One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by [JEE Main 16-4-2018]
A) \[\frac{25}{8}\frac{{{P}_{0}}{{V}_{0}}}{R}\]
B) \[\frac{25}{4}\frac{{{P}_{0}}{{V}_{0}}}{R}\]
C) \[\frac{25}{16}\frac{{{P}_{0}}{{V}_{0}}}{R}\]
D) \[\frac{5}{8}\frac{{{P}_{0}}{{V}_{0}}}{R}\]
Correct Answer: A
Solution :
Temperature at A = \[\frac{3{{P}_{0}}V}{nR}\] Temperature at\[b=\frac{{{P}_{0}}\times 2{{V}_{0}}}{nR}\] Maximum temperature can be between B and C P-V equation for process BC \[P-3{{P}_{0}}=\frac{{{P}_{0}}-3{{P}_{0}}}{2{{V}_{0}}-{{V}_{0}}}\times (V-{{V}_{0}})\] \[P-3{{P}_{0}}=\frac{-2{{P}_{0}}V}{{{V}_{0}}}+2{{P}_{0}}\] \[P=\frac{-2{{P}_{0}}V}{{{V}_{0}}}+2{{P}_{0}}\] Multiply by V \[PV=\frac{-2{{P}_{0}}V}{{{V}_{0}}}+5{{P}_{0}}\] \[RT=-\frac{2{{P}_{0}}}{{{V}_{0}}}{{V}^{2}}+5{{P}_{0}}V\] Make \[\frac{dT}{dV}=0\] This gives \[T=\frac{25{{P}_{0}}{{V}_{0}}}{8R}\]
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