Question #226ac

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Answer 1 · 2016-08-19T22:54:37

Given
$n_{Ar} \to No. of moles of Argon = 3 mols$ $n_"Ar"->"No. of moles of Argon"=3" mols"$

$n_{Kr} \to No. of moles of Krypton = 5 mols$ $n_"Kr"->"No. of moles of Krypton"=5" mols"$

$p_{Ar} \to Partial pressure of Argon = 210 torr$ $p_"Ar"->"Partial pressure of Argon"=210" torr"$

$Let p_{Kr} \to Partial pressure of Krypton = ?$ $"Let "p_"Kr"->"Partial pressure of Krypton"=?$

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

So

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

$\Rightarrow p_{Kr} = \frac{5}{3} \times 210 = 350 torr$ $=>p_"Kr"=5/3xx210=350" torr"$

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