- #1

Pendleton

- 20

- 3

- Homework Statement
- A clamped piston partitions a container of ideal gas into halves of volume V, the left one at pressure ##P_1## and right at pressure ##P_2 < P_1##, and is kept at temperature T by a reservoir. Were the piston unclamped, moved by δ, and reclamped, how would the entropy of each gas, the reservoir, and universe change?

- Relevant Equations
- $$dU = dQ + dW$$

$$dU = TdS - PdV$$

**Attempted Solution:**

__Gas Entropy__

This system is isothermal: the energy of each gas remains constant.

$$dU = 0$$

By the combined statement of the first and second laws,

$$dU = TdS - PdV$$

Therefore,

$$0 = TdS - PdV$$

$$dS = \frac {PdV}{T}$$

Therefore,

**$$dS_1 = \frac {P_1 dV_1}{T} = \frac {P_1 Aδ}{T}$$**

$$dS_2 = \frac {P_2 dV_2}{T} = \frac {P_2 A(-δ)}{T} = \frac {-P_2 Aδ}{T}$$

$$dS_2 = \frac {P_2 dV_2}{T} = \frac {P_2 A(-δ)}{T} = \frac {-P_2 Aδ}{T}$$

__Reservoir Entropy Change__

By the fundamental thermodynamic relation,

$$TdS - PdV = dQ + dW$$

The reservoir has constant volume and is not worked on.

$$TdS_R - 0 = dQ_R + 0$$

$$dS_R = \frac{dQ_R}{T}$$

The heat absorbed by the reservoir is emitted by the gases.

$$dQ_R = dQ_1 + dQ_2$$

By the first law,

$$dU = dQ + dW$$

$$0 = dQ + dW$$

$$dQ = -dW$$

The work done on each gas is the pressure-volume work done by the other.

$$dW_1 = P_2 dV_2 = P_2 A(-δ) = -P_2 Aδ$$

$$dW_2 = P_1 dV_1 = P_1 Aδ$$

Therefore,

$$dQ_1 = -(-P_2 Aδ) = P_2 Aδ$$

$$dQ_2 = -P_1 Aδ$$

$$dQ_R = P_2 Aδ + (-P_1 Aδ) = (P_2 - P_1)Aδ$$

**$$dS_R = \frac {P_2 - P_1}{T} Aδ$$**

__Universe Entropy Change__

The change of the entropy of the universe is the sum of the changes of the entropy of the gasses and reservoir.

$$dS_U = dS_1 + dS_2 + dS_R$$

$$ = \frac {P_1 Aδ}{T} + \frac {-P_2 Aδ}{T} + \frac {P_2 - P_1}{T} Aδ$$

$$ = \frac{P_1 - P_2}{T} Aδ + \frac {P_2 - P_1}{T} Aδ$$

**$$dS_U = 0$$**