ME2121Thermodynamics Lecture 7 Gaseous Mixtures (Non-Reacting) Sections 13.1 to 13.3 (Cengel & Boles) Definitions & Basics • A non-reacting gaseous mixture is a multi-component system in which the constituents do not react with one another. No chemical reaction! • Mass fraction The total mass of a mixture is the sum of the constituents: m mi i 00E1 The ratio mi/m is called the mass fraction of the component in the mixture. L7: Gaseous Mixtures (Non-reacting) 2 for O2. – The kilomole (kmol) is the amount of substance numerically equal to its molar mass in kg E.g. N2 & H2. molecules. 1 kmole of O2 has a mass of 32 kg. ions.) • Avogrado's hypothesis – – Equal volumes of all gases contain the same number of molecules when the volumes are measured under the same conditions of temperature and pressure. must be specified and may be atoms.g. electrons or other particles or groups.Definitions & Basics • Mole The mole is the amount of substance of a system containing as many elementary entities as there are atoms in 0. 1 mole contains 6. (compared to 12 for Carbon-12) 00E1 L7: Gaseous Mixtures (Non-reacting) 3 Definitions & Basics (contd.032 kg since O has a molar mass of 32.012 kg of C.022 x 1026 elementary entities Elementary entities could be atoms e.g.022 x 1023 elementary entities 1 kmole contains 6. 00E1 L7: Gaseous Mixtures (Non-reacting) 4 . for He & Ar. For any gas. • Example of a mole One mole of O2 is a mass of 0. or molecules e. n for a mixture is the sum of the moles of the constituents. the mole fraction is also a measure of the volume fraction! 00E1 L7: Gaseous Mixtures (Non-reacting) 5 Definitions & Basics (contd. mi ni M i where M i molar mass of the ith component M 1 ni M i i xi M i n i where overscore denotes “per mole or kmole” (as appropriate) 00E1 L7: Gaseous Mixtures (Non-reacting) 6 . – The ratio.) • Mole fraction – The total number of moles.: The number of atoms or molecules is directly related to the volume of the gas! i.B.e. M Mass of ith component.) • Molar mass of gaseous mixture. ni/n is termed the mole fraction. xi.Definitions & Basics (contd. xi 1 for a mixture i N. .. i i RT RT RT RT i RT V V1 V2 . n n1 n2 ...) • Amagat’s law of partial volumes (also known as Leduc’s Law) The volume of a mixture of gases is equal to the sum of the volumes of the individual constituents when each exists alone at the pressure and temperature of the mixture. Pi Pi i Also x i ni PV RT Pi i n PV RT P partial pressure of the ith component 00E1 L7: Gaseous Mixtures (Non-reacting) 7 Definitions & Basics (contd..... ni ni PV nRT where R = universal gas constant i PV PV PV PV1 PV2 .... The mixture and constituents are assumed to behave as ideal gases! This is proven by considering a specified volume V and temperature T...... ni ni PV nRT where R = universal gas constant i PV PV PV PV PV 1 2 . i i RT RT RT RT i RT P P1 P2 .Vi Vi Also x i ni PVi RT Vi n PV RT V i partial volume of the ith component 00E1 L7: Gaseous Mixtures (Non-reacting) Pi P 8 .Definitions & Basics (contd. n n1 n2 ...............) • Dalton’s law of partial pressures The pressure of a mixture of gases is equal to the sum of the pressures of the individual constituents when each occupies a volume equal to that of the mixture at the same temperature. Ideal gas behaviour is assumed for the mixture and constituents! Considering a specified pressure P and temperature T. . Total mixture internal energy: U U 1 U 2 .U i U i i The internal energy per mole of mixture may be calculated: U U n U i n 00E1 i n u