#3.06*g#, #4.53*g#, and #5.80*g# masses of dinitrogen, dihydrogen, and helium gases were collected and the total pressure of the mixture was #104.6*kPa#. What are the partial pressures of each gas?

2 Answers
anor277
Feb 7, 2018

Well, we use old #"Dalton's Law of Partial Pressures...."#

Explanation:

Which explicitly states that in a gaseous mixture, the partial pressure exerted by each component gas is the same as it would exert if it ALONE occupied the container. The total pressure is the sum of the sum of the individual partial pressures.

And thus...#P_"Total"=((n_(N_2)+n_(H_2)+n_(He))xx0.0821*(L*atm)/(K*mol)xxT)/(V)#

Now we don't know #T# and we don't know #V#...but this immaterial because we do know #P_"Total"#, and can calculate the mole fraction of each gas.....i.e.

#chi_(N_2)=((3.06*g)/(28.01*g*mol^-1))/((3.06*g)/(28.01*g*mol^-1)+(4.53*g)/(2.015*g*mol^-1)+(5.80*g)/(4.003*g*mol^-1))=0.0288#

#chi_(H_2)=((4.53*g)/(2.015*g*mol^-1))/((3.06*g)/(28.01*g*mol^-1)+(4.53*g)/(2.015*g*mol^-1)+(5.80*g)/(4.003*g*mol^-1))=0.591#

#chi_(He)=((5.80*g)/(4.003*g*mol^-1))/((3.06*g)/(28.01*g*mol^-1)+(4.53*g)/(2.015*g*mol^-1)+(5.80*g)/(4.003*g*mol^-1))=0.381#

And, as required, #Sigma*chi_(N_2)+chi_(H_2)+chi_(He)=1#

And the pressure exerted by each gas is simply the product...

#chi_"component gas"xxP_"total"#...and thus....

#P_(N_2)=141*kPaxxchi_(N_2)=141*kPaxx0.0228=3.2*kPa.#

#P_(H_2)=141*kPaxxchi_(H_2)=141*kPaxx0.591=83.3*kPa.#

#P_(He)=141*kPaxxchi_(He)=141*kPaxx0.381=53.7*kPa,#

And...#P_(N_2)+P_(H_2)+P_(He)=P_"Total"=141*kPa# as required....

Surya K.
Feb 7, 2018

Nitrogen: #2538Pa#
Hydrogen: #104622Pa#
Helium: #33699Pa#

Explanation:

Use Dalton's Law of Partial Pressures, which states that:

#P_i=chi_i*P_"total"#, where #P_i# is the pressure of one component of the gas, #chi_i# its mole fraction, and #P_"total"# the total pressure of the gases.

We must find the number of moles of #N_2#, #H_2# and #He#.

The formula for moles is #n=m/M#, where #m# is the given mass and #M# its molar mass.

Nitrogen:
#3.06/28.01=0.109"mol"#.

Hydrogen:
#4.53/1.01=4.485"mol"#

Helium:
#5.8/4=1.45"mol"#

Totally, we have #0.109+4.485+1.45=6.044"mol"#.

The mole fraction, #chi_i#, is calculated by #chi_i="number of moles of i"/"total number of moles"#.

For nitrogen:
#chi_"nitrogen"=0.109/6.044#

#=0.018#

For hydrogen:
#chi_"hydrogen"=4.485/6.044#

#=0.742#

For helium:
#chi_"helium"=1.45/6.044#

#=0.239#

Since #P_"total"=141kPa#, we can calculate the partial pressure of each.

Nitrogen:
#P_"nitrogen"=141000*0.018#

#=2538Pa#.

Hydrogen:
#P_"hydrogen"=141000*0.742#

#=104622Pa#

Helium:
#P_"helium"=141000*0.239#

#=33699Pa#

Note that the figure is not exact, as I did rounding on the calculations for moles and mole fractions.

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