Q. The change in internal energy, when a gas is cooled from $927℃$ to $27℃$
A) 300%
B) 400%
C) 200%
D) 100%
Explanation
(a) Here, $T_{1} = 927℃ = (927 + 273)K = 1200K$
$T_{2} = 27℃ = (27 + 273)K = 300K$
As $U \propto T$
$\therefore\ \frac{\mathrm{\Delta}U}{U_{2}} = \frac{U_{1} - U_{2}}{U_{2}} = \frac{1200 - 300}{300} \times 100 = 300\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.