Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    Enthalpy is equal to

    A) \[\int_{{}}^{{}}{\frac{1}{\sin \left( x-\frac{\pi }{3} \right)\cos x}dx}\]       

    B) \[2\log \left| \sin x+\sin \left( x-\frac{\pi }{3} \right) \right|+C\]

    C) \[2\log \left| \sin x.\sin \left( x-\frac{\pi }{3} \right) \right|+C\]

    D) \[2\log \left| \sin x-\sin \left( x-\frac{\pi }{3} \right) \right|+C\]

    Correct Answer: B

    Solution :

    The Gibbs-Helmholtz equation is as \[\frac{3!}{{{100}^{3}}}\] Dividing above equation by\[f(x)=\left\{ \begin{matrix}    \frac{\sin (\cos x)-\cos x}{{{(\pi -2x)}^{3}}}, & x\ne \frac{\pi }{2}  \\    k, & x=\frac{\pi }{2}  \\ \end{matrix} \right.\] \[x=\frac{\pi }{2},\] This on rearrangement becomes. \[-\frac{1}{6}\] Thus, enthalpy,\[-\frac{1}{24}\] Where,      H = enthalpy


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