Q. The pressure inside a tyre is $4\ atm$ at $27℃$. If the tyre burts suddenly, new temperature will be $(\gamma = 7/5)$
A) $300\ (4)^{7/2}$
B) $300(4)^{2/7}$
C) $300\ (2)^{7/2}$
D) $300\ (4)^{- 2/7}$
Explanation
(d) In adiabatic operation ($eg$, bursing of tyre)
$p_{2}^{(1 - \gamma)}T_{2}^{\gamma} = p_{1}^{(1 - \gamma)}T_{1}^{\gamma}$
$T_{2} = T_{1}\left( \frac{p_{1}}{p_{2}} \right)^{(1 - \gamma)/\gamma}$
$= 300\left( \frac{4}{1} \right)^{\left( \frac{1 - 7/5}{7/5} \right)} = 300(4)^{- 2/7}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.