JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    A gaseous substance \[A{{B}_{2}}(g)\] converts to \[AB\,(g)\] in the presence of solid \[A(s)\] as \[A{{B}_{2}}(g)+A(s)\,\,AB(g)\] The initial pressure and equilibrium pressure are 0.7 and 0.95 bar. Now the equilibrium mixture is expanded reversibly and isothermally till the gas pressure falls to 0.4 bar. The volume percentage of \[AB(g)\] the final equilibrium is

    A)  67.5                           

    B)  32.5              

    C)  45.3                           

    D)  63.1

    Correct Answer: A

    Solution :

     \[{{P}_{1}}=\frac{60}{4\pi \times {{4}^{2}}}\times \frac{\pi \times {{(2\times {{10}^{-3}})}^{2}}}{4}\] \[=9.375\times {{10}^{-7}}\,J/s\] \[n\]             Now, \[1\times \cos \,{{60}^{o}}+2\,\cos \,{{60}^{o}}+x-4\,\cos \,{{60}^{o}}=0\] \[\therefore \] or         \[x=0.5\,\,N\] \[\frac{M}{\pi {{R}^{2}}}\times \pi {{\left\{ \frac{R}{2} \right\}}^{2}}=\frac{M}{4}\] \[=M-\frac{M}{4}=\frac{3M}{4}\] \[{{X}_{CM}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] \[{{X}_{CM}}\]         Volume % of \[{{m}_{2}}{{x}_{2}}=-{{m}_{1}}{{x}_{1}}\]

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