matematikuvol
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If in some thermodynamics system preasure P and temperature T are constant then Gibbs potential has minimum.
G=U-TS+PV
Variation of G is
\delta G=\delta U-T \delta S+P \delta V
U=U(S,V)
\delta U=(\frac{\partial U}{\partial S})_V\delta S+(\frac{\partial U}{\partial V})_S\delta V+\frac{1}{2}(\frac{\partial^2 U}{\partial S^2})_V(\delta S)^2+\frac{1}{2}(\frac{\partial^2 U}{\partial V^2})_S(\delta V)^2+\frac{\partial^2 U}{\partial S\partial V}\delta S \delta V
If we use Maxwell relation we get
\delta G=\frac{1}{2}(\frac{\partial^2 U}{\partial S^2})_V(\delta S)^2+\frac{1}{2}(\frac{\partial^2 U}{\partial V^2})_S(\delta V)^2+\frac{\partial^2 U}{\partial S\partial V}\delta S \delta V
and from here
\delta G>0
Is it true from G is minimum. Or \delta G=0, \delta^2 G>0. Tnx for your answer.
G=U-TS+PV
Variation of G is
\delta G=\delta U-T \delta S+P \delta V
U=U(S,V)
\delta U=(\frac{\partial U}{\partial S})_V\delta S+(\frac{\partial U}{\partial V})_S\delta V+\frac{1}{2}(\frac{\partial^2 U}{\partial S^2})_V(\delta S)^2+\frac{1}{2}(\frac{\partial^2 U}{\partial V^2})_S(\delta V)^2+\frac{\partial^2 U}{\partial S\partial V}\delta S \delta V
If we use Maxwell relation we get
\delta G=\frac{1}{2}(\frac{\partial^2 U}{\partial S^2})_V(\delta S)^2+\frac{1}{2}(\frac{\partial^2 U}{\partial V^2})_S(\delta V)^2+\frac{\partial^2 U}{\partial S\partial V}\delta S \delta V
and from here
\delta G>0
Is it true from G is minimum. Or \delta G=0, \delta^2 G>0. Tnx for your answer.
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