Advanced General Physics 323/324
Thermal Physics Unit T3 Thermodynamic Potentials
Prerequisites: T1 and T2
Overview: This unit introduces the "thermodynamic potentials"
called the Helmholtz free energy F and the Gibbs free energy G which, together
with the internal energy U and the enthalpy H form a group of four related
thermodynamic quantities. Which of these will be used in a given situation
depends mainly upon which set of two independent variables one wants to use.
F. W. Sears and G. L. Salinger, Thermodynamics, Kinetic Theory, and
Statistical Thermodynamics (S&S) Chapt. 7 - Thermodynamic Potentials;
Chapt. 8 - Applications of Thermodynamics to Simple Systems.
Comment: This unit involves some rather subtle ideas with
great practical application. Make sure to get the definitions of the new
quantities clear, and the appropriate pair of independent variables in each
After completing this unit you should understand:
- U, the internal energy.
- H, the enthalpy.
- F, the Helmholtz free energy.
- G, the Gibbs free energy.
- What is wrong with this argument? In the Joule-Thomson effect we are told
that no heat enters or leaves the system. Thus dQ = 0 so dS = 0. We also
derive H2 = H1, or dH = 0. But Eq. (37) says that dH =
TdS +VdP, so we must have dP=0. But the pressure must change from one side of
the plug to the other in order that the gas is push through and to have an
effect. Therefore we can't have an effect!
- Find the fallacy in the following argument: The first law says dU = dQ -
PdV. If a system remains at constant volume no work will be done and PdV = 0,
so dU = dQ. According to the definition of specific heat at constant volume dU
= CV dT, so CV = (∂U/∂T)V. Then we
can re-write the first law as dU =CV dT - PdV. In general dU =
(∂U/∂V)T dV + (∂U/∂T)V dT, so
(∂U/∂V)T = -P. However, in the special case of an ideal
gas we know that the internal energy is independent of the volume so
(∂U/∂V)T = 0, not -P. Explain this contradiction.
Chapt. 7: Problem 6; Chapt. 8, Problems 1,6,10;
- What is the Joule-Thomson coefficient for an ideal gas?
- Obtain the expression for the Helmholtz free energy F for one mole
of ideal gas. From this expression find the pressure P in terms of V
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