Q. The temperature of a gas contained in a closed vessel of constant volume increases by 1$℃$ when the pressure of the gas is increased by 1%. The initial temperature of the gas is
A) 100 K
B) 273${^\circ}$C
C) 100${^\circ}$C
D) 200 K
Explanation
(a) According to Gay Lussac’s law $p \propto T$
$\therefore\ \ \ \ \ \ \ \ \ \frac{dp}{p} \times 100 = \frac{dT}{T} \times 100$
$1 = \ \frac{1}{T} \times 100$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = 100$ K
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.