1. An insulated 10.0-L tank containing compressed Ar( g ) at 20.0 atm and 400. K

Question

1. An insulated 10.0-L tank containing compressed Ar(g) at 20.0 atm and 400. K is connected to an un-insulated 20.0-L tank containing compressed Ne(g) at 10.0 atm and 300. K by a closed valve. When the valve is opened and the contents of the tanks allowed to mix and reach 300. K:

a) What is the mole fraction of Ar in the mixture?

b) What is the measured pressure (in atm)?

c) What is the partial pressure (in atm) of Ar in the mixture?

2. Consider the following two vessels containing pure gases. Vessel 1 contains 1.50L of pure He(g) at 250. K and 2.50 atm; Vessel 2 contains 1.00 L of pure O2(g) at 350. K and 1.50 atm. The contents of these two vessels are mixed at 300. K in a 2.00-L vessel. Determine the total pressure (in atm) and the partial pressure of O2 (in atm) in the third vessel.

Explanation / Answer

Q1.

PV = nRT

n = PV/(RT)

mol of Ar = PV/(RT) = 20*10/(0.082*400) = 6.097560

mol of Ne= PV/(RT) = 10*20/(0.082*300) = 8.13

total mol = 8.13+6.097560 = 14.22756

Ptotal at end

PV = nRT

P = nRT/V

Vtotal = 20+10 = 30

P = (14.22756)(0.082*300)/(10+20) = 11.66659 atm

c

x-ar = 6.097560/ (14.22756) = 0.4285738

P-Ar = x-ar * Ptotal = 0.4285738*11.66659 = 4.999 amt i.e. almost 5 atm

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