MHT CET · Chemistry · States of Matter
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.
- A \(1894 \mathrm{~K}\)
- B \(694 \mathrm{~K}\)
- C \(894 \mathrm{~K}\)
- D \(394 \mathrm{~K}\)
Answer & Solution
Correct Answer
(C) \(894 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
V_{1}=3.4 \mathrm{~L}, & T_{1}=25^{\circ} \mathrm{C}=25+273=298 \mathrm{~K} \\
V_{2}=10.2 \mathrm{~L}, & T_{2}=?
\end{array}\)
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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