Q. When one mole of monoatomic ideal gas at $T$ temperature undergoes adiabatic change under a constant external pressure of 1 atm change in volume is from 1 L to 2 L, the final temperature in Kelvin would be
A) $\frac{T}{2^{(2/3)}}$
B) $T + \frac{2}{3 \times 0.0821}$
C) $T$
D) $T - \frac{2}{3 \times 0.0821}$
Explanation
(a)
$\frac{T}{T_{final}} = \left( \frac{V_{2}}{V_{1}} \right)^{\gamma - 1}$
$\frac{T}{T_{final}} = \left( \frac{2}{1} \right)^{(5/3 - 1)} = 2^{(2/3)}$
$T_{final} = \frac{T}{2^{(2/3)}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.