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Thermodynamic Processes

5 g of a gase...

Question

5 g of a gaseous mixture of the $H e$ and $A r$ contained in a 10 l vessel at 27°C exert a pressure of 1 atm (absolute). Determine the mass percentage of $H e$ .
Take $R = \frac{1}{12}$ Latm/molK

10.8

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14.2

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24.4

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74.6

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Solution

The correct option is C 24.4
Let the mass of $H e$ be x g, mass of $A r$ will be (5 - x) g
Moles $= \frac{mass}{molas mass}$
Moles of $H e = \frac{x}{4}$
Moles of $A r = \frac{5 - x}{40}$
Total moles of gas $= \frac{x}{4} + \frac{5 - x}{40}$
$P V = n R T \Rightarrow n = \frac{P V}{R T} = \frac{1 \times 10}{1 / 12 \times 300} = 0.4 \Rightarrow \frac{x}{4} + \frac{5 - x}{40} = 0.4 \Rightarrow x = 1.22 g$
Mass percentage of Helium = $\frac{1.22}{5} \times 100 = 24.4 %, A r = 75.6 %$

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5 g of a gaseous mixture of the He and Ar contained in a 10 l vessel at 27°C exert a pressure of 1 atm (absolute). Determine the mass percentage of He. Take R=112 Latm/molK

The correct option is C 24.4Let the mass of He be x g, mass of Ar will be (5 - x) g Moles =massmolas mass Moles of He=x4 Moles of Ar=5−x40 Total moles of gas =x4+5−x40 PV=nRT⇒n=PVRT=1×101/12×300=0.4⇒x4+5−x40=0.4⇒x=1.22 g Mass percentage of Helium = 1.225×100=24.4%, Ar=75.6%

5 g of a gaseous mixture of the $H e$ and $A r$ contained in a 10 l vessel at 27°C exert a pressure of 1 atm (absolute). Determine the mass percentage of $H e$ .
Take $R = \frac{1}{12}$ Latm/molK