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A glass bulb of volume 200 cc is sealed ...

A glass bulb of volume 200 cc is sealed at `27^(@)`. If the pressure of the gas inside the bulb is `1.25xx10^(-4)cm` of Hg, calculate the number of molecules inside the bulb. (Avagadro's number = `6.0225xx10^(23),R=8.314xx10^(7)` erg/mole -K)

Text Solution

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Volume `V=200cm^(3)`, pressure `P=1.25xx10^(-4)cm` of Hg
= `1.25xx10^(-4)xx13.6xx980("dyne")/(cm^(2))`
`T=273+27=300K`, Avogadro.s number, N = `6.0225xx10^(23)`,
`R=8.314xx10^(7)erg//"mole"-K`
By ideal gas equation, PV = nRT
`n=(PV)/(RT)=(1.25xx10^(-4)xx13.6xx980xx200)/(8.314xx10^(7)xx300)"moles"`
Number of molecules inside the bulb = number of moles X Avogadro.s number
`(1.25xx10^(-4)xx13.6xx980xx200)/(8.314xx10^(7)xx300)xx6.0225xx10^(23)`
= `8.018xx10^(15)`
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