Hi
Originally Posted by
fardeen_gen 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 ?