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A gas has a volume of $3 \cdot 4 \mathrm{~L}$ at $25^{\circ} \mathrm{C}$. What is the final temperature if the volume increases to $10 \cdot 2 \mathrm{~L}$ at constant pressure.
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The correct answer is:
$894 \mathrm{~K}$
$\begin{array}{ll}
V_{1}=3.4 \mathrm{~L}, & T_{1}=25^{\circ} \mathrm{C}=25+273=298 \mathrm{~K} \\
V_{2}=10.2 \mathrm{~L}, & T_{2}=?
\end{array}$
According to Charle's Law,
$\begin{array}{l}
\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}} \text { (at constant } \mathrm{P} \text { and } \mathrm{n} \text { ) } \\
\therefore \mathrm{T}_{2}=\frac{\mathrm{V}_{2} \times \mathrm{T}_{1}}{\mathrm{~V}_{1}}=\frac{10.2 \mathrm{~L} \times 298 \mathrm{~K}}{3.4 \mathrm{~L}} \quad \therefore \mathrm{T}_{2}=894 \mathrm{~K}
\end{array}$
V_{1}=3.4 \mathrm{~L}, & T_{1}=25^{\circ} \mathrm{C}=25+273=298 \mathrm{~K} \\
V_{2}=10.2 \mathrm{~L}, & T_{2}=?
\end{array}$
According to Charle's Law,
$\begin{array}{l}
\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}} \text { (at constant } \mathrm{P} \text { and } \mathrm{n} \text { ) } \\
\therefore \mathrm{T}_{2}=\frac{\mathrm{V}_{2} \times \mathrm{T}_{1}}{\mathrm{~V}_{1}}=\frac{10.2 \mathrm{~L} \times 298 \mathrm{~K}}{3.4 \mathrm{~L}} \quad \therefore \mathrm{T}_{2}=894 \mathrm{~K}
\end{array}$
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