JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    The state of an ideal gas is changed through an isothermal process at temperature \[{{T}_{0}}\] as shown in figure. The work done by gas in going from state B to C is double the work done by gas in going from state A to B. If the pressure in the state B is \[{{P}_{0}}/2\] then the pressure of the gas in state C is 

    A) \[{{P}_{0}}/2\]

    B)        \[{{P}_{0}}/4\]

    C) \[{{P}_{0}}/6\]

    D)        \[{{P}_{0}}/8\]

    Correct Answer: D

    Solution :

    [d] Work done by gas in going isothermally from stated to 5 is \[\Delta {{W}_{AB}}=nRT\text{ In}\frac{{{P}_{A}}}{{{P}_{B}}}=nRT\text{ In 2}\]                  ....(i) Work done by gas in going isothermally from state B to C is \[\Delta {{W}_{BC}}=nRT\text{ In}\frac{{{P}_{B}}}{{{P}_{C}}}=nRT\frac{{{P}_{0}}}{2{{P}_{C}}}\]                       ...(ii) It is given that \[\Delta {{W}_{BC}}=2\Delta {{W}_{AB}}\] \[\text{In }\frac{{{P}_{0}}}{2{{P}_{C}}}=\text{In}{{\left( \text{2} \right)}^{2}}\text{        }\therefore {{\text{P}}_{C}}=\frac{{{P}_{0}}}{8}\]


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