The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for two different gases. If the same amount of both the gases has been taken, then the relationship between their molecular weights will be

M_{2} < M_{1}

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M_{1} < M_{2}

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M_{1} = M_{2}

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M_{1}^{3} = M_{2}

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Solution

The correct option is A $M_{2} < M_{1}$
From ideal gas equation, $P V = n R T$

And, $n = \frac{w}{M}$ where M is the molecular weight of the gas

Now, $P V = \frac{w}{M} R T$

So, we draw a vertical line at a volume say $V_{0}$ .

Then for this volume, we see that pressure is higher for the gas with molecular weight $M_{2}$ (from the graph) hence, this means that $M_{2}$ must be smaller than $M_{1}$ .

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