CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer The pressure and temperature of \[4\text{ }d{{m}^{3}}\] of carbon dioxide gas are doubled. Then the volume of carbon dioxide gas would be :

    A)  \[2\text{ }d{{m}^{3}}\]                

    B)  \[3\text{ }d{{m}^{3}}\]

    C)  \[4\text{ }d{{m}^{3}}\]                

    D)  \[8\text{ }d{{m}^{3}}\]

    Correct Answer: C

    Solution :

    Given \[\frac{{{P}_{2}}}{{{P}_{1}}}=2,\frac{{{T}_{2}}}{{{T}_{1}}}=2,\,{{V}_{1}}=4\,d{{m}^{3}},\,{{V}_{2}}=?\] From gas equation,                 \[\frac{{{P}_{1}}\,{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}\,{{V}_{2}}}{{{T}_{2}}}\] or            \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{P}_{2}}}{{{P}_{1}}}\times \frac{{{T}_{1}}}{{{T}_{2}}}\] \[\therefore \]  \[\frac{4}{{{V}_{2}}}=2\times \frac{1}{2}=1\] \[\therefore \]  \[{{V}_{2}}=4\,\,d{{m}^{3}}\]

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