A graph between volume V and absolute tempera | vedclass

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(B) From the ideal gas equation,$PV = nRT$,we can write $V = (\frac{nR}{P})T$. Comparing this with the equation of a straight line $y = mx$,where $y =

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decreased

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(B) From the ideal gas equation,$PV = nRT$,we can write $V = (\frac{nR}{P})T$. Comparing this with the equation of a straight line $y = mx$,where $y = V$ and $x = T$,the slope $m = \frac{nR}{P}$. Since $n$ and $R$ are constants,the slope is inversely proportional to pressure,i.e.,$	ext{slope} \propto \frac{1}{P}$. In the given $V-T$ graph,the line representing the process from state-$1$ to state-$2$ is steeper than the line passing through the origin to state-$1$. This means the slope increases as we move from state-$1$ to state-$2$. Since $	ext{slope} \propto \frac{1}{P}$,an increase in slope implies a decrease in pressure. Therefore,the pressure decreased.

$A$ graph between volume $V$ and absolute temperature $T$ is obtained when a given mass of a gas is heated. During the heating process from state-$1$ to state-$2$,the pressure

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