JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Processes

  • question_answer
    A diesel engine takes in 5 moles of air at \[20{}^\circ C\] and 1 atm, and compresses it adiabatically to \[\frac{1}{10}\text{th}\] of the original volume. If air is diatomic then work done and change in internal energy is

    A) - 46 kJ, 46 kJ

    B) 36 kJ, - 36 kJ

    C) 46 kJ, - 46 kJ

    D) - 36 kJ, 36 kJ

    Correct Answer: A

    Solution :

    [a]Let \[{{p}_{1}}=1\] atm, \[n=5\,mol,\text{ }293K\]
    Using   \[{{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}\]
    \[\Rightarrow \]            \[{{T}_{2}}={{T}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}\]
    Now, work done \[=\frac{nR({{T}_{1}}-{{T}_{2}})}{\gamma -1}\]
    \[=\frac{5\times 8.3\times (293-736)}{0.4}=-46kJ\]
    and \[\Delta U=\Delta Q-W=0-W=46kJ\]

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