# Question 226ac

Aug 19, 2016

Given
${n}_{\text{Ar"->"No. of moles of Argon"=3" mols}}$

${n}_{\text{Kr"->"No. of moles of Krypton"=5" mols}}$

${p}_{\text{Ar"->"Partial pressure of Argon"=210" torr}}$

"Let "p_"Kr"->"Partial pressure of Krypton"=?#

Now In a gas mixture partial pressures of constituent gas is proportional to their respective no. of moles present in the mixture.

So

${p}_{\text{Kr"/p_"Ar"=n_"Kr"/n_"Ar"=>p_"Kr}} / 210 = \frac{5}{3}$

$\implies {p}_{\text{Kr"=5/3xx210=350" torr}}$

$\text{Total Pressure"(P)=p_"Kr"+p_"Ar"=(350+210)=560" torr}$