Partial pressure
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?
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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
Answer: 0.0795860 atm
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)
© BrainMass Inc. brainmass.com December 24, 2021, 5:03 pm ad1c9bdddf>https://brainmass.com/chemistry/gas-laws/partial-pressure-25283