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BCECE Medical BCECE Medical Solved Papers-2012

  • question_answer
    Certain amount of an ideal gas of molecular mass M is contained in a closed vessel. If the vessel is moving with a constant velocity v, then the rise in temperature of the gas when the vessel is suddenly stopped will be \[\left( Take\,\,\gamma =\frac{{{C}_{p}}}{{{C}_{V}}} \right)\]

    A)  \[\frac{M{{v}^{2}}}{2\,R(\gamma +1)}\]

    B)  \[\frac{M{{v}^{2}}}{2\,R(\gamma -1)}\]

    C)  \[\frac{M{{v}^{2}}(\gamma -1)}{2\,R}\]

    D)  \[\frac{M{{v}^{2}}(\gamma +1)}{2\,R}\]

    Correct Answer: C

    Solution :

    If m is the total mass of the gas then its kinetic energy \[=\frac{1}{2}m{{v}^{2}}\] when the vessel is suddenly stopped, then the total kinetic energy will increase the temperature of the gas. Hence \[\frac{1}{2}m{{v}^{2}}=\mu {{C}_{V}}\Delta T\] or            \[\frac{1}{2}m{{v}^{2}}=\frac{m}{M}{{C}_{V}}\Delta T\] But                 \[{{C}_{V}}=\frac{R}{\gamma -1}\] \[\therefore \]  \[\frac{1}{2}m{{v}^{2}}=\frac{m}{M}\times \frac{R}{\gamma -1}\times \Delta T\] \[\Delta T=\frac{M{{v}^{2}}}{2R}(\gamma -1)\]


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