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NEET Sample Paper NEET Sample Test Paper-89

  • question_answer
    One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of \[27{}^\circ C.\] If the work done during the process is 3 kJ, then final temperature of the gas is \[({{C}_{v}}=20J/K)\]

    A) 100 K   

    B)        150 K

    C) 195 K   

    D)        255 K

    Correct Answer: B

    Solution :

    [b] Since the gas expands adiabatically (i.e., no change in enthalpy) so the heat is totally converted into work. For the gas, \[{{C}_{V}}=20\,J/K.\] Thus, \[20\,J\] of heat is required for \[1{}^\circ \] change in temperature of the gas. Heat change involved during the process i.e., work done = 3 kJ =3000 J. Change in temperature \[\frac{3000}{20}K=150\,K\] Initial temperature = 300 K Since, the gas expands so the temperature decreases and thus final temperature is \[300\text{ }-150=\text{ }150\text{ }K\]


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