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(D) The Van der Waals equation for $1$ mole of a real gas is given by $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$,where $V_m$ is the molar vol

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$\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$

Vedclass · Answer

(D) The Van der Waals equation for $1$ mole of a real gas is given by $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$,where $V_m$ is the molar volume $(V_m = V/n)$.Substituting $V_m = V/n$ into the equation:$\left( P + \frac{a}{(V/n)^2} \right) (V/n - b) = RT$$\left( P + \frac{n^2 a}{V^2} \right) \left( \frac{V - nb}{n} \right) = RT$Multiplying both sides by $n$ gives the expression for $n$ moles:$\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$

$Gas$	$a \ (atm \ L^2 \ mol^{-2})$	$b \ (L \ mol^{-1})$
$W$	$4.0$	$0.027$
$X$	$8.0$	$0.030$
$Y$	$6.0$	$0.032$
$Z$	$12.0$	$0.027$

Van der Waals equation of state is obeyed by real gases. For $n$ moles of a real gas,the expression will be

Similar Questions

Explain real gases showing deviations from ideal gas with correction of pressure and volume and derive van der Waals equation.

Derive the van der Waals equation.

At high temperature and low pressure,the Vander Waal's equation is reduced to

The Van der Waals parameters for gases $W$,$X$,$Y$,and $Z$ are given below. Which one of these gases has the highest critical temperature?

$Gas$ $a \ (atm \ L^2 \ mol^{-2})$ $b \ (L \ mol^{-1})$

$W$ $4.0$ $0.027$

$X$ $8.0$ $0.030$

$Y$ $6.0$ $0.032$

$Z$ $12.0$ $0.027$

Assertion : Compressibility factor $(Z)$ for non-ideal gases can be greater than $1$.
Reason : Non-ideal gases always exert higher pressure than expected.

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