JEE Main & Advanced Physics Thermodynamical Processes Question Bank Adiabatic Process

  • question_answer
    A gas is suddenly compressed to 1/4 th of its original volume at normal temperature. The increase in its temperature is \[(\gamma =1.5)\]                                                   [DCE 2004]

    A)            273 K                                       

    B)            573 K

    C)            373 K                                       

    D)            473 K

    Correct Answer: A

    Solution :

                       \[\because \ T{{V}^{\gamma -1}}=\]constantÞ \[{{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}\]                    Þ \[{{T}_{2}}={{T}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}={{T}_{1}}{{(4)}^{1.5-1}}=2{{T}_{1}}\]                    \ change in temperature            \[={{T}_{2}}-{{T}_{1}}=2{{T}_{1}}-{{T}_{1}}={{T}_{1}}=273\,K\]


You need to login to perform this action.
You will be redirected in 3 sec spinner