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Ideal Gas Equation

In Boyles exp...

Question

In Boyles experiment for a given gas at different temperatures the graph drawn between pressure and density are straight lines as shown then:

T_{1} > T_{2}

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T_{2} > T_{1}

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T_{1} = T_{2}

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T_{1}^{3} = T_{2}

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Solution

The correct option is A $T_{1} > T_{2}$
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
so, $P = \frac{d R T}{M}$
so pressure vs density graph will be a straight line with slope $=$ $\frac{R T}{M}$
Hence, steeper the line, more is slope and so more is temperature.
So here $T_{1} > T_{2}$

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Similar questions

The figure shows pressure versus density graph for an ideal gas at two temperature $T_{1}$ and $T_{2}$ :

T_{1}

T_{2}

T_{1}

T_{2}

T_{1}

T_{2}

T_{1}

T_{2}

\frac{P V}{T}

P

1 g

T_{1}

T_{2}

= 1.427 k g m^{- 3}

P V / T

A

T_{1}

T_{2}

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