 #1
Pendleton
 20
 3
 Homework Statement:

Setup: Consider a container of volume V separated bisected by a piston of crosssectional area A under a weight w. The piston is clamped midway, and the lower half of the container is filled with an ideal gas. The system (gas + weight) is in such thermal equilibrium with a reservoir at temperature T that P < w/A, where P is the gas pressure. The clamp is then released, and the piston descends a small distance δ, after which it is clamped again and the system equilibrates.
Question 1: During this process, what is the heat dQ transferred the reservoir to the ideal gas system? Correspondingly what is the entropy change of the system dS(s) and the entropy change for the universe dS(u)?
Question 2: What is the corresponding change dF of the Helmholtz Free Energy of the system?
 Relevant Equations:

First Law: dU = dQ + dW
Combined First and Second Laws: dU = TdS  PdV
Fundamental Relation: TdS  PdV = dQ + dW
Helmholtz Free Energy: dF = SdT  PdV + E(Xdx)
Attempt at a Solution:
Heat Absorbed By The System
By the first law of thermodynamics,
dU = dQ + dW
The system is of fixed volume and therefore mechanically isolated.
dW = 0
Therefore
dQ = dU
The change of energy of the system equals the change of energy of the gas plus the change of energy of the mass.
dU = dU(g) + dU(m)
The change of energy of the gas is zero because its temperature is constant.
dU(g) = 0
Therefore, the change of energy of the system equals the change of energy of the mass.
dU = dU(m)
The energy of the mass decreases by its weight w times its descent δ.
dU(m) = wδ
Therefore the change of energy of the system is
dU = wδ
Therefore, the heat absorbed by the system is
dQ = wδ
Change of Entropy of The System
By the combined first and second law,
dU = TdS  PdV
The volume of our system is fixed.
dU = TdS
Recall that the change of energy of our system is
dU = wδ
Therefore,
wδ = TdS
Therefore, the change of entropy of our system is
dS_{s} = wδ/T
Change of Entropy of The Universe
By the fundamental thermodynamic relation,
TdS  PdV = dQ + dW
The reservoir is mechanically isolated.
dV = 0, dW = 0
Therefore,
TdS = dQ
The heat absorbed by the reservoir is the opposite of that absorbed by the system.
dQ = (wδ) = wδ
Therefore,
TdS = wδ
Therefore, the change of entropy of the universe equals
dS_{u} = wδ/T
Change of Helmholtz Free Energy of The System
By the definition of change of Helmholtz free energy,
dF = SdT PdV + E(Xdx)
where E(Xdx) is the sum of the products of each non pressurevolume force X and its displacement dx. Here, that is just the weight of the mass times its descent δ.
E(Xdx) = wδ
Therefore,
dF = SdT PdV wδ
The temperature and volume of the system are constant.
dT = 0, dV = 0
Therefore, the change of the Helmholtz free energy of the system is
dF = wδ
Heat Absorbed By The System
By the first law of thermodynamics,
dU = dQ + dW
The system is of fixed volume and therefore mechanically isolated.
dW = 0
Therefore
dQ = dU
The change of energy of the system equals the change of energy of the gas plus the change of energy of the mass.
dU = dU(g) + dU(m)
The change of energy of the gas is zero because its temperature is constant.
dU(g) = 0
Therefore, the change of energy of the system equals the change of energy of the mass.
dU = dU(m)
The energy of the mass decreases by its weight w times its descent δ.
dU(m) = wδ
Therefore the change of energy of the system is
dU = wδ
Therefore, the heat absorbed by the system is
dQ = wδ
Change of Entropy of The System
By the combined first and second law,
dU = TdS  PdV
The volume of our system is fixed.
dU = TdS
Recall that the change of energy of our system is
dU = wδ
Therefore,
wδ = TdS
Therefore, the change of entropy of our system is
dS_{s} = wδ/T
Change of Entropy of The Universe
By the fundamental thermodynamic relation,
TdS  PdV = dQ + dW
The reservoir is mechanically isolated.
dV = 0, dW = 0
Therefore,
TdS = dQ
The heat absorbed by the reservoir is the opposite of that absorbed by the system.
dQ = (wδ) = wδ
Therefore,
TdS = wδ
Therefore, the change of entropy of the universe equals
dS_{u} = wδ/T
Change of Helmholtz Free Energy of The System
By the definition of change of Helmholtz free energy,
dF = SdT PdV + E(Xdx)
where E(Xdx) is the sum of the products of each non pressurevolume force X and its displacement dx. Here, that is just the weight of the mass times its descent δ.
E(Xdx) = wδ
Therefore,
dF = SdT PdV wδ
The temperature and volume of the system are constant.
dT = 0, dV = 0
Therefore, the change of the Helmholtz free energy of the system is
dF = wδ