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Thermodynamics - specific heats and electric susceptibility

  1. Oct 25, 2013 #1
    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!
     
  2. jcsd
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