A vessel of volume , `V = 5.0` litre contains `1.4 g` of nitrogen at a temperature `T = 1800 K`. Find the pressure of the gas if `30%` of its molecules are dissociated into atoms at this temperature.

Answer

`1.94 xx 10^(5)Nm^(-2)`

Explanation

Here, `V = 5 litre = 5 xx 10^(-3)m^(3)`
`m=1.4 g , T=1800 K`
Mass of atomic nitrogen = `1.4xx30/100 = 0.42g`
Mass of molecules nitrogen = `1.4 xx 70/100 = 0.98g`
Number of moles of atomic nitrogen,
`n_(1) = 0.42/14 =0.03`
Number of moles of molecular nitrogen
`n_(2) = 0.98/28 = 0.035`
As `P=P_(1)+P_(2)`
`n_(1)(RT)/(V) +n_(2)(RT)/V = (n_(1)+n_(2))(RT)/V`
`:. = ((0.03+0.035)xx8.31 xx 1800)/(5 xx 10^(-3))`
`=1.94 xx 10^(5)Nm^(-2)`