AFMC AFMC Solved Paper-2007

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

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

    B) \[3\,d{{m}^{3}}\]

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

    D) \[8\,d{{m}^{3}}\]

    Correct Answer: C

    Solution :

    \[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[\frac{{{P}_{1}}\times 4}{{{T}_{1}}}=\frac{2{{P}_{1}}\times {{V}_{2}}}{2{{T}_{1}}}\] \[2{{V}_{2}}=8\] \[2{{V}_{2}}=8\]


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