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# Partial pressure

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A 400 L container at temperature of 4 celsius initially contains 0.25 mol He gas. Molecular hydrogen, which is a gas at 4 celsius is added until total pressure is 9.38*10^-2 atm. What is the partial pressure of the molecular hydrogen in the container?

https://brainmass.com/chemistry/gas-laws/partial-pressure-25283

## SOLUTION This solution is FREE courtesy of BrainMass!

Note
* denotes multiplication
^ denotes power

Volume=V= 400 L
No of moles=n= 0.25 mol
Temperature= T= 4 C= 277.15 K =273.15+4

Ideal gas Law is PV=nRT
or P=nRT/V
R= Gas constant= 0.082058 L.atm.mol ^-1K ^-1
Therefore
P=Pressure= 0.0142140 atm =(0.25*0.082058*277.15)/400

Thus the calculation of pressure of Helium is correct

In a mixture of two gases the total pressure is the sum of partial pressures

Partial pressure of Helium= 0.0142140 atm
Total pressure= 0.0938 atm (given)
Therefore partial pressure of Hydrogen= 0.0795860 atm =0.0938-0.014214

This can also be obtained in another way

We know the total pressure of helium and Hydrogen
From this we can calculate the no of moles of Hydrogen

Volume=V= 400 L
Pressure= 0.0938 atm
No of moles=n= ? mol (To be determined)
Temperature= T= 4 C= 277.15 K =273.15+4

Ideal gas Law is PV=nRT
or n=PV/RT
R= Gas constant= 0.082058 L.atm.mol ^-1K ^-1
Therefore
n=Total no of moles= 1.65 mol =(0.0938*400)/(0.082058*277.15)

No of moles of Helium (originally)= 0.25 mol
Therefore no of moles of Hydrogen added= 1.40 mol =1.65-0.25

Now calculate the partial pressure of hydrogen using the no of moles of hydrogen

Volume=V= 400 L
No of moles=n= 1.40 mol
Temperature= T= 4 C= 277.15 K =273.15+4

Ideal gas Law is PV=nRT
or P=nRT/V
R= Gas constant= 0.082058 L.atm.mol ^-1K ^-1
Therefore
P=Pressure= 0.0795980 atm =(1.4*0.082058*277.15)/400

Which is close to the answer we obtained above (minor difference because of rounding off error)

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!