For a gaseous mixtue of `2.41g` of helium and `2.79g` of neon in an evacuated `1.04 dm^(3)` container at 298 K Calculate the partial pressure of each gas and hence find the total pressure of the mixture.

Explanation

Mass of He `=2.41g`
No. of moles of He `=("Mass")/("Molar Mass")=(2.41)/(4)`
`=0.6025` moles
Mass of Ne`=2.79g` ,
No. of moles of Ne `=("Mass")/("Molar Mass")=(2.79)/(20)`
Volume of the Total no . Of moles of the mixutre
`=0.6025+0.1395=0.7420` moles
Container `V=1.04 dm^(3)`
Temperature `T=298K`
Pressure `P=(1)/(V)=RT`
According to ideal gas equation PV =nRT
`P=(0.7420xx0.0821xx298)/(1.04)=17.45 atm`
Partial pressure P= molefraction `xx` Total pressure
`=(nA)/(nA+nB)xxP`
Partial Pressure of Helium `P_(H_(e))=(0.6025)/(0.7420)xx17.45`
Accorodign to Dalton's law of partial pressure
`=3.280 atm`.
`P=P_(1)+P_(2)+P_(3) ....`
`P_("Total") P_(He)+P_(Ne)=14.169+3.280=17.449 atm`