The pressure versus temperature graph of an i | vedclass

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(A) According to the ideal gas law,$PV = nRT$,which can be rearranged as $P = (nR/V)T$.For a constant volume process,the pressure-temperature graph is

Vedclass · Accepted Answer

$V_1 = V_2, V_3 = V_4$ and $V_2 > V_3$

Vedclass · Answer

(A) According to the ideal gas law,$PV = nRT$,which can be rearranged as $P = (nR/V)T$.For a constant volume process,the pressure-temperature graph is a straight line passing through the origin with a slope equal to $nR/V$.In the given graph,lines $1-2$ and $3-4$ represent processes at constant volumes. Since lines $1-2$ and $3-4$ are segments of lines passing through the origin,we have $V_1 = V_2$ and $V_3 = V_4$.The slope of the line is $m = nR/V$. Since the slope is inversely proportional to volume $(m \propto 1/V)$,a smaller slope corresponds to a larger volume.The slope of line $1-2$ is less than the slope of line $3-4$. Therefore,the volume $V_2$ must be greater than the volume $V_3$ $(V_2 > V_3)$.Thus,the correct relation is $V_1 = V_2, V_3 = V_4$ and $V_2 > V_3$.

The pressure versus temperature graph of an ideal gas with an equal number of moles at different volumes is plotted as shown in the figure. Choose the correct alternative.

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