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, $\large\frac{2}{5}$.