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KeithPhysics
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Homework Statement
Find the Equation of State of a fluid with Volume expansivity ##\alpha_P## and an isothermal compresibility ##\kappa_T## are given by
$$\alpha_P=\alpha_0 \Big(1-\frac{P}{P_0}\Big) \\ \: \: \: \: \kappa_T=\kappa_0[1+\beta_0(T-T_0)]$$
¿ What conditions should the constants ##\alpha_b,P_0,\kappa_0## y ##\beta_0## have for the problem to have a solution?
Homework Equations
I used this relation given in my book for ##PVT## systems
$$\Big(\frac{\partial P}{\partial T}\Big)_V=\frac{\alpha_P}{\kappa_T}$$
definition of isothermal compresibility ##\kappa_T##:
$$\kappa_T=-\frac{1}{V} \Big(\frac{\partial V}{\partial P}\Big)_{T}=\kappa_0[1+\beta_0(T-T_0)]$$
The Attempt at a Solution
I solve the equation with constant ##V##
$$\Big(\frac{\partial P}{\partial T}\Big)_V=\frac{\alpha_0 (1-\frac{P}{P_0})}{\kappa_0(1+\beta_0(T-T_0))}$$
$$=\Big(\frac{\alpha_0}{\kappa_0 P_0}\Big)\frac{P_0-P}{(1+\beta_0(T-T_0))}$$
$$\frac{d P}{P_0-P}=\Big(\frac{\alpha_0}{\kappa_0 P_0}\Big)\frac{dT}{(1+\beta_0(T-T_0))}$$
$$-\ln(P_0-P)=\Big(\frac{\alpha_0}{\kappa_0 P_0 \beta_0}\Big) \ln(1+\beta_0(T-T_0))+f(V) \tag{5}$$
The problem is that I find this ##f(V)## to be a function of ##T## and ##P## with the problem conditions any Help ?
> Derivation of the problem **(Not Necesary)**
Using the definition of isothermal compresibility ##\kappa_T##:
$$\kappa_T=-\frac{1}{V} \Big(\frac{\partial V}{\partial P}\Big)_{T}=\kappa_0[1+\beta_0(T-T_0)]$$
Now taking ##T=cte##
$$\Big(\frac{1}{P_0-P}\Big) \Big(\frac{\partial P}{\partial V}\Big)_{T}=f'(V)$$
$$\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{f'(V)}\Big)=\Big(\frac{\partial V}{\partial P}\Big)_{T}$$
$$\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{f'(V)}\Big)\Big(\frac{-1}{V}\Big)=-\frac{1}{V} \Big(\frac{\partial V}{\partial P}\Big)_{T}=\kappa_0[1+\beta_0(T-T_0)]=\kappa_T$$
$$\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{f'(V)}\Big)\Big(\frac{-1}{V}\Big)=\kappa_0[1+\beta_0(T-T_0)]$$
Now we obtain ##f(V)##
\begin{align*}
\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{f'(V)}\Big)\Big(\frac{-1}{V}\Big)&=\kappa_0[1+\beta_0(T-T_0)]\\
\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{\kappa_0[1+\beta_0(T-T_0)]}\Big)\Big(\frac{-1}{V}\Big)&=f'(V)\\
\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{\kappa_0[1+\beta_0(T-T_0)]}\Big)-\ln(V)&=f(V)\\
\frac{-\ln(V)}{\kappa_0(P_0-P)[1+\beta_0(T-T_0)]}&=f(V)
\end{align*}
$$\kappa_T=-\frac{1}{V} \Big(\frac{\partial V}{\partial P}\Big)_{T}=\kappa_0[1+\beta_0(T-T_0)]$$
Now taking ##T=cte##
$$\Big(\frac{1}{P_0-P}\Big) \Big(\frac{\partial P}{\partial V}\Big)_{T}=f'(V)$$
$$\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{f'(V)}\Big)=\Big(\frac{\partial V}{\partial P}\Big)_{T}$$
$$\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{f'(V)}\Big)\Big(\frac{-1}{V}\Big)=-\frac{1}{V} \Big(\frac{\partial V}{\partial P}\Big)_{T}=\kappa_0[1+\beta_0(T-T_0)]=\kappa_T$$
$$\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{f'(V)}\Big)\Big(\frac{-1}{V}\Big)=\kappa_0[1+\beta_0(T-T_0)]$$
Now we obtain ##f(V)##
\begin{align*}
\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{f'(V)}\Big)\Big(\frac{-1}{V}\Big)&=\kappa_0[1+\beta_0(T-T_0)]\\
\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{\kappa_0[1+\beta_0(T-T_0)]}\Big)\Big(\frac{-1}{V}\Big)&=f'(V)\\
\Big(\frac{1}{P_0-P}\Big)\Big(\frac{1}{\kappa_0[1+\beta_0(T-T_0)]}\Big)-\ln(V)&=f(V)\\
\frac{-\ln(V)}{\kappa_0(P_0-P)[1+\beta_0(T-T_0)]}&=f(V)
\end{align*}
That finished, any other idea of finding the equation of state or the constants conditions ?