MHT CET · Chemistry · States of Matter
The volume of a gas is \(4 \mathrm{dm}^3\) at \(0^{\circ} \mathrm{C}\). Calculate new volume at constant pressure when the temperature is increased by \(10^{\circ} \mathrm{C}\).
- A \(2.07 \mathrm{dm}^3\)
- B \(3.21 \mathrm{dm}^3\)
- C \(4.14 \mathrm{dm}^3\)
- D \(6.54 \mathrm{dm}^3\)
Answer & Solution
Correct Answer
(C) \(4.14 \mathrm{dm}^3\)
Step-by-step Solution
Detailed explanation
According to Charles' law,
\(
\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2} \text { (at constant } \mathrm{P} \text { and } \mathrm{n} \text { ) } \\
& \mathrm{T}_1=0{ }^{\circ} \mathrm{C}=273 \mathrm{~K} \\
& \mathrm{~T}_2=10^{\circ} \mathrm{C}=283 \mathrm{~K} \\
& \frac{4 \mathrm{dm}^3}{273 \mathrm{~K}}=\frac{\mathrm{V}_2}{283 \mathrm{~K}} \\
& \mathrm{~V}_2=\frac{4 \times 283}{273}=4.1465 \mathrm{dm}^3
\end{aligned}
\)
\(
\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2} \text { (at constant } \mathrm{P} \text { and } \mathrm{n} \text { ) } \\
& \mathrm{T}_1=0{ }^{\circ} \mathrm{C}=273 \mathrm{~K} \\
& \mathrm{~T}_2=10^{\circ} \mathrm{C}=283 \mathrm{~K} \\
& \frac{4 \mathrm{dm}^3}{273 \mathrm{~K}}=\frac{\mathrm{V}_2}{283 \mathrm{~K}} \\
& \mathrm{~V}_2=\frac{4 \times 283}{273}=4.1465 \mathrm{dm}^3
\end{aligned}
\)
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