Answer
$ 1.5 atm$
Work Step by Step
Given $T_{Ne}=30°C=303K$ $T_{N_{2}}=20°C=293K$
$P_{N_{2}}=1 atm$
Ideal gas constant$ R=0.0821 atm L/Kmol$
Molar mass of $ M_{Ne}=20.18 g/mol$
Molar mass of $M_{N_{2}}= 2\times14.007g/mol = 28.014 g/mol
$ Density of nitrogen $d=\frac{m}{V}$
PV=nRT ($n=\frac{m}{M}$)
So $d=\frac{PM}{RT}$
For nitrogen $d=\frac{1*28.014}{0.082*293} =1.2 g/L
$ Both gases have the same density. So pressure of neon gas is $P_{Ne}=\frac{dRT}{M}$ $P_{Ne}=\frac{1.2*0.082*30}{20.18}= 1.5 atm $
