A closed container of volume 0.02m^...
Question
A closed container of volume 0.02m^3 contains a mixture of neon and argon gases, at a temperature of 27∘C and pressure of 1×10^5Nm^−2. The total mass of the mixture
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A closed container of volume ( 0.02 mathrm{m}^{3} ) contains a mixture of neon and argon gases, at a temperature of ( 27^{circ} mathrm{C} ) and pressure of ( 1 times 10^{5} mathrm{Nm}^{-2} ). The total mass of the mixture is ( 28 mathrm{g} ). If the molar masses of neon and argon are 20 and 40 g mol ( ^{-1} ) respectively, find the masses of the individual gases in the container assuming them to be ideal (Universal gas constant ( R=8.314 mathrm{J} / mathrm{mol}-mathrm{K}) )

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A closed container of volume 0.02m^3 contains a mixture of neon and argon gases, at a temperature of 27∘C and pressure of 1×10^5Nm^−2. The total mass of the mixture is 28g. If the molar masses of neon and argon are 20 and 40 g mol^−1 respectively, find the masses of the individual gases in the container assuming them to be ideal (Universal gas constant R=8.314J/mol−K)
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Given, temper mass of the neon gas in the mixture. Then, mass Let ( m ) be the mald be ( (28-m) ) of argon Number of gram moles of neon, ( n_{1}=frac{m}{20} ) Number of gram moles of argon, ( n_{2}=frac{(28-m)}{40} ) From Dalton's law of partial pressures. Total pressure of the mixture ( (p)= ) Pressure due to neon ( left(p_{1}right) ) ( + ) Pressure due to argon ( left(p_{2}right) )
or
[
p=p_{1}+p_{2}=frac{n_{1} R T}{V}+frac{n_{2} R T}{V}
]
[
=left(n_{1}+n_{2}right) frac{R T}{V}
]
Substituting the values
[
1.0 times 10^{5}=left(frac{m}{20}+frac{28-m}{40}right) frac{(8.314)(300)}{0.02}
]
Solving this equation, we get
[
m=4.074 mathrm{g} text { and } 28-m=23.926 mathrm{g}
]
Therefore, in the mixture, 4.074 g neon is present and the rest
i.e., ( 23.926 mathrm{g} ) argon is present.

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