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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

  • question_answer
    One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of \[\text{27}{{\,}^{\text{o}}}\text{C}\text{.}\] If the work done during the process is 3 kJ, the final temperature will be equal to\[\text{(C}\upsilon \text{=20}\,\text{J}\,{{\text{K}}^{-1}}\text{)}\]

    A) \[~150\text{ }K\]                             

    B) \[~100\text{ }K\]

    C) \[26.85{{\,}^{o}}C\]       

    D) \[~295K\]

    Correct Answer: A

    Solution :

    We know that work done, \[W={{C}_{^{\upsilon }}}({{T}_{1}}-{{T}_{2}})\] \[W=3kJ,\,{{C}_{\upsilon }}=20\,J{{K}^{-1}},\] \[{{T}_{1}}=27+273=300\,K,\,{{T}_{2}}=?\] \[\therefore \]  \[3\times 1000=20(300-{{T}_{2}})\] \[\therefore \]  \[{{T}_{2}}=\frac{300}{20}=150\,K\]


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