An open vessel containing air is heated from 300K to 400K. The fraction of air, which goes out with respect to originally present is:

Answers (1)

 PV = nRT , T1 = 300k , T2 = 400k

Since, it is an open vessel, pressure is constant and there is no change in the volume of vessel.

\therefore   P1 = P2  and V1 = V2

\frac{n_{1}}{n_{2}} = \frac{T_{2}}{T_{1}} = \frac{400}{300} = \frac{4}{3}

n_{2} = \frac{3}{4}n_{1}

Therefore, fraction of air that goes out = 1 - \frac{3}{4} = \frac{1}{4}

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