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Ideal Gas Equation

500 mL of nit...

Question

500 mL of nitrogen at $27^{o} C$ are cooled to - $5^{o} C$ at the same pressure. Calculate the new volume.

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Solution

Given,
$V_{1} = 500 m L$
$T_{1} = 27^{0} C = 300 K$
$T_{2} = - 5^{0} C = 268 K$
Now using Charle's Law,
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$⟹ \frac{500}{300} = \frac{V_{2}}{268}$
$⟹ V_{2} = 446.66 m L$

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