A gas at $65^{\circ} C$ occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L?

${39.96}^{\circ} C$

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${309.96}^{\circ} C$

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${36.96}^{\circ} C$

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${319.96}^{\circ} C$

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Solution

The correct option is C
${36.96}^{\circ} C$

According to Charles' Law,

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

where $V_{1} = 4.22 L : T_{1} = 65^{\circ} C = (273 + 65) K = 338 K : V_{2} = 3.87 L$

$\Rightarrow T_{2} = \frac{V_{2} T_{1}}{V_{1}}$

$\Rightarrow T_{2} = \frac{3.87 \times 338}{4.22} = 309.96$

Hence, the final temperature $= 309.96 K = (309.96 - 273)^{\circ} C = {36.96}^{\circ} C$

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A gas at 65∘C occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L?

The correct option is C 36.96∘C According to Charles' Law, V1T1=V2T2 where V1=4.22 L :T1=65∘C=(273+65)K=338 K : V2=3.87 L ⇒T2=V2T1V1 ⇒T2=3.87×3384.22=309.96 Hence, the final temperature =309.96K=(309.96−273)∘C=36.96∘C

A gas at $65^{\circ} C$ occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L?