MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
Problem Set 3 Solutions
1. The Omega Potential is a Legendre transformation given by:
Ω = U - µN
Derive the differential equation, functionality (e.g., Ω = Ω (?,?,?)), the coefficient
relations and Maxwell relations.
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
2. Derive a Legendre transformation if electrical work is done on the system where:
= φ dq
dw
where φ is the voltage potential and q is the fundamental charge on an electron. Assume
that both mechanical work and chemical work are also present. Derive the differential
equation, functionality, the coefficient relations and Maxwell relations.
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
3. Extensive variables can be directly measured while intensive variables cannot be directly
measured. Intensive variables need to be determined indirectly by measuring an
extensive variable(s). Detail an example of determining the value for an intensive
variable.
A Hg thermometer is a common example of indirectly
measuring an intensive variable (temperature) by directly
measuring an extensive variable (volume). The system is
closed, but the volume of the thermometer is larger than
the volume of the system, so the volume of the Hg is
allowed to increase without increasing the pressure of the
system (at least to first order). Thus, we can assume that
the pressure is constant. The Hg volume change is directly
proportional to the Hg temperature change thus providing a
means to indirectly measure the temperature of the system's
surroundings.
4. Using equation 2.8 on page 24 of Gaskell, show for an ideal gas that:
c p − cv = nR
where R is the universal gas constant and n is the number of moles in a system.
For an ideal gas, dU = 0 (p. 25), and the equation of state
is given by:
pV = nRT
The first term on the right of equation 2.8 is just:
nR
⎛ ∂V ⎞
.
⎜
⎟ =
⎝ ∂T ⎠ p P
The 2nd term in the parentheses on the right side of
equation 2.8 is 0 since dU = 0.
Thus, the right side of equation 2.8 is:
nR
c p − cv =
( P + 0)
P
= nR
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
5. Using the state function of entropy as a function of T and P we developed in class,
determine the change of entropy of 3 moles of lead when it is initially at 590K to 600K at
1 atm. Note that Cp(s) = 9.75x10-3 T [J/(K mole)].
See page 63- 64 Gaskell. Cp(l) = 32.4 – 3.1x10-3 T (J/K mole)) Constant pressure thus:
600 c p
∆S = ∫
dT
590 T
Answer: