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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank First law of thermodynamics and Hess law

  • question_answer
    Work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 300 K is (Gas constant = 2)            [AIIMS 2000]

    A)                 938.8 cal.            

    B)                 1138.8 cal.

    C)                 1381.8 cal.          

    D)                 1581.8 cal.

    Correct Answer: C

    Solution :

           \[-\,W=+\,2.303\,nRT\,\]log \[\frac{{{p}_{1}}}{{{p}_{2}}}\]                                 \[-\,W=2.303\times 1\times 2\times 300\]log \[\frac{10}{1}\] \[=1381.8\,cal.\]


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