The pressure p of a gas is plotted against it | vedclass

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

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Curve for $V_2$ has a greater slope than that for $V_1$

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(C) From the ideal gas equation,$PV = nRT$,we get $P = (\frac{nR}{V})T$. Comparing this with the equation of a straight line $y = mx$,where $y = P$ and $x = T$,the slope $m = \frac{nR}{V}$. Since $n$ and $R$ are constants,the slope is inversely proportional to the volume,i.e.,$m \propto \frac{1}{V}$. Given $V_1 > V_2$,it follows that the slope for $V_2$ $(m_2)$ must be greater than the slope for $V_1$ $(m_1)$,i.e.,$m_2 > m_1$. Therefore,the curve for $V_2$ has a greater slope than that for $V_1$.

The pressure $p$ of a gas is plotted against its absolute temperature $T$ for two different constant volumes,$V_1$ and $V_2$. When $V_1 > V_2$,the

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