Q. A diatomic gas initially at $18℃$ is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be
A) $10℃$
B) $887℃$
C) $668\ K$
D) $144℃$
Explanation
(c) For an adiabatic process,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ TV^{\gamma - 1} = constant$
$\Rightarrow \ \ \ \ \ \ \ T_{2} = T_{1}\left\lbrack \frac{V_{1}}{V_{2}} \right\rbrack^{\gamma - 1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (273 + 18)\left\lbrack \frac{V}{V/8} \right\rbrack^{0.4} = 668\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.