A gas at 350 ; K and 15 ; bar has a molar vol | vedclass

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(C) The compressibility factor $Z$ is defined as the ratio of the molar volume of a real gas $(V_{m})_{real}$ to the molar volume of an ideal gas $(V_

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$Z < 1$ and attractive forces are dominant

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(C) The compressibility factor $Z$ is defined as the ratio of the molar volume of a real gas $(V_{m})_{real}$ to the molar volume of an ideal gas $(V_{m})_{ideal}$ at the same temperature and pressure: $Z = \frac{(V_{m})_{real}}{(V_{m})_{ideal}}$.Given that $(V_{m})_{real}$ is $20$ percent smaller than $(V_{m})_{ideal}$,we have $(V_{m})_{real} = 0.8 	imes (V_{m})_{ideal}$.Therefore,$Z = \frac{0.8 	imes (V_{m})_{ideal}}{(V_{m})_{ideal}} = 0.8$.Since $Z < 1$,the gas shows negative deviation from ideal behavior,which indicates that attractive intermolecular forces are dominant.

Gas	$Ar$	$Ne$	$Kr$	$Xe$
$a \ (atm \ dm^6 \ mol^{-2})$	$1.3$	$0.2$	$5.1$	$4.1$
$b \ (10^{-2} \ dm^3 \ mol^{-1})$	$3.2$	$1.7$	$1.0$	$5.0$

$A$ gas at $350 \; K$ and $15 \; bar$ has a molar volume $20$ percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor $(Z)$ is

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