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  4. 500 mL of nitrogen at 27° C is cooled to -5° C at the same pressure. The new volume becomes

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Q. $500\, mL$ of nitrogen at $27^{\circ} C$ is cooled to $-5^{\circ} C$ at the same pressure. The new volume becomes

AIPMTAIPMT 1995States of Matter

Solution:

Initial volume, $V_1 = 500\,mL$
Initial temperature,
$T_1 = 27^\circ C = 27 + 273 = 300\, K$
Final temperature,
$T_2 = - 5 + 273 = 268\, K$
$V_2=?$
$\frac{V_1}{T_1}=\frac{V_2}{T_2}$
$V_2=\frac{V_1 T_2}{T_1}=\frac{500\times268}{300}$
$=446.66\,mL$