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(B) For an ideal gas,the ideal gas equation is $pV = nRT$. Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $pV = \frac

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(B) For an ideal gas,the ideal gas equation is $pV = nRT$. Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $pV = \frac{m}{M} RT$. Rearranging this gives $p = \left( \frac{RT}{M} \right) \left( \frac{m}{V} \right)$. Since density $d = \frac{m}{V}$,the equation becomes $p = \left( \frac{RT}{M} \right) d$. This is in the form of a straight line $y = mx$,where the slope is $\frac{RT}{M}$. Since the slope is directly proportional to temperature $(T)$,a higher temperature will result in a steeper slope. Given $T_3 > T_2 > T_1$,the slope order will be $T_3 > T_2 > T_1$. Therefore,the plot with the steepest line for $T_3$ and the least steep for $T_1$ is correct.

Which amongst the given plots is the correct plot for pressure $(p)$ $vs$ density $(d)$ for an ideal gas?

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