J & K CET Medical J & K - CET Medical Solved Paper-2003

  • question_answer
    A certain mass of gas at NTP is expanded to three times its volume under adiabatic conditions. The resulting temperature of gas will be : (y for gas is 1.40)

    A) \[273\times {{\left( \frac{1}{3} \right)}^{1.4}}\]                 

    B) \[273\times {{(3)}^{0.4}}\]

    C) \[273\times {{\left( \frac{1}{3} \right)}^{0.4}}\]                 

    D)  \[273\times {{(3)}^{1.4}}\]

    Correct Answer: C

    Solution :

                    An adiabatic process is one in which no heat is transferred to or from working fluid. For a gas undergoing adiabatic process \[T{{V}^{\gamma -1}}=constant\] where T is temperature, V the volume and y the ratio of specific heat. \[\therefore \]  \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}\] Given, \[{{T}_{1}}=273\,K,\] \[{{V}_{2}}=3\,{{V}_{1}},\] \[\gamma =1.4\] \[\therefore \]  \[{{T}_{2}}=273\times {{\left( \frac{{{V}_{1}}}{3{{V}_{1}}} \right)}^{1.4-1}}=273\times {{\left( \frac{1}{3} \right)}^{0.4}}\]


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