35.0 L of helium gas with pressure 48.0 atm is added to 100.0 L tank containing

Question

35.0 L of helium gas with pressure 48.0 atm is added to 100.0 L tank containing gas at a pressure of 57.0 atm Calculate the partial pressure of helium in the 100.0L tank Calculate the total pressure of the mixture in the 100.0 L tank. Calculate the mole traction of helium in the tank. If the mixture contains 18.0 moles of neon, how many moles of helium are in the mixture? 1 atm = 760 mmHg = 14.7 psi = 101. kPa u_rms = Squareroot 3RT/W u_1/u_2 = Squareroot W_2/W_1 R = 0.08206 L atm/mol-k = 8.314 j/mol-k 1 j = 1 kg m^2/s^2

Explanation / Answer

1) partial pressure of He in 100L tank = inital V x initial p / final volume

= 35.0L x 48.0atm/100L

= 16.8 atm

2) total pressure = PHe + PNe according to Dalton's law

= 16.8 atm + 57 atm

= 73.8 atm

3) mole fraction of He

From Dalton 's law parital pressure of gas = mole fraction x total pressure

Thus mole fraction of He = Phe /total pressure

= 16.8/73.8

=0.2276

d)The mixture has 18 moles of Neon .

Let x be the moles of He

Then mole fraction of He = x /(18+x) = 0.2276

Solving for x , x=5.304moles

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