BCECE Engineering BCECE Engineering Solved Paper-2008

  • question_answer
    If two molecules of A and B having mass 100 kg and 64 kg and rate of diffusion of A is \[12\times {{10}^{-3}},\] then what will be the rate of diffusion of B?

    A) \[15\times {{10}^{-3}}\]               

    B)        \[64\times {{10}^{-3}}\]

    C)        \[~5\times {{10}^{-3}}\]               

    D)        \[~46\times {{10}^{-3}}\]

    Correct Answer: A

    Solution :

    Key Idea According to Grahams law of diffusion At constant pressure and temperature, the rate of diffusion of a gas is inversely proportional to the square root of its vapour density. Rate of diffusion \[(r)\propto \frac{1}{\sqrt{d}}\] Molecular weight \[(M)=2\times \]vapour density \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}\] \[{{m}_{A}}=\left( \frac{100}{2} \right)\text{kg/molecule}\] \[{{m}_{B}}=\left( \frac{64}{2} \right)\text{kg/molecule}\]           \[{{r}_{A}}=12\times {{10}^{-3}}\]and \[{{r}_{B}}=?\] \[\frac{{{r}_{A}}}{{{r}_{B}}}=\sqrt{\frac{{{d}_{B}}}{{{d}_{A}}}}=\sqrt{\frac{{{M}_{B}}}{{{M}_{A}}}}\] \[\frac{12\times {{10}^{-3}}}{{{r}_{B}}}=\sqrt{\frac{64/2}{100/2}}=\sqrt{\frac{64}{100}}=\frac{8}{10}\] \[{{r}_{B}}=\frac{12\times {{10}^{-3}}\times 10}{8}\] \[=15\times {{10}^{-3}}\]


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