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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Adiabatic Process

  • question_answer
    At N.T.P. one mole of diatomic gas is compressed adiabatically to half of its volume\[\gamma =1.41\]. The work done on gas will be                                                               [RPET 1997]

    A)            1280 J                                      

    B)            1610 J

    C)            1815 J                                      

    D)            2025 J

    Correct Answer: C

    Solution :

                       \[{{T}_{2}}={{T}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}=273{{(2)}^{0.41}}\]\[=273\times 1.328=363K\]                    \[W=\frac{R({{T}_{1}}-{{T}_{2}})}{\gamma -1}=\frac{8.31(273-363)}{1.41-1}\]\[=-\,1824\] Þ |W| » 1815 J


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