For one mole of a van der Waals gas when b=0 | vedclass

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(C) The van der Waals equation for $1 \ 	ext{mole}$ of gas is $(P + \frac{a}{V^2})(V - b) = RT$.Given $b = 0$,the equation simplifies to $(P + \frac{

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$1.5$

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(C) The van der Waals equation for $1 \ 	ext{mole}$ of gas is $(P + \frac{a}{V^2})(V - b) = RT$.Given $b = 0$,the equation simplifies to $(P + \frac{a}{V^2})V = RT$.Expanding this,we get $PV + \frac{a}{V} = RT$,which can be rearranged as $PV = RT - a(\frac{1}{V})$.Comparing this with the equation of a straight line $y = mx + c$,where $y = PV$,$x = \frac{1}{V}$,$m = -a$,and $c = RT$.The slope of the line is $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20.1 - 21.6}{3.0 - 2.0} = \frac{-1.5}{1.0} = -1.5$.Since the slope $m = -a$,we have $-a = -1.5$,which gives $a = 1.5 \ 	ext{atm} \cdot 	ext{liter}^2 \cdot 	ext{mol}^{-2}$.

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

For one mole of a van der Waals gas when $b=0$ and $T=300 \ K$,the $PV$ vs. $1/V$ plot is shown below. The value of the van der Waals constant $a$ (in $\text{atm} \cdot \text{liter}^2 \cdot \text{mol}^{-2}$) is:

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