question_answer

Two spherical vessels of equal volume are connected by a narrow tube. The apparatus contains an ideal gas at 1 atm and 300 K. Now, if one vessel is immersed in a bath of constant temperature 600K and other in a bath of constant temperature 300K, then common pressure will be

A) 1 atm

B) 4/5 atm

C) 4/3 atm

D) 3/2 atm

Correct Answer: C

Solution :

[c] Initial pressure, \[{{p}_{1}}={{p}_{2}}={{p}_{0}}=\frac{nR{{T}_{0}}}{{{V}_{0}}}\] when

\[{{p}_{0}}=1\] atm \[{{T}_{0}}=300K.\]

Finally, let common pressure is p.

So, \[\frac{p{{V}_{0}}}{R\times 2{{T}_{0}}}={{n}_{1}}\] and  \[\frac{p{{V}_{0}}}{R{{T}_{0}}}={{n}_{2}}\] and \[{{n}_{1}}+{{n}_{2}}=2n\]

\[\frac{p{{V}_{0}}}{2R{{T}_{0}}}+\frac{p{{V}_{0}}}{R{{T}_{0}}}=2\left( \frac{{{p}_{0}}{{V}_{0}}}{R{{T}_{0}}} \right)\]

\[\Rightarrow \]            \[p=\frac{4}{3}{{p}_{0}}=\frac{4}{3}atm\]