The compressibility factor (Z) of a gas at mo | vedclass

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(A) At moderate pressure,the volume $V$ is small enough that the term $\frac{a}{V^2}$ cannot be neglected,but the volume correction '$b$' can be negle

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Increases with an increase in temperature.

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(A) At moderate pressure,the volume $V$ is small enough that the term $\frac{a}{V^2}$ cannot be neglected,but the volume correction '$b$' can be neglected compared to $V$.The van der Waals equation becomes: $(P + \frac{a}{V^2})V = RT$.Expanding this: $PV + \frac{a}{V} = RT$,which simplifies to $PV = RT - \frac{a}{V}$.Dividing by $RT$: $\frac{PV}{RT} = 1 - \frac{a}{RTV}$,so $Z = 1 - \frac{a}{RTV}$.Since $Z = 1 - \frac{a}{RTV}$,as temperature $T$ increases,the term $\frac{a}{RTV}$ decreases,causing $Z$ to increase towards $1$.

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

The compressibility factor $(Z)$ of a gas at moderate pressure 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:

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