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NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    One mole of solid iron was vaporized in an oven at its boiling point of 3433 K and enthalpy of vaporization of iron is \[344.3\text{ }kJ\text{ }mo{{l}^{-1}}\]. The value of entropy vaporization (in J \[mo{{l}^{-1}}\]) of iron is

    A) 100                  

    B) 10   

    C) \[-100\]             

    D) 110

    Correct Answer: A

    Solution :

    [a] \[\Delta S=\frac{\Delta H}{T}=\frac{334.3\times {{10}^{3}}}{3443}=100Jmo{{l}^{-1}}\]


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