T 1 = 350 K and T i
(2) = T 2 = 250 K. The adiabatic wall separated the two
subsystems is then replaced with a diathermic wall. Consider a quasi-static
process, i.e., after a given amount of heat (e.g., ΔU
(2) ) transferred from
subsystem
(1) to subsystem
(2) through the diathermic wall, the original adiabatic wall is placed back as separation wall so that the two subsystems
individually will “relax” into thermal equilibrium internally within each
subsystem. The “replacement with diathermic wall and the placing-back of
adiabatic wall” quasi-static step is then repeated a sufficient number of times
until the two systems reach thermal equilibrium at the final equilibrium state.
Construct the quasi-static process toward thermal equilibrium of T f
(1) = T f
(2) ;
determine the equilibrium states of both subsystems.
9:6 Callen [1:53–54] wrote
The hypothetical problem of equilibrium in a closed composite system with
an internal movable adiabatic wall is a unique indeterminate problem.
Physically, the release of the piston would lead it to perpetual oscillation in
the absence of viscous damping. With viscous damping, the piston would
eventually come to rest at such a position that the pressures on either side
would be equal, but the temperatures in each subsystem would then depend
on the relative viscosity in each subsystem. The solution to this problem
depends on dynamical considerations [underline added]. Show that the
application of the entropy maximum formalism is correspondingly indeterminate with respect to the temperatures (but determinate with respect to the
pressures).
Consider one such composite system with He gas in subsystem 1, and N 2 gas
in subsystem 2. The initial conditions of the two subsystems are given as
N
(1) = 0.5 gm-mole and N
(2) = 0.75 gm-mole, T i
(1) = 200 K and T i
(2) = 300
K, V i
(1) = 12 L and V i
(2) = 8 L. Remove the stopper to the piston and the
composite system will move towards mechanical equilibrium.
Viscous damping is the key to understanding the process. Instead permitting
the piston to oscillate, one can construct a quasi-equilibrium process similar
to Problem 9.5: imagine a set of large number of stoppers to the piston;
remove the first stopper; permitting the piston to move only a small distance
to the next stopper; then permitting the two subsystems to relax to internal
thermodynamic equilibrium individually; determine the individual equilibrium states; then repeat the next sequence of steps until p f
(1) = p f
(2) . Determine
the final pressures and temperatures of both subsystems.
HINT: Each quasi-equilibrium step is a two-phase step: the first phase leads
to isentropic heating and cooling in subsystems 1 and 2, respectively; then
the eddy kinetic energy in each subsystem will dissipate into internal energy
in the second phase; the consideration of the second phase requires the
auxiliary assumption of how the eddy kinetic energy is distributed in the two
subsystems (you can make your own arbitrary auxiliary assumption).
9:7 Two kmols of H 2 O are enclosed in a rigid vessel and heated to a temperature
of 2000 K and a pressure of 1 MPa.
272
9 Applications to Special States of Thermodynamic Equilibrium …
(2) = T 2 = 250 K. The adiabatic wall separated the two
subsystems is then replaced with a diathermic wall. Consider a quasi-static
process, i.e., after a given amount of heat (e.g., ΔU
(2) ) transferred from
subsystem
(1) to subsystem
(2) through the diathermic wall, the original adiabatic wall is placed back as separation wall so that the two subsystems
individually will “relax” into thermal equilibrium internally within each
subsystem. The “replacement with diathermic wall and the placing-back of
adiabatic wall” quasi-static step is then repeated a sufficient number of times
until the two systems reach thermal equilibrium at the final equilibrium state.
Construct the quasi-static process toward thermal equilibrium of T f
(1) = T f
(2) ;
determine the equilibrium states of both subsystems.
9:6 Callen [1:53–54] wrote
The hypothetical problem of equilibrium in a closed composite system with
an internal movable adiabatic wall is a unique indeterminate problem.
Physically, the release of the piston would lead it to perpetual oscillation in
the absence of viscous damping. With viscous damping, the piston would
eventually come to rest at such a position that the pressures on either side
would be equal, but the temperatures in each subsystem would then depend
on the relative viscosity in each subsystem. The solution to this problem
depends on dynamical considerations [underline added]. Show that the
application of the entropy maximum formalism is correspondingly indeterminate with respect to the temperatures (but determinate with respect to the
pressures).
Consider one such composite system with He gas in subsystem 1, and N 2 gas
in subsystem 2. The initial conditions of the two subsystems are given as
N
(1) = 0.5 gm-mole and N
(2) = 0.75 gm-mole, T i
(1) = 200 K and T i
(2) = 300
K, V i
(1) = 12 L and V i
(2) = 8 L. Remove the stopper to the piston and the
composite system will move towards mechanical equilibrium.
Viscous damping is the key to understanding the process. Instead permitting
the piston to oscillate, one can construct a quasi-equilibrium process similar
to Problem 9.5: imagine a set of large number of stoppers to the piston;
remove the first stopper; permitting the piston to move only a small distance
to the next stopper; then permitting the two subsystems to relax to internal
thermodynamic equilibrium individually; determine the individual equilibrium states; then repeat the next sequence of steps until p f
(1) = p f
(2) . Determine
the final pressures and temperatures of both subsystems.
HINT: Each quasi-equilibrium step is a two-phase step: the first phase leads
to isentropic heating and cooling in subsystems 1 and 2, respectively; then
the eddy kinetic energy in each subsystem will dissipate into internal energy
in the second phase; the consideration of the second phase requires the
auxiliary assumption of how the eddy kinetic energy is distributed in the two
subsystems (you can make your own arbitrary auxiliary assumption).
9:7 Two kmols of H 2 O are enclosed in a rigid vessel and heated to a temperature
of 2000 K and a pressure of 1 MPa.
272
9 Applications to Special States of Thermodynamic Equilibrium …