The figure shows the volume versus temperatur | vedclass

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(A) For an ideal gas,the equation of state is $PV = nRT$.Rearranging this for volume $V$ as a function of temperature $T$,we get $V = (nR/P)T$.This re

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$P_1 > P_2$

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(A) For an ideal gas,the equation of state is $PV = nRT$.Rearranging this for volume $V$ as a function of temperature $T$,we get $V = (nR/P)T$.This represents a straight line passing through the origin with a slope $m = nR/P$.Since the mass of the gas is the same,$n$ is constant. Therefore,the slope $m$ is inversely proportional to the pressure $P$,i.e.,$m \propto 1/P$.From the given graph,the slope of the line corresponding to $P_2$ is greater than the slope of the line corresponding to $P_1$ (i.e.,$m_2 > m_1$).Since $m \propto 1/P$,a larger slope implies a smaller pressure.Therefore,$P_2  P_2$.

The figure shows the volume versus temperature graph for the same mass of an ideal gas corresponding to two different pressures $P_1$ and $P_2$. Then,

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