At 240.55 K,for one mole of an ideal gas,a g | vedclass

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(B) For an ideal gas,the kinetic energy $(KE)$ is given by $KE = \frac{3}{2} PV$. Since the graph of $P$ versus $V^{-1}$ is a straight line passing th

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$3000$

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(B) For an ideal gas,the kinetic energy $(KE)$ is given by $KE = \frac{3}{2} PV$. Since the graph of $P$ versus $V^{-1}$ is a straight line passing through the origin,we have $P = m 	imes V^{-1}$,where $m$ is the slope. Comparing this with the ideal gas equation $PV = RT$,we get $P = RT 	imes V^{-1}$,so the slope $m = RT$. However,the problem defines the slope $m$ in terms of $KE$ as $P = \frac{2 \ KE}{3} 	imes V^{-1}$. Thus,the slope $m = \frac{2 \ KE}{3} = 2000 \ J \ mol^{-1}$. Solving for $KE$: $KE = \frac{2000 	imes 3}{2} = 3000 \ J \ mol^{-1}$.

At $240.55 \ K$,for one mole of an ideal gas,a graph of $P$ (on $y$-axis) and $V^{-1}$ (on $x$-axis) gave a straight line passing through the origin. Its slope $(m)$ is $2000 \ J \ mol^{-1}$. What is the kinetic energy (in $J \ mol^{-1}$) of the ideal gas?

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