Question

Suppose that Daniel has a 2.50 L bottle that contains a mixture of O2, N2, and CO2 under a total pressure of 5.40 atm. He knows that the mixture contains 0.290 mol N2 and that the partial pressure of CO2 is 0.250 atm. If the temperature is 273 K, what is the partial pressure of O2?

PO2=

atm

Answer 1

Given, Protal = 5.40 atm > Post Post Peod 5.40 atm. to Total volume, Vtotal = 2.50 Li Pog=0.250 atm moles of Ne, nag=0.290 mol, T = 273k since, [PV=nRT Ideal gas equation. * (PV rotal = rotal TRXT 5.40 atm x 2.50 K = N Total * 0.0821K stom, ktorent x 273K 13.50 = P Total * 29.4133 mot7 ► nostal = 13.50 = 0.602 mol. 22:4133 mchtě natal = noot Net nog = 0.602 mot a partial pressure of No IPN = XNqX Porotal (mole fraction) since , XN) = - 0.290 mol Total 0.602 mol. » XHg = 0.4817. of Pris = 0.4817 *5.40 atm = 2.60 atm. since, Protal = Post Priz+ Pcos a Pogo Protal - Png Pos = 5.40 atm – 2.60 atm = 0.250 am n Pon = 2.55 aton raction) niz a Since