Consider the following table: Gas a / | vedclass

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(D) The constant $a$ represents the magnitude of intermolecular forces of attraction. $A$ higher value of $a$ implies stronger attraction,which pulls

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Gas $C$ will occupy more volume than gas $A$; gas $B$ will be more compressible than gas $D$.

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(D) The constant $a$ represents the magnitude of intermolecular forces of attraction. $A$ higher value of $a$ implies stronger attraction,which pulls molecules closer,resulting in a smaller volume occupied.For gases $A$ and $C$,$b$ is the same $(0.05196)$. Since $a_A (642.32) > a_C (431.91)$,gas $A$ has stronger intermolecular forces and occupies less volume than gas $C$. Thus,gas $C$ occupies more volume than gas $A$.For gases $B$ and $D$,$a$ is the same $(155.21)$. The constant $b$ represents the excluded volume per mole. $A$ smaller value of $b$ means the molecules occupy less space,making the gas more compressible.Since $b_B (0.04136) < b_D (0.4382)$,gas $B$ is more compressible than gas $D$.

Gas	$a / (kPa \cdot dm^6 \cdot mol^{-2})$	$b / (dm^3 \cdot mol^{-1})$
$A$	$642.32$	$0.05196$
$B$	$155.21$	$0.04136$
$C$	$431.91$	$0.05196$
$D$	$155.21$	$0.4382$

Consider the following table:

Gas $a / (kPa \cdot dm^6 \cdot mol^{-2})$ $b / (dm^3 \cdot mol^{-1})$

$A$ $642.32$ $0.05196$

$B$ $155.21$ $0.04136$

$C$ $431.91$ $0.05196$

$D$ $155.21$ $0.4382$

$a$ and $b$ are van der Waals constants. The correct statement about the gases is:

Similar Questions

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Consider the following table: Gas $a / (kPa \cdot dm^6 \cdot mol^{-2})$ $b / (dm^3 \cdot mol^{-1})$ $A$ $642.32$ $0.05196$ $B$ $155.21$ $0.04136$ $C$ $431.91$ $0.05196$ $D$ $155.21$ $0.4382$ $a$ and $b$ are van der Waals constants. The correct statement about the gases is: