Question

QUESTION 1 [ 10 MARKS] The gravimetric analysis of a gas mixture is 50% N2, 40% CO2 and 10% CH4 with a temperature of 40°C and pressure of 1 bar. Calculate, 1.1 the partial pressure of each component, in bar. 1.2 the molar mass for the gas mixture, in kg/kmol. 1.3 the mass by 4 m of mixture, in kg. QUESTION 2 [ 10 MARKS A vessel having a volume of 3 m’initially contains a mixture of Triatomic at 21°C, 2 bar consisting of 23% SO2 and 77% CO2 on a molar basis. Triatomic is allowed to flow into the vessel until the pressure is 4 bar. If the final temperature of the mixture within the vessel is 300 K, calculate the mass of each component present, in kg.

Answer 1

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Sol). Here, gravimetric analysis_is_done. 50_the_percentages Of gascs. givenoce_the_bercentages_of Their_mass: 50, let's_assume that_the_mass of gas_mixture_is_100g. mass_of_N2 Then 59, X1002509 100 si_My_no-of moles of N2 =_50/28 = 1.785 moles {n-m/ Mozzeg/mol Mass of C02 - 4900* X1003 40g (Mraz: W9q/ molt - 12p0-of_moles of co2 = 40/44 20:91_moles * mass of_CHY_ID mixture 900X100 = 109 .: 73, no-of moles of CH4 = 10/16 Menu = 169 - 169/mel? = 0.625 moles N = Foral no. of moles = n, thatn3=33?imole's a): Partial bressure (mole fraction of Pressure of mixture Of component component . [ mnole componente PN₂ = niXP 1.785x1 : 0.537 bar N 3.32 P: 1 bac {Given mi. mole fraction * Peo2 = 22 xD=0.91 = 0:274_bab N 3:32 Pchy - 73. XP N 0'625 x1 : 0'188_bat 3.32

Molat mass ipverage) Mass 01 given mixture No or moles(total - m N m nithatha 100 3.32 fimul means 102103 23:32 moles Mayg 30-12 g/moll c)= Nou). dels. suppose_that mass of ym3 of myture ism, then we know the for ideal mixture PV = nRT PV: MRT (0) Mang V: Volumes PEpressure d mixture1bar 4m2 M=30 12 g/mol T=407 = 33K R=0:0821. Latm_mol'k! Now, P:1bar: 0.987_alm V-4m3 = 40001. Putting. Ihese_voluss_.in 0.we_get m (in grams). Py Mayg. RT 0.987 X 4000 x 3012 0.0821x313 gm = 4627.4_gm ... m: 4:627 kg