First slide

laws

Question

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:

Easy

A

1 : 2
B

3 : 2
C

5 : 2
D

14 : 1

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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