The density of a gas at $27^{\circ} C$ and 760 $m m$ pressure is 24 . Calculate the temperature (in $K$ ) at which it will be 18, the pressure remaining constant. (as nearest integer)

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Solution

Given
$d_{1} = 24$
$T_{1} = 27^{\circ} C$ or $300.15 K$
$P_{1} = 760 m m$ or 1 $a t m$
$d_{2} = 18$
$T_{2} = ?$
$P_{2} = 760 m m$ or 1 $a t m$
$W e k n o w d e n s i t y = \frac{P M}{R T}$
$∴ \frac{d_{1}}{d_{2}} = \frac{P_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} = \frac{T_{2}}{T_{1}} (∵ P_{1} = P_{2} = 1 a t m)$
$T_{2} = \frac{d_{1}}{d_{2}} \times T_{1}$
$T_{2} = \frac{24}{18} \times 300.15 = 400.2 K$
so nearest integer value is 400.

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