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A closed vessel A having volume V contains N₂ at pressure P and temperature T. Another closed vessel B having the same volume V contains He at the same pressure P but temperature 2T. The ratio of masses of N₂ and He in the vessels A and B is:

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5 : 2

3 : 2

14 : 1

1 : 2

answer is D.

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Detailed Solution

PV = nRT

$\Rightarrow PV = \left( {nM} \right)\frac{{RT}}{M} = m\frac{{RT}}{M}$

$\Rightarrow m = \frac{{PV}}{R}\frac{M}{T}$

So at same pressure and volume, $m \propto \frac{M}{T}$

$\therefore \frac{{^m{N_2}}}{{^mHe}} = \frac{{^M{N_2}}}{{{M_{He}}}} \times \frac{{{T_{He}}}}{{{T_{{N_2}}}}}$

$= \frac{{28}}{4} \times 2 = 14$

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