i i i n xi ui i L7: Gaseous Mixtures (Non-reacting) 9 Total Mixture Properties (contd..) • Enthalpy Total mixture enthalpy: H U PV The enthalpy per mole of mixture may be calculated: H U V Pi 1 1 U i PiV ni ui PiV n n i n i i. H xi hi i 00E1 L7: Gaseous Mixtures (Non-reacting) 10 ..Total Mixture Properties • Internal energy For a constituent. U i ni ui where ui is the internal energy per mole of the ith constituent at the mixture temperature.e... ) • Entropy Total mixture entropy: S S i ni S i i i Entropy per mole of mixture: S 1 S i xi S i n i i Question: How are specific properties of a mixture related with those of its constituents? e.g. i dT i The relation with universal gas constant: R C P CV xi R i 00E1 L7: Gaseous Mixtures (Non-reacting) 11 Total Mixture Properties (contd.) • Specific heats For ideal gases. show that s mf i si i 00E1 L7: Gaseous Mixtures (Non-reacting) 12 .Total Mixture Properties (contd. the specific heat quantities per mole: CV dU xi CV . i dT i CP dH xi C P . ) But dS Mds and xi ni mi n Mi m M ds mi Ri dPi dS 1 mi C P i dTi M m i M i Ti i M i Pi ds dPi dTi 1 mi C P i mi Ri Pi Ti i m i The entropy change resulting from mixing may be calculated from the temperature and partial pressure changes and constituent masses! 00E1 L7: Gaseous Mixtures (Non-reacting) 14 . Tds dh vdP and Pv RT C dT R ds P dP T P dS i • C P dTi R dPi where Ti temperature of component Ti Pi Therefore for the mixture.Entropy of Mixing • For each component (ideal gas behaviour assumed). dT dP dS xi C P i xi Ri i Ti i Pi i 00E1 L7: Gaseous Mixtures (Non-reacting) 13 Entropy of Mixing (contd. 14 Ar 0.44 kPa 16 .Example 1 The percentage volumetric composition of air is as follows: nitrogen (N ) 78.0 Partial pressure of O2 = vol fraction of O2 x total pressure = 0. determine the partial pressure of the oxygen.87 75. calculate the % gravimetric composition.55 O2 0.09.31 TOTAL 28.2095 32 6.2095 x 150 = 31.70 23. If the mixture is maintained at 150 kPa and 25°C.96.7809 28 21.38 1. oxygen (O ) 20.95 100. (a) (b) (c) (d)=(b)x(c) (e)=(d)/(d) N2 0. 32 and 40. Given that the molar masses of the constitutients are 28. and hence the % gravimetric composition may be calculated as shown in the following table: 15 Example 1 (contd) Constituent Volume fraction Molar mass Mass per kmole % grav comp. argon (Ar) 0.95. respectively. Solution: The mass per kmole of mixture for each constituent.0096 40 0. P and temperature. x O2 = x N2 = 0. T of the mixture. Calculate the end state of the mixture and the entropy change. no. the other N2. each at 7 bar and 90°C.87 kJ/K 18 . One contains O2.Example 2 Two containers. PO2 PN 2 0.7 m3.5 P where P is the pressure of the mixture Therefore change of entropy in the mixing process may be calculated: Tds dh vdP and Pv RT ds C P dT R dP T P PV nR T dT = 0 P S nO2 R ln PO2 P n N 2 R ln PN 2 =1. nN2 Therefore mole fraction of both O2 and N2 . of moles of O2. there will be no changes of pressure. are isolated from each other by a partition. nO2 = no. Considering Avogrado's Law.) Partial pressures of O2 and N2 after removal of the partition. Solution: Assuming adiabatic conditions.5 after removal of partition 17 Example 2 (contd. each of volume 0. The partition is removed and mixing occurs. of moles of N2 . Assignment 5 Please attempt: Problem Set 3: Q 1 (problem-solving class) Q 2 (Imparo online tutorial) (Please attempt problems before referring to the Outline Solutions for maximum benefit in learning) Enjoy!! 00E1 L7: Gaseous Mixtures (Non-reacting) 19 .