[Thermo] Equation of State Given, Find Internal Energy and Specific Heat

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SUMMARY

The discussion focuses on deriving the internal energy and specific heat from the given equation of state: P = RT/(v-b) - a/(T(V+c)^2). The user successfully differentiates pressure with respect to temperature and attempts to integrate to find the change in internal energy (u2-u1) using the equation u2-u1 = Integral(Cv dT) + Integral(T * (dP/dT)v - P) dV. The final expression for the heat capacity is established as Cv(T,V) = Cv^{IG}(T) + 2a/(T^2(V+C)), indicating a relationship between the internal energy, specific heat, and the parameters of the equation of state.

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Homework Statement




P=RT/(v-b) - a/(T(V+c)^2)


Homework Equations



u2-u1= Integral ( Cv dT) + Integral ( T * (dP/dT)v - P) dV

The Attempt at a Solution



I've differentiated P with respect to T to get R/(v-b)+a/(T^2(V+C)^2 and plugged the relevants back into the equation for u2-u1. I'm confused as to how to derive both Cv and U2-u1 from the equation. At first I thought Cv would drop out of the first equation and I could just solve for my internal energy solving to get 2* Integral(a / T(v+c)^2) dv but then I'm stuck with just an equation for internal energy and no way to go back and get the specific heat. Any pointers?

Thanks
 
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As in the OP, I also get $$du=C_vdT+\frac{2a}{T(V+C)^2}dV$$
Assuming I know ##u(T_1,V_1)##, I can integrate the differential equation from ##V_1## to infinite V at constant temperature to obtain $$u(T_1,\infty)=u(T_1,V_1)+\frac{2a}{T_1(V_1+C)}$$At infinite volume, ##C_v=C_v^{IG}##, where ##C_v^{IG}(T)## is the ideal gas heat capacity (a function only of T). Therefore, if I next integrate the differential equation at infinite volume between ##(T_1,\infty)## and ##(T,\infty)##, I obtain: $$u(T,\infty)=u(T_1,\infty)+\int_{T_1}^T{C_v^{IG}(T')dT'}$$Next, if I hold the temperature constant at T and integrate from infinite volume to finite arbitrary volume V, I obtain: $$u(T,V)=u(T,\infty)-\frac{2a}{T(V+C)}$$
Finally, combining the previous three equations then yields: $$u(T,V)=u(T_1,V_1)+\frac{2a}{T_1(V_1+C)}+\int_{T_1}^T{C_v^{IG}(T')dT'}$$$$-\frac{2a}{T(V+C)}$$From this, it follows that the heat capacity of the gas at (T,V) is given by: $$C_v(T,V)=C_v^{IG}(T)+\frac{2a}{T^2(V+C)}$$
 
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