BHU PMT BHU PMT Solved Paper-2002

  • question_answer
    A balloon contains 1500m3 of a helium at 27°C and 4 atmospheric pressure. The volume of helium at -3°C temperature and 2     atmospheric pressure will be :                                [BHU PMT-2002]

    A)                  \[{{v}_{2}}\]                     

    B)                  \[\frac{{{v}_{1}}{{v}_{2}}}{2}\]

    C)                  \[\frac{{{v}_{1}}{{v}_{2}}}{4}\]                                 

    D)                  \[\frac{{{v}_{1}}+{{v}_{2}}}{2}\]

    Correct Answer: A

    Solution :

                     The general gas equation is                                 \[\frac{{{v}_{d}}}{4}\]= gas constant                 Where \[\frac{{{v}_{d}}}{2}\] are pressure and volume at temperatures \[{{v}_{d}}\] respectively.                 Given,\[{{\omega }_{1}},{{\omega }_{2}},{{\omega }_{3}}\] \[{{A}_{1}},{{A}_{2}},{{A}_{3}}\]                 \[A_{1}^{2}\omega _{1}^{2}=A_{2}^{2}\omega _{2}^{2}=A_{3}^{2}\omega _{3}^{2}\]                   \[A_{1}^{2}{{\omega }_{1}}=A_{2}^{2}{{\omega }_{2}}=A_{3}^{2}{{\omega }_{3}}\]                                                 \[{{A}_{1}}\omega _{1}^{2}={{A}_{2}}\omega _{2}^{2}={{A}_{3}}\omega _{3}^{2}\]                                                 \[{{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}\]


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