You have a closed container filled with 1L of pure air at STP. Assume that this air is made of only nitrogen (79%) and oxygen (21%). What are the partial pressures of nitrogen and oxygen in the box?Hint: XN = PN/PTotal= NN/NTotal

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Quizplus · Accepted Answer

XS1U1B1NS1U1B0=mole fraction of nitrogen
NS1U1B1NS1U1B0=moles of nitrogen
NTotal=moles total
XS1U1B10S1U1B0=mole fraction of oxygen
NS1U1B1oS1U1B0=moles of oxygen
Since 79% of the air is nitrogren, the mole fraction (xN) is .79. (XS1U1B1NS1U1B0=NS1U1B1NS1U1B0/NS1U1B1TotalS1U1B0). Likewise, the mole fraction for oxygen is .21. This is calculated by realizing that 79% of the air is nitrogen, therefore 79% of the molecules (or moles) are nitrogen. So, you don't need to know NS1U1B1NS1U1B0 or NS1U1B1TotalS1U1B0 to calculate this.
Partial pressure equations are as follows:
XS1U1B1NS1U1B0 = PS1U1B1NS1U1B0/PS1U1B1TotalS1U1B0 NS1U1B1NS1U1B0/NS1U1B1TotalS1U1B0
.79 = Psub> NS1U1B1/PS1U1B1TotalS1U1B0
.79*PS1U1B1TotalS1U1B0= Psub>NS1U1B1
.79 * 1.0 atm = Psub> NS1U1B1 = .79 atm
Psub> 0S1U1B1 = Plsub>TotaS1U1B1 - Psub>NS1U1B1 = 1.0 atm - .79 atm = .21 atm

You Have a Closed Container Filled with 1L of Pure