Thermodynamics - specific heats and electric susceptibility

[skrat]

Homework Statement

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 $c_P=1700J/KgK$

Homework 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 $0=dS=(\frac{\partial S}{\partial T})_PdT+(\frac{\partial S}{\partial P})_TdP$

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!

 

[mark726]

Thank you for your post. It seems like you have made some progress in your attempt at solving this problem. However, there are a few areas where you may have made some mistakes.

Firstly, in your attempt to find the equation of state, you used the relation $P=\varepsilon_0 \chi E$. However, this relation is only valid for a linear dielectric material, which may not be the case for the oil in question. Instead, you may want to use the relation $P=\varepsilon_0 \chi E + \frac{1}{2}\varepsilon_0 \chi^2 E^2$, which is valid for a nonlinear dielectric material.

Secondly, in your equations for $dW$ and $dQ$, you are correct in saying that $W=EPV$, but remember that this is only valid for an ideal gas. For a liquid or solid, you will need to use a different equation. Additionally, you have not accounted for the heat transfer term $dQ$, which may be important in this problem.

Finally, to find the difference between isothermal and adiabatic susceptibility, you will need to use the Maxwell relations that you have derived, along with the given expression for the susceptibility as a function of temperature. You may want to consider using the equations you have derived and the given expression for the susceptibility to find the difference between the two susceptibilities.

I hope this helps guide you in the right direction. Good luck with your calculations!