Question

Help!! Do I need to get the partial pressure of all of the noble gases? If so, How?

You have a mixture of the noble gases Xe, Ne, and Kr at 6.0 atm. The partial pressure of Xe is 2.1 atm and the mole fraction of Ne is 0.3500. What is the partial pressure of Kr in this mixture?

Answer 1

The partial pressure exerted by each gas equals the total pressure times the mole fraction of that particular gas

\[P_{total} = n_{gasA}*P_{total}+n_{gasB}*P_{total} + ... \]

Answer 2

\[P_{total} = 6.0 atm \]

we can figure out the mole fraction of xenon easily because we know that the pressure exerted by xenon is equal to the total pressure times the mole fraction of xenon.

\[p_{Xenon} = n_{Xenon}*P_{total} \]

we need the molar fraction of xenon.

so we solve.

\[\frac{ P_{xenon} }{ P_{total} } = n_{mole fraction Xe}\]

plugging in the numbers

\[\frac{ 2.1 }{ 6.0 } = 0.35_{mole fraction of xenon}\]

this tells us that our molar fraction of xenon is 0.35.

now molar fractions add up to 1, why? because the key here is fraction it's kind of like a percentage.

\[molar fraction = Xe + Ne + Kr = 1 \]

we know that

Xenon is 0.35
and
Neon is 0.35

\[0.35+0.35 + Kr = 1 \]

so The molar fraction of Kr is equal to 0.3

now

\[Pressure_{Kr} = n_{Kr}*P_{total} \]

So this translates to
0.3*(6.0atm} = 1.8 atm


The sum of each of these partial pressures must equal the total pressure exerted on the system. so we add them up

Pt = total pressure and n = mole fraction for that particular component.

\[P_{total} = nXe*Pt+nNe*Pt + nKr*Pt \]



\[1.8+2.1+2.1 = 6.0 atm \]

we find that when we add up all the partial pressures we get the total pressure of 6 atm that's how we know our answer is correct.

so the partial pressure of Kr is 1.8 atm. and it's molar fraction is 0.30