A mixture of Nitrogen, Oxygen, and Helium gas is collected in a container at 1 atmosphere, if the partial pressure of oxygen is 320 mm and helium's pressure is 112, what is the partial pressure of the nitrogen?

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Answer 1 · 2018-05-10T04:23:20

$The sum of the partial pressures is equal to the total pressure....$ $"The sum of the partial pressures is equal to the total pressure...."$

And so $P_"Total"=SigmaP_"n"$

And we know $1*atm-=760*mm*Hg$ ....here a unit of length is used as a unit of pressure.

$1*atm={((320+112+P_(N_2))*mm*Hg)/(760*mm*Hg*atm^-1)}$

$P_(N_2)=(760-320-112)*mm*Hg=328*mm*Hg$

$P_(N_2)=(328*mm*Hg)/(760*mm*Hg)~=1/2*atm$

Answer 2 · 2018-05-10T04:28:50

328 mmHg

Assumptions made: Partial pressures given for oxygen and helium are in mmHg and required answer is in mm Hg.

First, convert the total pressure of system 1 atm to mm Hg:

$P_"total" = 1 atm * (760 mm Hg)/(1 atm)=760 mm Hg$

According to Dalton's Law of partial pressure:

$P_"total"=P_"oxygen"+P_"helium"+P_"nitrogen"$

$760 = 320 + 112 + P_"nitrogen"$

$P_"nitrogen" = 760 - 320 - 112 = 328 mm Hg$

If you need the value in atmosphere:
$P_"nitrogen"=328 mm Hg*(1 atm)/(760 mm Hg) = 0.432 atm$