# Thermodynamics - specific heats and electric susceptibility

1. Oct 25, 2013

### skrat

1. The problem statement, all variables and given/known data
Calculate the difference of specific heats $c_E-c_P$ of oil with density $800 kg/m^3$ at $T=27 °C$ in electric field with $E=10^7V/m$. In this conditions the electric susceptibility is 2 and is a function of T given as $\frac{\chi }{\chi +3}\propto 1+\frac{C}{T}$ where $C=30K$. What is the difference between isothermal and adiabatic susceptibility?
Take the volume of oil as constant and it's $c_P=1700J/KgK$

2. Relevant equations
Two important equations:
1. $S=S(T,E)$ thereby $0=dS=(\frac{\partial S}{\partial T})_EdT+(\frac{\partial S}{\partial E})_TdE$
2.$S=S(T,P)$ thereby $0=dS=(\frac{\partial S}{\partial T})_PdT+(\frac{\partial S}{\partial P})_TdP$

3. The attempt at a solution
$dU=dW+dQ$ (I am not sure how to calculate $dW$ for given system, yet i found something on the internet saying that $W=EPV$)
Using entropy the energy is than $dU=EVdP+TdS$
Free energy than $F=U-TS$
$dF=EVdP+TdS-SdT-TdS=EVdP-SdT$
Since I want to use Maxwell's relations: $(\frac{\partial (EV)}{\partial T})_P=-(\frac{\partial S}{\partial P})_T$
The second Maxwell relation comes from $G=F-VEP$
$dG=-SdT-VPdE$ and therefore relation $(\frac{\partial S}{\partial E})_T=(\frac{\partial (VP)}{\partial T})_E$

Now in the first two equations it is obvious that $(\frac{\partial S}{\partial T})_EdT=\frac{c_Em}{T}$ and $(\frac{\partial S}{\partial T})_PdT=\frac{c_Pm}{T}$. Now using this and Maxwell relations I get:
1. $\frac{c_Em}{T}+V(\frac{\partial (P)}{\partial T})_EdE=0$ and
2. $\frac{c_Pm}{T}-V(\frac{\partial (E)}{\partial T})_PdP=0$

I SERIOUSLY hope that so far everything is legit! :D

Now the next step would be to find the equation of state... Everything written below is probably completely wrong:
Again I used some google and found out that $P=\varepsilon _0\chi E$. In this equation i inserted $\chi$ from $\frac{\chi }{\chi +3}=1+\frac{C}{T}$ which gives me $P=-3\varepsilon _0E(1+T/C)$.

Now $(\frac{\partial (P)}{\partial T})_E=-\frac{\varepsilon _0E3}{C}$ and $(\frac{\partial (E)}{\partial T})_P=\frac{P}{\varepsilon _03C(1+t/C)^2}$

Which now gives me completely useles equations:

1. $\frac{c_Em}{T}-V\frac{\varepsilon _0E3}{C}dE=0$ and
2. $\frac{c_Pm}{T}-V\frac{P}{\varepsilon _03C(1+t/C)^2}dP=0$

I apologize for all the formulas and all the writing but I just want you to see that I am trying yet I am not smart enough...

The first problem is that I don't even know if the equation for W is right and I am pretty sure that the equation of state is completely wrong. I would be really happy if somebody could guide me on how to get the equation of state.

Cheers and thanks!

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