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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2000

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

    A)  \[150K\]        

    B)  \[100K\]

    C)  \[26-{{85}^{o}}C\]       

    D)  \[295K\]

    Correct Answer: A

    Solution :

    We know that work done, \[W={{C}_{\upsilon }}({{T}_{1}}-{{T}_{2}})\] \[W=3kJ,\,\,C\upsilon =20j{{K}^{-1}}\] \[{{T}_{1}}=27+273=300K,{{T}_{2}}=?\] \[3\times 1000=20(300-{{T}_{2}})\] \[\therefore \] \[3000=6000-20{{T}_{2}}\] \[\therefore \] \[{{T}_{2}}=\frac{3000}{20}=150K\]


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