J & K CET Engineering J and K - CET Engineering 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 (7 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 \[\gamma \]the ratio of specific heat. \[\therefore \] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}\] Given,   \[{{T}_{1}}=273K,\,\,\,\,\,{{V}_{2}}=3{{V}_{1}},\,\,\,\,\,\,\gamma =1.4.\] \[\therefore \] \[{{T}_{2}}=273\times {{\left( \frac{{{V}_{1}}}{3{{V}_{1}}} \right)}^{1.4-1}}=273\times {{\left( \frac{1}{3} \right)}^{0.4}}\]


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