question_answer 34)

                  The container shown in Fig. has two chambers, separated by a partition, of volumes \[{{V}_{1}}=2.0\] litre and \[{{V}_{1}}=3.0\] litre. The chambers contain \[{{\mu }_{1}}=4.0\] and \[{{\mu }_{2}}=\ 5.0\] moles of a gas at pressures \[{{p}_{1}}=1.00\] atm and \[{{p}_{2}}=2.00\] atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

Answer:

                  We know, \[PV=\frac{2}{3}\] E for 1 mole                 \[\therefore \] \[{{P}_{1}}\,{{V}_{1}}=\,\frac{2}{3}\,{{\mu }_{1}}{{E}_{1}}\]                 and  \[{{P}_{2}}{{V}_{2}}=\frac{2}{3}\,{{\mu }_{2}}{{E}_{2}}\]                 \[\therefore \] Total energy \[=({{P}_{1}}{{V}_{1}}+\text{ }{{P}_{2}}{{V}_{2}})\]                 \[=\frac{2}{3}\,\,({{\mu }_{1}}{{E}_{1}}+\,{{\mu }_{2}}{{E}_{2}})\]                             or \[({{\mu }_{1}}\,{{E}_{1}}+\,{{\mu }_{2}}{{E}_{2}})=\,\frac{3}{2}\,({{P}_{1}}{{V}_{1}}+\,{{P}_{2}}{{V}_{2}})\]            ?.. (i)                 When partition is removed, let pressure                 \[=P\] and volume \[=({{V}_{1}}+{{V}_{2}})\]                 It \[E=\] Total energy per mole                 \[\therefore \] \[P\,({{V}_{1}}\,+{{V}_{2}})=\frac{2}{3}\,\mu E=\,\frac{2}{3}\,({{\mu }_{1}}+\,{{\mu }_{2}})\,E\]                 Using eqn. (i)                 \[P({{V}_{1}}+{{V}_{2}})=\frac{2}{3}\times \frac{3}{2}({{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}})\]                 \[=({{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}})\]                 \[\therefore \] \[P\,=\,\frac{({{P}_{1}}{{V}_{1}}+\,{{P}_{2}}{{V}_{2}})}{{{V}_{1}}+\,{{V}_{2}}}\,\,=\,\left( \frac{1\times \,2+2\times \,3\,}{2+3} \right)\]                 \[=\frac{8}{5}=\,1.6\,\,atm\].