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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    In an adiabatic expansion of a gas initial and final temperatures are \[{{T}_{1}}\] and \[{{T}_{2}}\] respectively, then the change in internal energy of the gas is

    A) \[R\text{/ }\!\!\gamma\!\!\text{ }-1({{T}_{2}}-{{T}_{1}})\]                     

    B) \[R\text{/ }\!\!\gamma\!\!\text{ }-1({{T}_{1}}-{{T}_{2}})\]

    C)  \[R({{T}_{2}}-{{T}_{1}})\]                            

    D)  zero

    Correct Answer: A

    Solution :

    In an adiabatic expansion \[dQ=0\] So,        \[d=-dW\] Work done in adiabatic process                 \[dW=\frac{R}{1-\gamma }({{T}_{2}}-{{T}_{1}})\] \[\therefore \]Change in internal energy                 \[dU=\frac{-R}{1-\gamma }({{T}_{2}}-{{T}_{1}})\] or            \[dU=\frac{R}{1-\gamma }({{T}_{2}}-{{T}_{1}})\]


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