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JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    During an adiabatic compression, 830 J of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50%. The change in its temperature is nearly: \[(R=8.3J{{K}^{-1}}mo{{l}^{-1}})\]   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) 40 K                       

    B) 33 K

    C) 20 K

    D)

    E) 14 K

    Correct Answer: D

    Solution :

                    Given : work done, W = 830 J No. of moles of gas, \[\mu =2\] For diatomic gas \[\gamma =1.4\] Work done during an adiabatic change \[W=\frac{\mu R({{T}_{1}}-{{T}_{2}})}{\gamma -1}\] \[\Rightarrow \]\[830=\frac{2\times 8.3(\Delta T)}{1.4-1}=\frac{2\times 8.3(\Delta T)}{0.4}\] \[\Rightarrow \]\[DT=\frac{830\times 0.4}{2\times 8.3}=20K\]                                                


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