JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    A perfect gas at \[27{}^\circ C\] is heated at constant pressure so as to double its volume. The final temperature of the gas will be, close to   JEE Main Online Paper (Held On 07 May 2012)

    A) \[327{}^\circ C\]                                   

    B) \[200{}^\circ C\]

    C) \[54{}^\circ C\]

    D) \[300{}^\circ C\]

    Correct Answer: A

    Solution :

    Given,\[{{V}_{1}}=V\] \[{{V}_{2}}=2V\] \[{{T}_{1}}={{27}^{o}}+273=300K\] \[{{T}_{2}}=?\]From charle's law \[\frac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}\](\[\because \]Pressure is constant) or,\[\frac{V}{300}=\frac{2V}{{{T}_{2}}}\] \[\therefore \]\[{{T}_{2}}=600K=600-273={{327}^{o}}C\]


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