A helium-filled balloon has a volume of 2.75 L at $20^{\circ}$ C. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in $^{\circ}$ C?

${12.9}^{\circ} C$

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${10.9}^{\circ} C$

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${20.9}^{\circ} C$

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${8.9}^{\circ} C$

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Solution

The correct option is B
${10.9}^{\circ} C$

According to Charles' Law,

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

where $V_{1} = 2.75 L : T_{1} = 20^{\circ} C = (273 + 20) K = 293 K : V_{2} = 2.46 L$

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

$\Rightarrow T_{2} = \frac{2.46 \times 293}{2.75} = 262.1$

Hence, the final temperature $= 262.1 K = (262.1 - 273)^{\circ} C = {10.9}^{\circ} C$

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A helium-filled balloon has a volume of 2.75 L at 20∘C. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in ∘ C?

The correct option is B 10.9∘C According to Charles' Law, V1T1=V2T2 where V1=2.75 L :T1=20∘C=(273+20)K=293 K : V2=2.46 L ⇒T2=V2T1V1 ⇒T2=2.46×2932.75=262.1 Hence, the final temperature =262.1K=(262.1−273)∘C=10.9∘C

A helium-filled balloon has a volume of 2.75 L at $20^{\circ}$ C. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in $^{\circ}$ C?