The given figure represents two isobaric proc | vedclass

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(D) For an ideal gas,the equation of state is $PV = nRT$.Rearranging for $V$ as a function of $T$,we get $V = \left(\frac{nR}{P}\right)T$.Comparing th

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$P_1 > P_2$

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(D) For an ideal gas,the equation of state is $PV = nRT$.Rearranging for $V$ as a function of $T$,we get $V = \left(\frac{nR}{P}ight)T$.Comparing this with the equation of a straight line $y = mx$,where $y = V$ and $x = T$,the slope of the $V-T$ graph is given by $	ext{Slope} = \frac{nR}{P}$.Since $n$ and $R$ are constants,the slope is inversely proportional to the pressure,i.e.,$	ext{Slope} \propto \frac{1}{P}$.From the given figure,the slope of the line corresponding to $P_2$ is greater than the slope of the line corresponding to $P_1$,i.e.,$(	ext{Slope})_2 > (	ext{Slope})_1$.Since $	ext{Slope} \propto \frac{1}{P}$,a higher slope implies a lower pressure. Therefore,$P_2  P_2$.

The given figure represents two isobaric processes for the same mass of an ideal gas. Then:

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