The pressure of a van der Waals gas is less t | vedclass

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(C) The ideal gas law assumes that there are no intermolecular forces of attraction or repulsion between gas molecules.However,real gases (van der Waa

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Intermolecular attraction

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(C) The ideal gas law assumes that there are no intermolecular forces of attraction or repulsion between gas molecules.However,real gases (van der Waals gases) exhibit intermolecular forces of attraction,known as van der Waals forces.These attractive forces pull the molecules inward,away from the walls of the container.Consequently,the molecules strike the container walls with less force than they would in the absence of these attractions.Therefore,the observed pressure of a real gas is lower than the pressure predicted for an ideal gas.

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

The pressure of a van der Waals gas is less than the pressure of an ideal gas because of

Similar Questions

Consider the van der Waals constants,$a$ and $b,$ for the following gases.

Gas $Ar$ $Ne$ $Kr$ $Xe$

$a \ (atm \ dm^6 \ mol^{-2})$ $1.3$ $0.2$ $5.1$ $4.1$

$b \ (10^{-2} \ dm^3 \ mol^{-1})$ $3.2$ $1.7$ $1.0$ $5.0$

Which gas is expected to have the highest critical temperature?

Dominance of strong repulsive forces among the molecules of the gas ($Z =$ compressibility factor)

The van der Waals equation of state for a real gas is given. For $n$ moles of a real gas,the equation can be expressed as:

Assertion : Compressibility factor for hydrogen varies with pressure with positive slope at all pressures.
Reason : Even at low pressure,repulsive forces dominate hydrogen gas.

$A$ gas described by van der Waal's equation:

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