Question

- Question 1 (a) In an ideal gas mixture, the partial pressure of a constituent gas is: 25 points inversely proportional to the mole fraction inversely proportional to the square of the mole fraction equal to the mole fraction directly proportional to the mole fraction (b) The value of the universal molar gas constant is: 8.3145 J/(kmol) 8.3145 kJ/(kg) 8.3145 J/(kg) 8314.5 J/(kmol K) (c) A mixture of ideal gases consists of 4.42 kg of carbon monoxide (CO) and 5.91 kg of nitrogen (N2) at a pressure of 4 bar and temperature of 650 K Using Rogers and Mayhew (1995) Thermodynamic and Transport Properties of Fluids, determine the following: (i) Mole fractions Ej of the constituent gases: Ico= Number IN2 = Number (in) Mean molar mass of the mixture: in = Number Units (iii) Specific gas constant for the mixture: R = Number Units (iv) Volume and density of the mixture: V = Number Units p = Number Units

Answer 1

THIS QUESTION IS FROM GAS MIXTURE LAWS AND RELATED FORMULAS ARE APPLIED TO SOLVE THE QUESTION

solution) (a) In an ideal gas, the partial pressure of constituent gas is directly proportional to molefraction. (6) The value of universal gas constant is 8.314 J/mol K. (c) Criven, mass of co, mi= 4.42 kg Pressure = 4 bar Mass of N21 m2 = 5.91 kg Temperature = 650k y Mole fractions of constituent gases - - Mole fraction = n . Mole fraction = n2 for co nitha I oa where n = Moles of = mass of co = 4.42=0.1578 co Molon Mass of co 28 na= Moles of = Mass of N2 = 5.91 Cl.all They - W2 Molar mass of Nz 28 Mole fraction = 0.1578 Mole lonaction = 0.02 of co 0.1578 to all I of N2 0.1578 +0.GR - Hà 09378 = 0.5721 Auswen (11) Mean on Average molar mass of mixture Mean molar mass of IM, M2 are molar = n, x Mitnx M₂ masses of co mixture hit na ( 8 N2 = 0.1578 X 28+0.211 X 28 0.1578 +0.211 = 4.4184+ 5.908 = 27.999 0.3688 Mean molar mass = 27.999 ~ 28 kg of mixture asto Auswer Answer

het R Il Specific has constant for mix mixture Ri= Cras constant = R - 8.314 = 0.2969 Il malk for co R2= Gras constant - l = 8.314 = 0.2969 Il mal-k. = Cras cons m= Gas constant = m. Ritm, R = 4.42x0.2969 +5.918 0.2969 for mixture mitme 4.42 + 5.91 2 (Rm= 0.2969 Il mal K Answer oliv volume of the mixture - Volume of co= m, RT = 64.42X0.2969 650 = 2.132x10-3 m. P 4x los 2- Volume of Ny = m₂ RT - 5.91x0.29 69x650 -2.851x10 m² 4X105 volume of mixture = mvitm, v = (4.42X 2.132 + 5.91x2851x103 mitm 4.42 +5.91 = 2.5433 x 103 m3 Augwen Density of mixture - meitm, la - Density of co= m = 4.42 - 2013.17 kg/m3 V 21B8X103- & Density of N₂ = my - S.91 - 2012.95 kg/m3 V2 2.851 X10 Density of mixture - 4.426 2073.17+ 5.91x2072.95 чча 4 см - 2013.048 kg/m3 Answer. () Partial pressure of water constituent gases k = 0.4278 partial pressure = Total pressure X Mole fraction = 4 x of co = 1.7112 bar a of co - Partial pressure - Total pressure x role fraction = uxo-5721 Answer of N = 2.2884 bar ) mitm₂ wi) The partial volume 1, 8 V2 are calculated above in part