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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- Jun 29th 2008, 05:31 AMfardeen_gen[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? - Jun 29th 2008, 05:41 AMflyingsquirrel
Hi

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 ? - Jun 29th 2008, 05:59 AMfardeen_gen
I cant. I started with calculus just two days ago after finishing trigo two days ago. Pls conclude.

- Jun 29th 2008, 06:02 AMfardeen_gen
Is d ln u = ln u / u a property of derivatives?

- Jun 29th 2008, 06:03 AMfardeen_gen
I couldnt understand ur second step at all! Please explain!

- Jun 29th 2008, 06:08 AMfardeen_gen
How is d ln K = zero?

P.S. I dont seem to know a thing. Sigh... - Jun 29th 2008, 06:38 AMflyingsquirrel
I've edited my previous post.

You don't know how to do a division ? :D

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 $\displaystyle u(V)=V$ )

See my previous post. (edited)

Quote:

Originally Posted by**fardeen_gen**

$\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$