A vessel of volume 0.02 , m^3 contains a mixt | vedclass

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(A) Given: Volume $V = 0.02 \, m^3$,Temperature $T = 20 + 273 = 293 \, K$,Pressure $P = 2 \, atm = 2 	imes 1.013 	imes 10^5 \, Pa \approx 2.026 	im

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$1 : 2$

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(A) Given: Volume $V = 0.02 \, m^3$,Temperature $T = 20 + 273 = 293 \, K$,Pressure $P = 2 \, atm = 2 	imes 1.013 	imes 10^5 \, Pa \approx 2.026 	imes 10^5 \, Pa$. Total mass $M = 5 \, g$.Let the mass of hydrogen be $m_H$ and the mass of helium be $m_{He}$. Then $m_H + m_{He} = 5 \, g$.Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M_{mol}}$:$PV = \left( \frac{m_H}{2} + \frac{m_{He}}{4} ight) RT$ (using molar masses $2 \, g/mol$ for $H_2$ and $4 \, g/mol$ for $He$ in $g/mol$ units).$2.026 	imes 10^5 	imes 0.02 = \left( \frac{m_H}{2} + \frac{5 - m_H}{4} ight) 	imes 8.314 	imes 293$.$4052 = \left( \frac{2m_H + 5 - m_H}{4} ight) 	imes 2436$.$4052 = \left( \frac{m_H + 5}{4} ight) 	imes 2436$.$16208 = 2436m_H + 12180$.$4028 = 2436m_H \implies m_H \approx 1.65 \, g$.$m_{He} = 5 - 1.65 = 3.35 \, g$.Ratio $m_H : m_{He} = 1.65 : 3.35 \approx 1 : 2$.

$A$ vessel of volume $0.02 \, m^3$ contains a mixture of hydrogen and helium at $20 \, ^oC$ and $2$ atmospheric pressure. The mass of the mixture is $5 \, g$. Find the ratio of the mass of hydrogen to that of helium in the mixture.

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