Two different masses m and 3m of an ideal gas | vedclass

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(A) From the ideal gas equation, $PV = nRT$, where $n = \frac{m}{M}$ ($M$ is the molar mass).So, $PV = \frac{m}{M} RT$.Rearranging for pressure $P$, w

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$3:1$

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(A) From the ideal gas equation, $PV = nRT$, where $n = \frac{m}{M}$ ($M$ is the molar mass).So, $PV = \frac{m}{M} RT$.Rearranging for pressure $P$, we get $P = \left( \frac{mR}{MV} ight) T$.Comparing this with the equation of a straight line $P = (	ext{slope}) \cdot T$, the slope is given by $S = \frac{mR}{MV}$.Since $R$, $M$, and $V$ are constants for both cases, the slope $S$ is directly proportional to the mass $m$ $(S \propto m)$.For curve $A$, the mass is $m$, so $S_A \propto m$.For curve $B$, the mass is $3m$, so $S_B \propto 3m$.The ratio of the slopes of curves $B$ to $A$ is $\frac{S_B}{S_A} = \frac{3m}{m} = \frac{3}{1}$.Thus, the ratio is $3:1$.

Two different masses $m$ and $3m$ of an ideal gas are heated separately in a vessel of constant volume. The pressure $P$ versus absolute temperature $T$ graphs for these two cases are shown in the figure. The ratio of the slopes of curves $B$ to $A$ is

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