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

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(C) The van der Waals equation for $1 \ mol$ of gas is: $(P + a/V^2)(V - b) = RT$.Given $b = 0$,the equation simplifies to: $(P + a/V^2)V = RT$.Expand

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$1.5$

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

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

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