A perfect gas at $27^{\circ} \mathrm{C}$ is heated at constant pressure so as to double its volume. The…
A perfect gas at $27^{\circ} \mathrm{C}$ is heated at constant pressure so as to double its volume. The final temperature of the gas will be, close to
-
$327^{\circ} \mathrm{C}$
-
$200^{\circ} \mathrm{C}$
-
$54^{\circ} \mathrm{C}$
-
$300^{\circ} \mathrm{C}$
Solution
Given, $V_1=V$
$
\begin{aligned}
& V_2=2 V \\
& T_1=27^{\circ}+273=300 \mathrm{~K} \\
& T_2=?
\end{aligned}
$
From charle's law
$
\frac{V_1}{T_1}=\frac{V_2}{T_2}(\because \text { Pressure is constant })
$
or, $\frac{V}{300}=\frac{2 V}{T_2}$
$
\therefore \quad T_2=600 \mathrm{~K}=600-273=327^{\circ} \mathrm{C}
$
Asked in: JEE Main 2012 (07 May Online)
Practice more Thermodynamics questions on Aicharya