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EAMCET Medical EAMCET Medical Solved Paper-2003

  • question_answer
    The kinetic energy of 4 moles of nitrogen gas at \[\text{127}{{\,}^{\text{o}}}\text{C}\] is ...... cals. \[\text{(R= 2 cal mo}{{\text{l}}^{-1}}{{\text{K}}^{-1}}\text{)}\]

    A)  4400                                     

    B)  3200

    C)  4800                                     

    D)  1524

    Correct Answer: C

    Solution :

                     \[E=\frac{3}{2}nRT\] Where, E = Kinetic energy n = number of moles = 4 R = Gas constant =2cal/mol/K T = Temperature in Kelvin = 127 + 273 = 400K \[E=\frac{3}{2}\times 4\times 2\times 400\] = 4800 calorie


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