Question

A mixture of He, Ar, and Xe has a total pressure of 2.80 atm . The partial pressure of He is 0.300 atm , and the partial pressure of Ar is 0.300 atm . What is the partial pressure of Xe?

Learning Goal: To use partial pressure in gas law calculations. In a mixture of gases, the total pressure of the gas mixture is equal to the sum of the partial pressures of the individual gases. For example, if you have a mixture of helium at 2 atm and argon at 4 atm , then the total pressure of the gas inside the cylinder is 6 atm . (Figure 1) Similarly, if you know the total pressure of a gas mixture, you can use the number of moles to find the partial pressures, or the partial pressures to find the number of moles. Example Problem: A mixture of N2, O2, and He gases has a total pressure of 1 atm. The partial pressures of N2 and O2 are 0.7 atm and 0.25 atm , respectively. What is the partial pressure of He? The sum of the partial pressures is equal to the total pressure: Ptotal = PN, + Po, + Pue Rearranging this equation, the partial pressure of He is PHe = Ptotal - PN, - Po, = 1.00 atm – 0.70 atm – 0.25 atm = 0.05 atm

A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He. The temperature of the mixture is 0 ∘C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

Answer 1

Partial pressure of Xe is 2.20 atm

Details calculation given below.....
Given to ta presur P 2.80afn Panta essu of He, te P0.300 atu Partio fresnu of Ar, and et = 0.300 afm partiol pressure of xe, e ? we kno tur Aumaton of part'ord Pressur is epa to tota pressuru AY fu Valu w epuotlon NOw put ati get 2.80 ad 0.300 adu t 0.200 od +re 2.g0-0. 300 0.300) atu P 2.20 a 2.20 afu. P .Patfa Pressuru of xeo is Gven tote volumi of tuu mixtwuzl8.0 L 4atu ressUne O.250 mo o 2.50 mO Let t uncnesn vahumaamount oFHe be nHe As we know At STP(o'c and 1otm) 22.4 L Voluima of n A STP (0'c ad iat) 18.0L Voum an - X18.0 O804 mo , 18.0L ofmixturu confeivw=n 0804 mo to to Nok: STP= standard temperatou and pressusu le, T 0 anal P 1otm

of we know that toted qmount AS mixtusu esual to jum of tu cmount in male) of individuo an Nnom ratetu O.804 He 0. 250 0 250 He O-250 mol -250 C0.04 O.304 mo moleculan weighl of te s 40/mat AS . Amunt of He Present= O304 ma 1216