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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Critical Thinking

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    Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is \[{{m}_{A}}\] and that in B  is \[{{m}_{B}}\]. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be \[\Delta P\] and 1.5 \[\Delta P\] respectively. Then                                          [IIT 1998]

    A)            \[4{{m}_{A}}=9{{m}_{B}}\]    

    B)            \[2{{m}_{A}}=3{{m}_{B}}\]

    C)            \[3{{m}_{A}}=2{{m}_{B}}\]    

    D)            \[9{{m}_{A}}=3{{m}_{B}}\]

    Correct Answer: C

    Solution :

                       Process is isothermal. There fore, T = constant,                    \[\left( P\propto \frac{1}{V} \right)\] volume is increasing, therefore pressure will decreases.                    In chamber A :                    \[\Delta P={{P}_{i}}-{{P}_{f}}=\frac{{{\mu }_{A}}RT}{V}-\frac{{{\mu }_{A}}RT}{2V}=\frac{{{\mu }_{A}}RT}{2V}\]                               ?..(i)                    In chamber B :                    \[1.5\Delta P={{P}_{i}}-{{P}_{f}}=\frac{{{\mu }_{B}}RT}{V}-\frac{{{\mu }_{B}}RT}{2V}=\frac{{{\mu }_{B}}RT}{2V}\]                               ?..(ii)                    from equations (i) and (ii) \[\frac{{{\mu }_{A}}}{{{\mu }_{B}}}=\frac{1}{1.5}=\frac{2}{3}\]            Þ \[\frac{{{m}_{A}}/M}{{{m}_{B}}/M}=\frac{2}{3}\]Þ \[3{{m}_{A}}=2{{m}_{B}}.\]


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