Q. Two spherical vessel of equal volume, are connected by a $a$ narrow tube. The apparatus contains an ideal gas at one atmosphere and $300K.$ Now if one vessel is immersed in a bath of constant temperature $600K$ and the other in a bath of constant temperature $300K.$ Then the common pressure will be
A) $1\ atm$
B) $\frac{4}{5}atm$
C) $\frac{4}{3}atm$
D) $\frac{3}{4}atm$
Explanation
(c) $\mu = \mu_{1} + \mu_{2}$
$\frac{P(2V)}{RT_{1}} = \frac{P^{'}V}{RT_{1}} + \frac{P^{'}V}{RT_{2}} \Rightarrow \frac{2P}{RT_{1}} = \frac{P^{'}}{R}\left\lbrack \frac{T_{2} + T_{1}}{T_{1}T_{2}} \right\rbrack$
$P^{'} = \frac{2PT_{2}}{(T_{1} + T_{2})} = \frac{2 \times 1 \times 600}{(300 + 600)} = \frac{4}{3}atm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.