2. Solved Papers
  3. WB JEE Medical

WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    The pressure and density of a diatomic gas\[\left( \gamma =\frac{7}{5} \right)\]  change adiabatically from \[({{P}_{1}},{{\rho }_{1}})\] to\[({{P}_{2}}{{\rho }_{2}}).\] IF \[\frac{{{\rho }_{2}}}{{{\rho }_{1}}}=32,\]then \[\frac{{{P}_{2}}}{{{P}_{1}}}\]should be:

    A)  16                                         

    B)  32

    C)  64                                         

    D)  128

    Correct Answer: D

    Solution :

                     In an adiabatic process. \[P{{V}^{\gamma }}=\text{constant}\]                 or            \[{{P}_{1}}V_{1}^{\gamma }={{P}_{2}}V_{2}^{\gamma }\]                 or            \[\frac{{{P}_{2}}}{{{P}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}\]                                      ?(i)                 Volume of gas   \[=\frac{\text{Mass}}{\text{Density}}\] ie.,          \[V=\frac{M}{\rho }\]or \[V\propto \frac{1}{\rho }\] \[\therefore \]  \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{\rho }_{2}}}{{{\rho }_{1}}}=32\] Thus, form Eq.(i), we have \[\frac{{{P}_{2}}}{{{P}_{1}}}={{(32)}^{\gamma }}={{(32)}^{7/5}}={{2}^{7}}=128\]


You need to login to perform this action.
You will be redirected in 3 sec spinner