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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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A student while performing an experiment forgot to add the reaction mixture and started heating the open round bottom flask on the flame. If he found the temperature to be $227 ^{\circ} C$ after heating, which was $27 ^{\circ} C$ earlier, what fraction of air would have been expelled out?

(A) $\large\frac{2}{5}$ (B) $\large\frac{4}{5}$ (C) $\large\frac{3}{5}$ (D) $\large\frac{1}{5}$
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For an open vessel containing gas inside, pressure and volume corresponding to the gas remains the same.
So $nT = $ constant for the gas inside.
Therefore $n_1 T_1 = n_2 T_2$
$\Rightarrow n_1 (273+27) = n_2 (227+23) \rightarrow 300 n_1 = 500 n_2 \rightarrow n_2 = \large\frac{3}{5} $$n_1$
So, 60% of the original amount of air is still there. Therefore the fraction of air expelled is two-fifths, $\large\frac{2}{5}$.
answered Mar 11, 2014 by balaji.thirumalai

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