§2. The Thermodynamics Properties of Homogeneous Substance

§2. The Thermodynamics Properties of Homogeneous Substance §2.1 The complete differential of internal energy, enthalpy, free energy and Gibbs function dU=TdS-pdV (2.1.1) According to the definition of enthalpy H=U+pV, then dH=TdS+VdP (2.1.2) According to the definition of free energy F=U-TS,then dF= -SdT-pdV (2.1.3) For Gibbs function G=U-TS+pV, by the same wayitgets dG= -SdT+Vdp (2.1.4)

The complete differential of U= U (S, V ) Comparing with expression (2.1.1) , we have Since so that Based on the expression (2.1.2), it gets that

Based on the expression (2.1.3), it gets that similarly The expression (2.1.6), (2.1.8), (2.1.10) and (2.1.12) called Maxwell equations

§2.2 The Simple Application of Maxwell Relations

2.2.1 T, V as independent variable The complete differential of U= U (T, V ) follows that and Thus it gets Comparing with above two expressions, it gets

For example: for the ideal gas For Van der Waals’ gas

2.2.2 T, p as independent variable The complete differential of H= H (T, V ) follows as and Thus it gets Comparing with above two expressions, it gets

From (2.2.5) and (2.2.8), one obtains that From S(T, p)=S(T, V(T, p)), One obtains that

For example: for ideal gas For any simple system Here α is the coefficient of expansion，kT is the compressibilityas constant temperature.

Example 1: Show that the ratio of adiabatic compressibility kS to isothermal compressibility kT equals to the ratio of isopyknic heat capacity to isotonic heat capacity. Solution The kS and kT are just So

Example 2: Show Solution

§2.3 The Processes of Throttle Expansion and Adiabatic Expansion 2.3.1 The process of throttle expansion Porous plug Based on U2-U1=p1V1-p2V2 H1=H2 (2.3.1) Remake! The processes of throttle expansion is a irreversible process. suppose that the pressure change is very small, then

Let μcalled Joule –Tomson coefficient From We obtain Joule-Thomson coefficient

For ideal gas so For practical gas a curve called inversion curve corresponding to T is called reverse-temperature, μ >0, corresponding to refrigeration region μ <0, corresponding to heat region • The complete differential of State function enthalpy H=H（T， p） • And • With expressions (2.2.8) and (2.2.10), then • Or

2.3.2 The process of adiabatic expansion For quasi-static adiabatic process it follows that So The decreasing of pressure leads to the decreasing of temperature.

§2.4 The determine of Basic Thermodynamics Function 2.4.1 T, V as independent variable , p = p(T, V) From We obtain which is called integral expression of internal energy. From We obtain the integral expression of entropy

2.4.2 T, p as independent variable, V=V (T, p) With it follows that with Similarly, we obtain

Example 1: Calculate the enthalpy , entropy and Gibbs function of a ideal gas as a function of temperature and pressure . Solution The equation of state for one mol ideal gas reads pv=RT we obtain So mol enthalpy follows that h=cpT+h0 (2.4.11’) and mol entropy s= cpln p- T ln p+s0 (2.4.12’)

with (2.4.11) and (2.4.12) the mol Gibbs function holds that and thus g = cpT-cpT ln T+R T ln p+h0 –T s0 (2.4.12’) There is another expression. Based on let it follows that Generally g is written by g=RT(φ+ln p)(2.4.15) Here

§2.5 Speciality Function Massieu proved that the basic thermodynamics function could be calculated with one thermodynamics function called speciality function. Based on dF=-SdT – pdV (2.5.1) so The expression (2.5.3) is called Gibbs-Helmholtz equation.

The complete differential of Gibbs function is dG=-SdT+VdP (2.5.4) So it gets With H=U+pV, we obtain that The expression (2.5.7) is also called Gibbs-Helmholtz equation.

Example : Calculate the thermodynamics function of a surface system Solution The state equation state of surface system f(σ, A, T)=0 (2.5.8) which can be simply written by σ = σ(T) The complete differential of free energy follows that dF=-SdT –σdA (2.5.9) and thus

Since so that F = σA (2.5.11) The entropy of the system and the free energy of the system

The transformation of all variables Thermodynamics rectangle

§2.6 Thermodynamics Theory of Thermal Radiation Two cavums Remark! The density of energy and the distribution of density via frequency are only the function of temperature. In 1901, the relation between radiation pressure p and density of energy u was given as follows Based on U( T, V ) = u(T)V (2.6.2) and

We obtain that We integrate and obtain With we have After integral we obtain Since S = constant in reversible adiabatic process, so that If one insert this into G=U-TS+pV , one gets G=0 (2.6.6)

Absolute darkbody; cavum ; blackbody radiation Radiant flux density Ju: the total radiant energy sideward through unit area in unit of time. Show the relation between radiant flux density Ju and density of radiant energy u as follows Solution The radiant energy sideward through area dA in unit of time holds that .

the total radiant energy sideward through area dA in unit of time follows that It is in a good agreement with the expression (2.6.7) Insert the expression (2.6.7) with the expression (2.6.3), it gets This is the Stefan-Boltzmann law. called Stefan- Boltzmann constant

§2.7 Thermodynamics in Magnetic Medium When magnetic field intensity and magnetization change in magnetic medium, work is performed by environment When thermodynamics system does not include magnetic field, the work follows that Here M=mV called the total magnetic moment of medium. Suppose the volume change of the system is neglected , thus Gibbs function

From the complete differential condition, it gets that Based on S=S(T, H), then or

When magnetic field does not change, heat capacity holds as follows Suppose medium obeying the Courier law and with formula (2.7.7), we obtain that

Adiabatic magnetic reversal refrigeration effect Suppose volume change, then Gibbs function is From the complete differential condition, it gets that This is one Maxwell relation, denotes the relation of magnetostriction effect with pressure magnetism effect.

One sample move from infinitude far x=∞ to x=a in magnetic field introducing by permanent magnet, work is performed by environment The first term is potential，the second term is work. Internal energy holds that

§2. The Thermodynamics Properties of Homogeneous Substance