A closed container of volume 0.02 m^3 contai | vedclass

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(A) Given: $V = 0.02 \ m^3$,$T = 27^{\circ}C = 300 \ K$,$P = 1 	imes 10^5 \ N/m^2$,$M_{total} = 28 \ g$. Let $m$ be the mass of neon $(Ne)$ and $(28

Vedclass · Accepted Answer

$m_{Ne} = 4 \ g, m_{Ar} = 24 \ g$

Vedclass · Answer

(A) Given: $V = 0.02 \ m^3$,$T = 27^{\circ}C = 300 \ K$,$P = 1 	imes 10^5 \ N/m^2$,$M_{total} = 28 \ g$. Let $m$ be the mass of neon $(Ne)$ and $(28 - m)$ be the mass of argon $(Ar)$. The total number of moles $\mu$ is given by $\mu = \frac{m}{20} + \frac{28 - m}{40} = \frac{2m + 28 - m}{40} = \frac{28 + m}{40} \dots (i)$. Using the ideal gas equation $PV = \mu RT$,we have $\mu = \frac{PV}{RT} = \frac{1 	imes 10^5 	imes 0.02}{8.314 	imes 300} \approx 0.801 \ mol$. For simplicity in calculation,using $R \approx 8.33$ or standard approximation,$\mu = \frac{2000}{2494.2} \approx 0.8$. Equating $(i)$ and $(ii)$: $\frac{28 + m}{40} = 0.8 \implies 28 + m = 32 \implies m = 4 \ g$. Thus,mass of neon is $4 \ g$ and mass of argon is $28 - 4 = 24 \ g$.

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