.0% O2, and 80.0% N2. It is at 400°F and 765.0 mmHg pressure.

Calculate the partial pressure of each component.

Solution:

P_tot= 765.0 mmHg y_CO2= 0.14

y_O2= 0.06 y_N2= 0.80

P_i = y_i ∙ P_tot

P_C_{𝑂2 =} y_CO2 ∙ P_tot =0.14 (765.0)=107.1 mmHg P_{𝑂2 =} y_O2 ∙ P_tot =0.06 (765.0)=45.9 mmHg P_N2 = y_N2 ∙ P_tot =0.80 (765.0)=612.0 mmHg

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Chapter 7: Ideal & Real Gases

Example 7.4 : Material Balance involving combustion

To evaluate the use of renewable resources, an experiment was carried out to pyrolize rice hulls. The product gas analyzed 6.4 % CO₂, 0.1 % O₂, 39% CO, 51.8% H₂, 0.6% CH₄, and 2.1 % N₂. It entered the combustion chamber at 90°F and a pressure of 35.0 in.

Hg, and was burned with 40% excess air (dry), which was at 70°F and an atmospheric pressure of 29.4 in. Hg; 10% of the CO remained unburned. How many cubic feet of air were supplied per cubic foot of entering gas? How many cubic feet of product gas were produced per cubic foot of entering gas if the exit gas was at 29.4 in. Hg and 400°F?

Solution:

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Chapter 7: Ideal & Real Gases

1

2

Combs. 3

T= 90 °F P= 35 in.Hg n₁=100 Ibmol y_1,CO2=0.064 y_1,O2=0.0010 y_1,CO=0.3900 y_1,H2=0.5180 y_1,CH4=0.006 y_1,N2=0.0210

T= 70°F

P= 29.4 in.Hg n_2,Air= Ibmol y_2,N2=0.79

y_2,O2=0.21 40% excess air

T= 400 °F P= 29.4 in.Hg n₃= Ibmol y_3,CO2

y_3,H2O y_3,CO y_3,O2 y_3,N2 CO +½ O₂  CO₂

H₂ + O₂  H₂0

CH₄ + 2O₂  CO₂ + 2H₂O

10% of the CO

remained unburned.

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Chapter 7: Ideal & Real Gases

Material Balances Involving Ideal Gases:

Example 7.5 : Material Balance without Reaction

Gas at 15°C and 105 kPa is flowing through an irregular duct. To determine the rate of flow of the gas, CO2 from a tank is passed into the gas stream.

The gas analyzes 1.2% CO2 by volume before and 3.4% CO2 by volume after the addition.

As the CO2 that was injected left the tank, it was passed through a rotameter, and found to flow at the rate of 0.0917 m³/min at 7°C and 131 kPa.

What was the rate of flow of the entering gas in the duct in cubic meters per minute?

Solution:

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Chapter 7: Ideal & Real Gases

1

2

tank

3

V₁(m³/min) y_1,CO2=0.012 y_1,I=0.988 T= 15 °C P= 105 kPa

V_2,CO2=0.0917 m³/min T= 7°C

P= 131 kPa

V₃(m³/min) y_3,CO2=0.034 y_3,I=0.966 T= 15 °C P= 105 kPa

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Chapter 7: Real Gases

• Real Gases are gases whose behavior does not conform to the assumptions underlying ideality.

• The ideal gas equation of state(PV=nRT) is not sufficient to describe the P,V, and T behaviour of most real gases.

Real gases depart from ideal behavior at:

• High pressure.

• Low temperature

Ideal gas gases :

• Low pressure.

• High temperature

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Chapter 7: Real Gases: Compressibility

Corresponding States

• The law of corresponding states expresses the idea that in the critical state all substances should behave alike.

Critical state (point)

• For a pure component it means: the maximum temperature and corresponding pressure at which liquid and vapor can coexist.

• The critical state for the gas-liquid transition is the set of physical conditions at which the density and other properties of the liquid and vapor become identical.

• A supercritical fluid is a compound in a state above its critical point (P_c,T_c)

Water T_c=373.946 °C (647.096 K) P_c= 217.7 atm (22.06 MPa)

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Chapter 7: Real Gases: Compressibility

Compressibility Factor, Z:

is a factor that introduced into the ideal gas law to compensate for non-ideality of gas.

