QUESTION THREE Amature of gases consists of 30 percent hydrogen, 40 percent helium, and 30 percent...

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201 Tirupati. Date: Page: sal) PARTA Let's_suppose_that_e total of 3 moles are present in the mixture. Then, as_ber as volumetric analysis: Moles of bydrogen : 0:31 Moss Of Hydrogen = 0.63 am Moles of Helium = 0.430 , Mass of Helium) = (04) x (0) = 1.69m Moles of Nitrogen = 0.32_,Mass el nitrogen = (0-31) 28 - B.UI em Total. Mass_of-mixture= M = 0.63 +1.630 +8.456 = 10:62 Mass frachon of Hydrogen = 0:63/10.636 to 056 Mass fraction of Helium 1.GX 10 60 = 10 0.151 Mass Fraction of Nitrogen = 8:43 110-6210.792 Abbaranet Mol. Mass Tatal Mass Total No. of moles 10.630 / Mavo - 10.69/mol PART B sol) | Here, V= 100 m2, P= 130kpa = 1.283 atm, T= 200°C= 523K 3105 Here, initially the tank was evacuated, the bressure to which ethane is added is given. by: (pe), such that ne: moles of ethane bez he „RXI__O For= total motes po pressure total Now, P=(MRT)/V n PU/RT no = (:283) X (105) / 0.0021 X523 = 2988 moles ie As done in the part A, broceed sames Ils done on the next page

201 Tirupati. Date: 11 Page: X of Subbase_1010LMASS_4_mixture is mee_as..bec_the_gravimetric analysis, mass of ethane = 0:31 Mass of metane: 0 0:79 moles of eihane : Mass(m) Mol. Massim = 0.32 = 0.011 20 methane : Moles of 0:0432 0.71. 16 males - Total no. of moles for 0:31 +0.72 16 in mixture (0.013 +0.043 Now, from earn o we get total moles = 2988 moles Eau ating 04 and 6 we act 10.053. = 2988) 2988 - 56327135 0.053 :دوم ' Total Total mass of mass of mixture mixture I gms = 56:38 kq i' ne moles of ethane = 0.012=563.77 moles ANIS Putting this value of ne in cruation as we geti pe: ne RT (563.77 X0.0821X523)/105 am - 0:242_ato - (0.242 x 101-325) kPa be 24.5 kPa V ANS