To what temperature must a gas at 300 K be cooled in order to reduce its volume to 1/3rd of its original volume, pressure remaining constant.

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Solution

Pressure is constant means Charles' law can be applied.

$Let, final temperature is TK (T_{2}), initial temperature is 300 K (T_{1}); initial volume is V (V_{1}) and final volume is \frac{V}{3} (V_{2}) Then from Charles' law we have, \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} or, \frac{V}{300} = \frac{V}{3} \times \frac{1}{T} or, T = \frac{300}{3} = 100 K ∴ Final temperature is 100 K .$

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To what temperature must a gas at 300 K be cooled in order to reduce its volume to 1/3rd of its original volume, pressure remaining constant.

Pressure is constant means Charles' law can be applied. Let, final temperature is TK (T2), initial temperature is 300 K(T1); initial volume is V(V1) and final volume isV3 (V2) Then from Charles' law we have,V1T1=V2T2 or, V300=V3×1T or, T=3003=100 K∴Final temperature is 100 K.