• Ideal gas law PV=nRT

• Real gas law:

PV=ZnRT or

P 𝑉=ZRT

𝑉 = ^𝑉_𝑛 [m³/mol]

Reduced Variables T_r,P_r and V_r:

Are corrected or normalized conditions of temperature, pressure, and volume, normalized (divided) by their respective critical conditions.

𝑇_𝑟 = _𝑇^𝑇

𝑐

𝑃_𝑟 = _𝑃^𝑃

𝑐

𝑉_𝑟 = _𝑉^𝑉

𝑐

Reduced Pressure, P_r

Compressibility factor, Z

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Chapter 7: Real Gases: Compressibility

Compressibility Charts: a) for lower reduced pressures, P_r

Compressibility factor, Z=PV/RT

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Chapter 7: Real Gases: Compressibility

Compressibility Charts: b) for higher reduced pressures, P_r

Pressure-volume factor, P rV ri=ZT r

𝑉_𝑟𝑖 = 𝑉 𝑉_𝑐𝑖

𝑉_𝑐𝑖 = 𝑅𝑇_𝐶 𝑃_𝐶 𝑉 = 𝑉

𝑛

𝑉_𝑟𝑖 :Dimensionless ideal reduced volume.

𝑉_𝑐𝑖:ideal critical molar-volume.

𝑉:molar-volume.

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Chapter 7: Real Gases: Compressibility

Example 7.7: Use of Z factor to calculate a mass of real gas

A tank of ammonia NH₃ (in gas phase) of 3.4 m³, temperature of the tank is 324.8 k and pressure of the tank is 2115.8 kPa.

Estimate the mass of the ammonia tank assuming:

A) Ammonia is ideal gas B) Ammonia is real gas

Solution:

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Chapter 7: Real Gases: Compressibility

Example 7.8: Use of Z factor to calculate a pressure

A tank of Liquid oxygen of 0.0284 m³ volume is filled with 3.500 kg of liquid O₂ that will vaporize at -25°C.

Will the pressure in the tank exceed the safety limit of the tank specified as l0⁴ kPa?

Solution:

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Chapter 7: Real Gases: Compressibility

Compressibility Factors, z⁰ and z¹

𝑍 = 𝑍⁰ + 𝑍¹𝜔

Z⁰ & Z¹ are function of T_r and P_r, find them from tables in Appendix C.

, Pitzer acentric factor:

The acentric factor  indicates the degree of acentricity or nonsphericity of a molecule. For helium and argon, co is equal to zero. For higher molecular weight hydrocarbons and for molecules with increased polarity, the value of  increases.

Compound Acentric Factor (𝜔) Compound Acentric Factor (𝜔)

Acetone 0.309 Water vapor 0.344

Benzene 0.212 Methane 0.008

Ammonia 0.25 Methanol 0.559

Argon 0 n butane 0.193

Carbon dioxide 0.225 n-pentane 0.251

Carbon monoxide 0.049 Nitric oxide 0.607

Chlorine 0.073 Nitrogen 0.04

Ethane 0.098 Oxygen 0.021

Ethanol 0.635 Propane 0.152

Ethylene 0.089 Propylene 0.148

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Chapter 7: Real Gases: Compressibility

𝑍

⁰

Table

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𝑍

¹

Table

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Real Gas Mixtures, Kay's method

Critical pressure and temperature for gas mixtures :

𝑃_𝑟 = ^𝑃

𝑃_𝑐, 𝑇_𝑟 = ^𝑇_𝑇

𝑐

𝑃_𝑐 = 𝑃_𝑐,𝐴𝑦_𝐴 + 𝑃_𝑐,𝐵𝑦_𝐵+𝑃_𝑐,𝐶𝑦_𝐶+ ……

𝑇_𝑐 = 𝑇_𝑐,𝐴𝑦_𝐴 + 𝑇_𝑐,𝐵𝑦_𝐵+𝑇_𝑐,𝐶𝑦_𝐶+ ……

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Example 7.9: Calculation of P-V-T Properties for a Real Gas Mixture

A gaseous mixture has the following composition (in mole percent):

Methane, CH4, 20%, Ethylene. C2H4, 30% Nitrogen, N2, 50%

at 90 atm pressure and 100°C. Compare the volume per mole as computed by the methods of: (a) the ideal gas law and (b) the pseudo- reduced technique (Kay's method).

Solution:

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Chapter 7: Real Gases: Compressibility

Reduced Pressure, P_r Pressure-volume factor, P rV ri=ZT r