- #1

Pendleton

- 20

- 3

- Homework Statement
- Setup: Consider a container of volume V separated bisected by a piston of cross-sectional 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

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

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

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 pressure-volume 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

dF = -wδ