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4 litre $C_2H_4(g)$ burns in oxygen at $27^{\large\circ}C$ and 1atm to produce $CO_2(g)$and $H_2O(g)$. Calculate the volume of $CO_2$ formed at $27^{\large\circ}C$ and 4 atm

$(a)\;2litre\qquad(b)\;3 litre\qquad(c)\;5 litre\qquad(d)\;7 litre$

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$C_2H_4(g)+3O_2(g) \rightarrow2CO_2(g)+2H_2O(g)$
Since 1 mole or volume of $C_2H_4$ gives = 2 volume of $CO_2$
Since Volume of $C_2H_4$ gives = $2\times4=8\;litre\;of\;CO_2$
Now at $27^{\large\circ}C$ and 4 atm
$\large\frac{P_1V_1}{T_1} = \large\frac{P_2V_2}{T_2}$
$\Rightarrow\large\frac{1\times8}{300} = \large\frac{4\times V_2}{300}$
$\Rightarrow\large\frac{1\times8}{4} = V_2$
$\Rightarrow V_2 = 2 \;litre$
Hence answer is (a)
answered Feb 15, 2014 by sharmaaparna1

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