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Thread: [SOLVED] Proof using calculus?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Proof using calculus?

    Bulk modulus of a material is given by B = -dP/(dV/V). Prove that Bulk modulus for adiabatic process = γP. (For adiabatic process PV^γ = K)

    Can anybody explains all the steps?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by fardeen_gen View Post
    Bulk modulus of a material is given by B = -dP/(dV/V). Prove that Bulk modulus for adiabatic process = γP. (For adiabatic process PV^γ = K)

    Can anybody explains all the steps?
    The hint is to work with $\displaystyle \ln \left(PV^{\gamma}\right)$ :


    $\displaystyle
    \ln \left(PV^{\gamma}\right)=\ln K \Longleftrightarrow \ln P+\gamma\ln V=\ln K$

    Let's compute the differential of $\displaystyle \ln P+\gamma\ln V$ :

    $\displaystyle
    \begin{aligned}
    \mathrm{d} \left(\ln P+\gamma\ln V\right) &= \frac{\partial}{\partial P}\left(\ln P+\gamma\ln V\right)\mathrm{d}P + \frac{\partial}{\partial V}\left(\ln P+\gamma\ln V\right)\mathrm{d}V \\
    &= \left(\frac{\partial \ln P}{\partial P}+\gamma\frac{\partial \ln V}{\partial P}\right)\mathrm{d}P +\left( \frac{\partial \ln P}{\partial V}+\gamma \frac{\partial \ln V}{\partial V}\right)\mathrm{d}V \\
    \end{aligned}

    $

    Since $\displaystyle \frac{\partial \ln V}{\partial P}=\frac{\partial \ln P}{\partial V}=0$ and $\displaystyle \mathrm{d}\ln u=\frac{\mathrm{d}u}{u}$

    $\displaystyle
    \begin{aligned}
    \mathrm{d} \left(\ln P+\gamma\ln V\right)&=\frac{\mathrm{d}\ln P}{\mathrm{d}P}\mathrm{d}P+\gamma\frac{\mathrm{d}\ ln V}{\mathrm{d}V}\mathrm{d}V\\
    &=\frac{\mathrm{d}P}{P}+\gamma\frac{\mathrm{d}V}{V }\\
    \end{aligned}
    $

    hence $\displaystyle \frac{\mathrm{d}P}{P}=-\gamma\frac{\mathrm{d}V}{V}$. Can you conclude ?
    Last edited by flyingsquirrel; Jun 29th 2008 at 06:41 AM.
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  3. #3
    Super Member fardeen_gen's Avatar
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    I cant. I started with calculus just two days ago after finishing trigo two days ago. Pls conclude.
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  4. #4
    Super Member fardeen_gen's Avatar
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    Is d ln u = ln u / u a property of derivatives?
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  5. #5
    Super Member fardeen_gen's Avatar
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    I couldnt understand ur second step at all! Please explain!
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  6. #6
    Super Member fardeen_gen's Avatar
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    How is d ln K = zero?

    P.S. I dont seem to know a thing. Sigh...
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  7. #7
    Super Member flyingsquirrel's Avatar
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    I've edited my previous post.

    Quote Originally Posted by fardeen_gen View Post
    I cant. I started with calculus just two days ago after finishing trigo two days ago. Pls conclude.
    You don't know how to do a division ?

    Quote Originally Posted by fardeen_gen View Post
    Is d ln u = ln u / u a property of derivatives?
    Yes. For $\displaystyle t\mapsto u(t)$, $\displaystyle \mathrm{d}\ln u$ is defined by $\displaystyle \mathrm{d}\ln u=\frac{\mathrm{d}(\ln u)}{\mathrm{d}t}\cdot \mathrm{d}t$ and as $\displaystyle \frac{\mathrm{d}(\ln u)}{\mathrm{d}t}=\frac{1}{u}\cdot \frac{\mathrm{d}u}{\mathrm{d}t}$ you get the expected equality. (Note that in this exercise we're in the particular case where $\displaystyle u(t)=t$ : we're working with $\displaystyle u(P)=P$ and $\displaystyle u(V)=V$ )

    Quote Originally Posted by fardeen_gen View Post
    I couldnt understand ur second step at all! Please explain!
    See my previous post. (edited)

    Quote Originally Posted by fardeen_gen
    How is d ln K = zero?
    $\displaystyle K$ is a constant which depends neither on $\displaystyle V$ nor on $\displaystyle P$ : $\displaystyle \frac{\partial K}{\partial V}=\frac{\partial K}{\partial P}=0$ hence

    $\displaystyle
    \mathrm{d}\ln K = \frac{\partial K}{\partial P}\mathrm{d}P+\frac{\partial K}{\partial V}\mathrm{d}V=0\cdot \mathrm{d}P+0\cdot\mathrm{d}V=0$
    Last edited by flyingsquirrel; Jun 29th 2008 at 07:20 AM.
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