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BHU PMT BHU PMT (Screening) Solved Paper-2005

  • question_answer
    A perfect gas is found to obey the relation\[P{{V}^{3/2}}=\] constant during an adiabatic process, if such a gas initially at a temperature T, is compressed to half of its initial volume, then its final temperature will be:

    A)  2T                                         

    B)  4T

    C) \[{{(2)}^{1/2}}T\]                            

    D) \[2{{(2)}^{1/2}}T\]

    Correct Answer: C

    Solution :

                      From gas equation we have \[PV=RT\]where R is gas constant. Also given \[P{{V}^{3/2}}=\] constant \[\therefore \]Putting the value of \[P=\frac{RT}{V},\]we have \[\frac{RT}{V}{{V}^{3/2}}=\]constant \[T{{V}^{1/2}}=\]constant                 \[{{T}_{1}}V_{1}^{1/2}={{T}_{2}}V_{2}^{1/2}\] Given, \[{{T}_{1}}=T,{{V}_{1}}=V,{{V}_{2}}=\frac{V}{2'}\] \[\therefore \]  \[T{{V}^{1/2}}={{T}_{2}}{{\left( \frac{V}{2} \right)}^{1/2}}\] \[\Rightarrow \]               \[{{T}_{2}}={{(2)}^{1/2}}T\]


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