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