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 :

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 \[{{n}_{1}}+{{n}_{2}}=2n\] \[\frac{p{{V}_{0}}}{2R{{T}_{0}}}+\frac{p{{V}_{0}}}{R{{T}_{0}}}=2\left( \frac{{{p}_{0}}{{V}_{0}}}{R{{T}_{0}}} \right)\] \[\Rightarrow \] \[p=\frac{4}{3}{{p}_{0}}=\frac{4}{3}\text{atm}\]