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Standard XII

Physics

Ideal Gas Equation

The graph dra...

Question

The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for different masses of same gas at same temperature then

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 $m_{1} > m_{2}$
From ideal gas equation, $P V = n R T$
and n = $\frac{w}{M}$ where M is the molecular weight of the gas
Now PV = $\frac{w}{M}$ RT
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 as $M_{2}$ (from the graph) hence, this means that $M_{2}$ must be smaller than $M_{1}$

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