Search any question & find its solution
Question:
Answered & Verified by Expert
A perfect gas at $27^{\circ} \mathrm{C}$ is heated at constant pressure so as to double its volume. The increase in temperature of the gas will be
Options:
Solution:
1691 Upvotes
Verified Answer
The correct answer is:
$300^{\circ} \mathrm{C}$
We know for perfect gas
Here, $\quad \frac{V}{V_{2}}=\frac{T_{1}}{T_{2}}$
According to question,
$V_{1}=V$ then $V_{2}=2 V$ and $T_{1}=300 \mathrm{~K}$
$$
\therefore \quad \begin{aligned}
\frac{1}{2} &=\frac{300}{T_{2}} \\
T_{2} &=600 \mathrm{~K} \\
T_{2} &=327^{\circ} \mathrm{C}
\end{aligned}
$$
So, $\Delta t=327-27=300^{\circ} \mathrm{C}$
Here, $\quad \frac{V}{V_{2}}=\frac{T_{1}}{T_{2}}$
According to question,
$V_{1}=V$ then $V_{2}=2 V$ and $T_{1}=300 \mathrm{~K}$
$$
\therefore \quad \begin{aligned}
\frac{1}{2} &=\frac{300}{T_{2}} \\
T_{2} &=600 \mathrm{~K} \\
T_{2} &=327^{\circ} \mathrm{C}
\end{aligned}
$$
So, $\Delta t=327-27=300^{\circ} \mathrm{C}$
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.