At $300 \ K$ temperature,$20 \ g \ H_2$,$220 \ g \ CO_2$,and $140 \ g \ N_2$ are filled in a vessel having volume $2 \ L$. Find the total pressure in $bar$ unit and which gas is removed from the vessel so that pressure can be reduced by $50 \%$. $(R = 8.34 \times 10^{-2} \ L \ bar \ K^{-1} \ mol^{-1})$

(A) $1$. Calculate moles of each gas: $n(H_2) = 20/2 = 10 \ mol$,$n(CO_2) = 220/44 = 5 \ mol$,$n(N_2) = 140/28 = 5 \ mol$.
$2$. Total moles $n_{\text{total}} = 10 + 5 + 5 = 20 \ mol$.
$3$. Using ideal gas equation $PV = nRT$: $P = (nRT)/V = (20 \times 8.34 \times 10^{-2} \times 300) / 2 = 249.42 \ bar$.
$4$. To reduce pressure by $50 \%$,total moles must be reduced to $10 \ mol$.
$5$. Since $n(H_2) = 10 \ mol$,removing all $H_2$ gas will reduce the total moles to $5 + 5 = 10 \ mol$,thereby reducing the pressure by $50 \%$.