JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Process (2021)

  • question_answer
    A diatomic gas initially at 18°C is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be

    A) \[{{10}^{o}}C\]

    B) \[{{887}^{o}}C\]

    C) \[668K\]

    D) \[{{144}^{o}}C\]

    Correct Answer: C

    Solution :

    [c]\[T{{V}^{\gamma -1}}=\]constant
    \[\Rightarrow {{T}_{2}}={{T}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}=(273+18)\ {{\left( \frac{V}{V/8} \right)}^{0.4}}=668K\]

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