Search any question & find its solution
Question:
Answered & Verified by Expert
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 $10^{\circ} \mathrm{C}$.
Options:
Solution:
1264 Upvotes
Verified Answer
The correct answer is:
$4.14 \mathrm{dm}^3$
According to Charles' law,
$$
\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2} \text { (at constant } \mathrm{P} \text { and } \mathrm{n} \text { ) } \\
& \mathrm{T}_1=0{ }^{\circ} \mathrm{C}=273 \mathrm{~K} \\
& \mathrm{~T}_2=10^{\circ} \mathrm{C}=283 \mathrm{~K} \\
& \frac{4 \mathrm{dm}^3}{273 \mathrm{~K}}=\frac{\mathrm{V}_2}{283 \mathrm{~K}} \\
& \mathrm{~V}_2=\frac{4 \times 283}{273}=4.1465 \mathrm{dm}^3
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2} \text { (at constant } \mathrm{P} \text { and } \mathrm{n} \text { ) } \\
& \mathrm{T}_1=0{ }^{\circ} \mathrm{C}=273 \mathrm{~K} \\
& \mathrm{~T}_2=10^{\circ} \mathrm{C}=283 \mathrm{~K} \\
& \frac{4 \mathrm{dm}^3}{273 \mathrm{~K}}=\frac{\mathrm{V}_2}{283 \mathrm{~K}} \\
& \mathrm{~V}_2=\frac{4 \times 283}{273}=4.1465 \mathrm{dm}^3
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.