question_answer 33)

                  Consider a \[P-V\] diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig. (a) Find the work done when the gas is taken from state 1 to state 2. (b) What is the ratio of temperature \[{{T}_{1}}/{{T}_{2}},\], if \[{{V}_{2}}=2{{V}_{1}}\]? (b) Given the internal energy for one mole of gas at temperature T is (3/2) RT, find the heat supplied to the gas when it is taken from state 1 to 2, with \[{{V}_{2}}=2{{V}_{1}}\].                

Answer:

                  (a) Work done to change the volume by dV is given by                 \[dW=PdV\]                 Total work done to change the volume from \[{{V}_{1}}\]  to \[{{V}_{2}}\] is given by \[\int{dW\,=\,\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{P\,dV}\,\,\text{or}\,\,W=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{P\,dV}\,\,}\]            ?? (i)                 Since \[P{{V}^{1/2}}=\] Constant,                                            \[\therefore \] \[\text{P = }\,\frac{\text{Constant}\,\text{(K)}}{{{\text{v}}^{\text{1/2}}}}\]                 Hence \[W=\,\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{K\,{{V}^{-1/2}}\,dV=\,K\,\left[ \frac{{{V}^{1/2}}}{1/2} \right]}_{{{V}_{1}}}^{{{V}_{2}}}\]                 \[=2K\,[V_{2}^{1/2}\,-V_{1}^{1/2}]\]                 \[=2\,{{P}_{1}}V_{1}^{1/2}\,[V_{2}^{1/2}-V_{1}^{1/2}]\]                 (b) We know, \[PV=\,\mu \,RT\]                 or \[T=\frac{PV}{\mu R}\]                                     ?.. (i)                 Also \[P{{V}^{1/2}}=K\] (constant)                 or \[P{{V}^{1/2}}=K\] (constant)                 or \[P=K{{V}^{1/2}}\]                 \[\therefore \] \[T=\,\frac{K{{V}^{1/2}}}{\mu R}\] or \[T\alpha \sqrt{V}\]                 Now \[\frac{{{T}_{1}}}{{{T}_{2}}}\,=\sqrt{\frac{{{V}_{1}}}{{{V}_{2}}}}\,=\,\sqrt{\frac{{{V}_{1}}}{2{{V}_{1}}}}\,=\,\sqrt{\frac{1}{2}}\] (a) According to Ist law thermodynamics                 \[dQ=dU+dW\]                        ?.. (i)                 Now \[{{U}_{1}}\,=\,\frac{3}{2}\,RT,\] and \[{{U}_{2}}=\,\frac{3}{2}\,R{{T}_{2}}\]                 \[\therefore \] \[dU=\,{{U}_{2}}-{{U}_{1}}=\,\frac{3}{2}\,R\,({{T}_{2}}-{{T}_{1}})\]                 \[=\frac{3}{2}\,R{{T}_{1}}\,\left[ \frac{{{T}_{2}}}{{{T}_{1}}}-1 \right]\]                 \[=\frac{3}{2}\,R{{T}_{1}}\,\left[ \sqrt{2}-1 \right]\,\left( \because \,\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{1}{\sqrt{2}} \right)\]                 and \[dW=\,2{{P}_{1}}V_{1}^{1/2}\,\left[ V_{1}^{1/2}-V_{1}^{1/2} \right]\]                 \[=\,2{{P}_{1}}\,{{V}_{1}}\,\left[ {{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{1/2}}-1 \right]\]                 But \[{{P}_{1}}{{V}_{1}}=\text{ }R{{T}_{1}}\]                 \[\therefore \] \[dW\,=2R{{T}_{1}}\,\left[ \sqrt{2}-1 \right]\,\left[ \because \,\sqrt{\frac{{{V}_{1}}}{{{V}_{2}}}}=\frac{1}{\sqrt{2}} \right]\]                 Put values of \[dU\]and \[dW\] in eqn. (i) we get                 \[dQ=\,\frac{3}{2}\,R{{T}_{1}}\,(\sqrt{2}-1)\,+2R{{T}_{1}}\,(\sqrt{2}-1)\]                 \[=\left( \sqrt{2}-1 \right)\,R{{T}_{1}}\,\left( \frac{3}{2}\,+2 \right)\]                 \[=\frac{7}{2}\,\left( \sqrt{2}-1 \right)\,R{{T}_{1}}\,=1.449\,\,R{{T}_{1}}\]                 Hence net work done in one cycle,                 \[W=\,{{W}_{AB}}+\,{{W}_{BC}}+\,{{W}_{CD}}+\,{{W}_{DA}}\]                 \[=\,\frac{{{P}_{B}}{{P}_{B}}}{1-\gamma }\,\left[ {{2}^{-\gamma +1}}-1 \right]\,+\frac{{{P}_{A}}{{V}_{A}}}{\gamma -1}\,\,[{{2}^{-\gamma +1}}-1]\]                 \[=\,\frac{{{P}_{B}}{{P}_{B}}}{(1-\gamma )}\,\left[ {{2}^{-\gamma +1}}-1 \right]\,-\frac{{{P}_{A}}{{V}_{A}}}{(1-\gamma )}\,\,[{{2}^{-\gamma +1}}-1]\]                 Since \[{{V}_{A}}={{V}_{B}}\]                 \[\therefore \] \[W=\,\frac{{{V}_{A}}}{1-\gamma }\,[{{2}^{-\gamma +1}}-1]\,[{{P}_{B}}-{{P}_{A}}]\]                 Since \[\gamma =\,5/3\]                 \[\therefore \] \[W=\,\frac{-3{{V}_{A}}}{2}\,\left[ {{2}^{-2/3}}-1 \right]\,[{{P}_{B}}-{{P}_{A}}]\]                 \[=0.555\text{ }\left[ {{P}_{B}}\text{ }{{P}_{A}} \right]\text{ }{{V}_{A}}\].