- #1
matematikuvol
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When people do Legendre transforms they suppose that [tex]U=U(S,V)[/tex]. But you can see in some books that heat is defined by:
[tex]dQ=(\frac{\partial U}{\partial P})_{V}dP+[(\frac{\partial U}{\partial V})_P+P]dV[/tex]
So they supposed obviously that [tex]U=U(V,P)[/tex].
In some books you can that internal energy is function of [tex]T,P[/tex], and in some books function of [tex]V,T[/tex]. Why then in definition of Legendre transforms of thermodynamics potential we use [tex]U=U(S,V)[/tex]. Tnx for the answer.
[tex]dQ=(\frac{\partial U}{\partial P})_{V}dP+[(\frac{\partial U}{\partial V})_P+P]dV[/tex]
So they supposed obviously that [tex]U=U(V,P)[/tex].
In some books you can that internal energy is function of [tex]T,P[/tex], and in some books function of [tex]V,T[/tex]. Why then in definition of Legendre transforms of thermodynamics potential we use [tex]U=U(S,V)[/tex]. Tnx for the answer.