JEE Main & Advanced Physics Thermodynamical Processes Question Bank Graphical Questions

  • question_answer
    A cyclic process for 1 mole of an ideal gas is shown in figure in the V-T, diagram. The work done in AB, BC and CA respectively

    A)            \[0,\,R{{T}_{2}}\ln \left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)\,,\,R\,({{T}_{1}}-{{T}_{2}})\]       

    B)            \[R({{T}_{1}}-{{T}_{2}}),\,0,\,R{{T}_{1}}\ln \frac{{{V}_{1}}}{{{V}_{2}}}\]       

    C)            \[0,\,R{{T}_{2}}\ln \left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\,,\,R\,({{T}_{1}}-{{T}_{2}})\]

    D)            \[0,\,R{{T}_{2}}\ln \left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\,,\,R\,({{T}_{2}}-{{T}_{1}})\]

    Correct Answer: C

    Solution :

                       Process AB is isochoric,     \ \[{{W}_{AB}}=P\,\Delta V=0\]                    Process BC is isothermal  \ \[{{W}_{BC}}=R{{T}_{2}}.\ln \left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\]                    Process CA is isobaric \ \[{{W}_{CA}}=-\,P\Delta V\]\[=-\,R\Delta T\]\[=-\,R({{T}_{1}}-{{T}_{2}})\]\[=R({{T}_{2}}-{{T}_{1}})\] (Negative sign is taken because of compression)


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