# 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)

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