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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank 2nd and 3rd law of thermodynamics and Entropy

  • 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 \[({{C}_{v}}=20\,J{{K}^{-1}})\] [KCET 2000; AFMC 2000; AIIMS 2001]

    A)                 150 K    

    B)                 100 K

    C)                 \[{{26.85}^{o}}C\]           

    D)                 295 K

    Correct Answer: A

    Solution :

           We know that work done, \[W={{C}_{v}}({{T}_{1}}-{{T}_{2}})\]                    \[3\times 1000=20\,(300-{{T}_{2}})\]; \[\therefore 3000=6000-20\,{{T}_{2}}\]                                 \[\therefore {{T}_{2}}=\frac{3000}{20}=150\,K\].


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