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

  • question_answer
    The latent heat of vapourization of \[\varepsilon \] liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy \[\left( \Delta U \right)\] of 3 moles of liquid at the same temperature

    A) \[13.0\text{ }kcal/mol\]              

    B) \[-13.0\text{ }kcal/mol\]

    C) \[27.0\text{ }kcal\]         

    D) \[-7.0\text{ }kcal/mol\]

    Correct Answer: C

    Solution :

    \[3{{H}_{2}}O(l)\to 3{{H}_{2}}O(g);\] \[\Delta n=3,\Delta E=\Delta H-\Delta nRT\] \[=30-3\times \frac{2}{1000}\times 500=27kcal\]


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