Question

A mixture of helium and neon gases at 335 K contains 3 times the number of helium atoms as neon atoms. The concentration (n/V) of the mixture is found to be 0.150 mol/L. Assuming ideal behavior, what is the partial pressure of helium? kPa

Answer 1

page:ot Arxwe):- According to the given information, to find the following problems. Given, * wokai is ponttal pressure of hellum : - ne knows that, ideal gas equation, (Pv=nRT - teve, P = pressure of the gas. ve volume of the gas no no.of moles R = gas constanta (0.0821 4 alm. mol.k). to temperature at Kelvin. Now, mixture have. Plot a Pue + Pnel - Here, Plot - total prevare.

page:ca Pue a partial pressure of Helium. Pue a partial pressure of Heon. and, fone tot = Whet nive] - Here, mtot = total noof moi Plite - No.of mol of Helium rine a No. of mol of Neon. NOW, at 335k in the mixture applying Equation -- 0 Plot & Neon = ntot RT :. According to problem, nlus not Veon = 0.150 molld. vcom o volume of concentration. then, Plot 1 Hot ) RT . \ncon

page:03 Ptot - 0.150 mot fik * 0.0891 Katm.onet. X X 335 K 0.150 X 0.0821 x 335 atm Phot-4.195 atm Again, nie = 3x nne puitting this value into Equration - 12 tot = (3x Anel + nne. Intot unnel .: mol fraction of Nel Xne) - no. of mol of the total no. of mol - noe not on ne unte at 1 mol fraction of we lyned = o. as.

page:ou -* mole -fraction of tle in the mixture will be, (1-XNve) 1-0.85 = 0.75 .: XHe = 0.45 x mepresents mole ( fraction). NOWO, partial pressure of Helium :- het, pantial pressure of He = mole frackon (x) x total pressure. (Ptet) PHe = He x Prot 2 0.75 4.195 atm - 3.093 atm. 23.093 ator X 101.385 kpa Tatim 1. Phe = 313.4 kpa. 1. partial pressure of Helium (PHe} = 313.4 kpa

Page:os * what is tolal pressure at 5°cir het, ideal gas law, pranRT * moles of He transferred. na Pixui / R XT1) 211858 18.17) kpa RX (95+273) = 2387.5 kPa Rx(248) ni = figuu/R. Kpa * moles of oa transfered na - Pexve (lRxT2) = (1958 4199) kPa RX625 +213) 5237.5 kpa RX (298) n2 = 17.536/rkpa

page:06 * -tank pressure in tank pressure , P = (nitng) Rrlu Here, NE 5.506 then, P = (7. 844 + 14.546)\Pa R (534 943)/s.so - ( 45. 42 ) xpa R ( 326) /s150 R 1 15.42 x 326 kpa 5.50 8186.92 kpa S:50 - 1506.71 kPa les total pressure 1506.41 kpa