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Ideal Gas Equation

Question

In Boyles law experiment the graph drawn between pressure and density for a given temperature of different gases are the straight lines then the molecular weights have the relation

$M_{1} > M_{2}$
$M_{1} = M_{2}$
$M_{2} > M_{1}$
$M_{2}^{2} = M_{1}$

A

M_{1} > M_{2}

B

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

C

M_{2} > M_{1}

D

M_{1} = M_{2}

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Solution

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First we derive an expression for density of a gas.
Density = $\frac{m a s s (w)}{v o l u m e (V)}$
but we know that PV = nRT
now n = $\frac{w}{M}$ where M is molecular weight
So, PV = $\frac{w}{M} R T$
this gives $\frac{w}{V} = \frac{P M}{R T}$ = density = d
clearly graph of pressure vs density is a straight line with slope = $\frac{R T}{M}$
steeper the line, more is the slope and so less is the mass (at a given temperature)
so here $M_{1} > M_{2}$

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