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A balloon filled with an air sample occupies $3 \mathrm{~L}$ volume at $35^{\circ} \mathrm{C}$. On lowering the temperature to $T$, the volume decreases to $2.5 \mathrm{~L}$. The temperature $T$ is?
[Assume $p$-constant]
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[Assume $p$-constant]
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2116 Upvotes
Verified Answer
The correct answer is:
$-16^{\circ} \mathrm{C}$
$$
\begin{array}{rlrl}
V_1 & =3 \mathrm{~L} & & T_1=273+35=308 \mathrm{~K} \\
V_2 & =2.5 \mathrm{~L} & & T_2=? \\
\frac{V_1}{T_1} & =\frac{V_2}{T_2} & & (\text { Charle's law }) \\
\frac{3}{308} & =\frac{2.5}{T_2} & & \\
T_2 & =\frac{2.5 \times 308}{3} & & \\
& =256.6 \mathrm{~K} \Rightarrow 256.6-273=-16.3^{\circ} \mathrm{C}
\end{array}
$$
At $-16^{\circ} \mathrm{C}$, the volume decreases to $2.5 \mathrm{~L}$.
\begin{array}{rlrl}
V_1 & =3 \mathrm{~L} & & T_1=273+35=308 \mathrm{~K} \\
V_2 & =2.5 \mathrm{~L} & & T_2=? \\
\frac{V_1}{T_1} & =\frac{V_2}{T_2} & & (\text { Charle's law }) \\
\frac{3}{308} & =\frac{2.5}{T_2} & & \\
T_2 & =\frac{2.5 \times 308}{3} & & \\
& =256.6 \mathrm{~K} \Rightarrow 256.6-273=-16.3^{\circ} \mathrm{C}
\end{array}
$$
At $-16^{\circ} \mathrm{C}$, the volume decreases to $2.5 \mathrm{~L}$.
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