Kelvin said:
Is "the partial derivative of internal energy by entropy" is a more fundamental definition of temperature? Is the expression
KE = 3/2 kT
meaningful when the system is in equilibrium? thermodynaimcs parameters are pressure, volume & temperature only?
Q_Goest:
Is that means my teacher was wrong or probably I made some mistakes when jotting notes?
Yes,u can bring into discussion temperature only in equilibrium states,or in states not far from the equlilibrium states.That T in the eq.
T=\frac{2}{3}\frac{E}{k}
is the same temperature with the one dicussed in the thermodynamics of equilibrium processes.It is called KINETIC TEMPERATURE AND IS DENOTED BY
\Theta =:\frac{2}{i}\frac{E}{2}
.However,because this kinetic temperature,in the case of statistical systems in equilibrium,coincides with the absolute termodynamic temperature (denoted by T and measured in Kelvin),it is denoted like the latter,viz.with T.
Temperature is a statistical quantity.That's because entire thermodynamics of reversible/equilibrium processes (in either formulation,but the neogibbsian is more easy to use) contains the same results as a subtheory of statistical physics of equlibrium processe called 'statistical thermodynamics'.
In the neogibbsian formulation of thermodynamics,temperature is defined implicitely by:
\frac{1}{T}(U,\{X_{i}\})=:(\frac{\partial S(U,\{X_{i}\})}{\partial U})_{\{X_{i}\} (1)
,which is bsically the same with its definition within the microcanonical ensemble (classical or quantum) of statistical mecanics of equilibrium processes:
\frac{1}{T}(E,\{X_{i}\})=:(\frac{\partial S(E,\{X_{i}\})}{\partial U})_{\{X_{i}\} (2)
,where E is the value of the Hamiltonian,assumed fixed at macroscopical level.It is actually the internal energy from thermodynamics.
Daniel.
