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In Boyle's experiment for a given gas at different temperatures, the graph drawn between pressure and density are straight lines as shown, then

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T₁³ = T₂

T₁ > T₂

T₁ = T₂

T₂ > T₁

answer is A.

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Detailed Solution

We know, PV = nRT

$\Rightarrow PV = (nM) \frac{RT}{M} = m \frac{RT}{M}$

where M = molar mass and m = Mass of gas.

$∴ P = (\frac{m}{V}) \frac{RT}{M} = d (\frac{RT}{M})$

where RT/M = Slope of P-d graph.

$∴ \frac{{RT}_{2}}{M} < \frac{{RT}_{1}}{M}$

$\Rightarrow T_{1} > T_{2}$

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