JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    An open vessel \[27{}^\circ C\] at is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) 750 K                           

    B) \[750{}^\circ C\]

    C) \[500{}^\circ C\]                      

    D)   500 K

    Correct Answer: D

    Solution :

    From ideal gas equation,\[\frac{{{P}_{1}}{{V}_{1}}}{{{n}_{1}}R{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{n}_{2}}R{{T}_{2}}}\] \[\because \]\[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}};\therefore {{n}_{1}}{{T}_{1}}={{n}_{2}}{{T}_{2}}\] \[{{n}_{1}}\times 300K=\frac{3}{5}{{n}_{1}}\times {{T}_{2}}\] \[{{T}_{2}}=500K\]


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