In the flask diagrammed below, the left bulb contains 3.0L of neon at a pressure of 2.0 atm. The right bulb contains 2.0L of argon at a pressure of 1.0 atm. What is the mole fraction of neon and argon in the final mixture after the valve is opened and a uniform pressure is reached at constant temperature (25 ?C)?

I've determined the final pressure to be 1.6atm using P1V1+P2V2 = PfVf, but I have no idea how to get the mole fractions.

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-11-11T21:40:35+01:00

Answer:

Mole fraction of Neon = 0.75 = 3/4

Mole fraction of Argon = 0.25 = 1/4

Explanation:

Step 1: Given data

Volume of the left bulb = 3.0 L of Neon

Pressure of the left bulb = 2.0 atm

Volume of the right bulb = 2.0 L of argon

Pressure of the right bulb = 1.0 atm

Temperature = 25°C

Step 2: Calculate nimber of moles

Via the ideal gas law P*V = n*R*T

n = P*V / R*T

nNeon = (2*3)/ (0/08206* 298)

nNeon : 6/ 24.45 = 0.245

nArgon = (1*2)/(0/08206*298)

nArgon = 2/24.45 = 0.0818

nTotal = nNeon + nArgon = 0.245 + 0.0818 = 0.3268

Step 3: Calculate mole fraction

Mole fraction of Neon = 0.245 / 0.3268 = 0.75 = 3/4

Mole fraction of Argon = 0.0818 / 0.3268 = 0.25 = 1/4