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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2002

  • question_answer
    In a adiabatic process, initial pressure is 2 atm and final volume is half of initial volume, then final pressure will be (\[\gamma \] = 1.3)

    A)  \[{{2}^{2.3}}\]atm          

    B)  4 atm

    C)  6 atm            

    D)  None of these

    Correct Answer: A

    Solution :

     Poisson's ratio \[p{{V}^{\gamma }}=\]constant \[{{p}_{1}}{{V}_{1}}^{\gamma }={{p}_{2}}{{V}_{2}}^{\gamma }\] \[2V={{p}_{2}}{{\left( \frac{V}{2} \right)}^{\gamma }}\] \[{{p}_{2}}={{2}^{1+\gamma }}={{2}^{2.3}}\]atm


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