A cubical box of side 5 m contains helium gas at a pressure 320 N/m². During an observation time of 1 second, an atom travelling with root mean square speed parallel to one of the sides of the cube was found to make 1000 hits with a particular wall, without any collision with the other atoms. (take $R = \frac{25}{3} J / m o l - K$ ). Then the total mass of helium gas in box is(in grams)

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Detailed Solution

Time between two consecutive collision $= \frac{1}{1000} \sec$
So, $\frac{2 l}{v_{r m s}} = \frac{1}{1000}$
$v_{r m s} = 2000 \times 5 = 10000 m / s$
$v_{r m s} = \sqrt{\frac{3 R T}{M}} = 10000$
So, $T = \frac{10^{8} \times 4 \times 10^{- 3}}{3 \times (\frac{25}{3})} = 16 \times 10^{3} K$
Mass of helium gas $\Rightarrow P V = n R T$
$P V = \frac{m}{M} R T$
$m = \frac{P V M}{R T} = \frac{320 \times 125 \times (4 \times 10^{- 3})}{(\frac{25}{3}) \times 16 \times 10^{3}} = 1.2 g m$

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