Q. Two thermally insulated vessels 1 and 2 are filled with air at temperatures $(T_{1},T_{2})$ volume $(V_{1},V_{2})$ and pressure $(P_{1},P_{2})$ respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be
A) $T_{1} + T_{2}$
B) $(T_{1} + T_{2})/2$
C) $\frac{T_{1}T_{2}(P_{1}V_{1} + P_{2}V_{2})}{P_{1}V_{1}T_{2} + P_{2}V_{2}T_{1}}$
D) $\frac{T_{1}T_{2}(P_{1}V_{1} + P_{2}V_{2})}{P_{1}V_{1}T_{1} + P_{2}V_{2}T_{2}}$
Explanation
(c) The number of moles of the system remains same,
$\frac{P_{1}V_{1}}{RT_{1}} + \frac{P_{2}V_{2}}{RT_{2}} = \frac{P(V_{1} + V_{2})}{RT} \Rightarrow T = \frac{P\left( V_{1} + V_{2} \right)T_{1}T_{2}}{(P_{1}V_{1}T_{2} + P_{2}V_{2}T_{1})}$
According to Boyle’s law,
$P_{1}V_{1} + P_{2}V_{2} = P\left( V_{1} + V_{2} \right)\ \therefore T = \frac{\left( P_{1}V_{1} + P_{2}V_{2} \right)T_{1}T_{2}}{(P_{1}V_{1}T_{2} + P_{2}V_{2}T_{1})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.