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NEET AIPMT SOLVED PAPER 2001

  • 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?    

    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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