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NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी NEET PYQ-Thermodynamics

  • question_answer
    When 1 mole gas is heated at constant volume, temperature is raised from 298 to 308K. Heat supplied to the gas is 500 J. Then which statement is correct?                                                                                               [AIPMT 2001]

    A)  \[q=w=500\,J,\,\Delta U=0\]

    B)  \[q=\Delta U=500\,J,\,w=0\]

    C)  \[q=w=500\,J,\,\Delta U=0\]

    D)  \[\Delta U=0,\,\,q=w=-500\,J\]

    Correct Answer: B

    Solution :

    As we know that \[\Delta H=\Delta E+P\Delta V\]
                When    \[\Delta V=0\]    \[\therefore \] \[\Delta H=\Delta E\]
                From first law of thermodynamics
                            \[\Delta E=q-W\]
                In given problem \[\Delta H=500\,J\]
                            \[-W=-p\Delta V,\]          \[\Delta V=0\]
                So,       \[\Delta E=q=500\,J\]


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