- #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,
$$dS_1 = \frac {P_1 dV_1}{T} = \frac {P_1 Aδ}{T}$$
$$dS_2 = \frac {P_2 dV_2}{T} = \frac {P_2 A(-δ)}{T} = \frac {-P_2 Aδ}{T}$$
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.
$$dS_U = dS_1 + dS_2 + dS_R$$
$$ = \frac {P_1 Aδ}{T} + \frac {-P_2 Aδ}{T} + \frac {P_2 - P_1}{T} Aδ$$
$$ = \frac{P_1 - P_2}{T} Aδ + \frac {P_2 - P_1}{T} Aδ$$
$$dS_U = 0$$
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,
$$dS_1 = \frac {P_1 dV_1}{T} = \frac {P_1 Aδ}{T}$$
$$dS_2 = \frac {P_2 dV_2}{T} = \frac {P_2 A(-δ)}{T} = \frac {-P_2 Aδ}{T}$$
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.
$$dS_U = dS_1 + dS_2 + dS_R$$
$$ = \frac {P_1 Aδ}{T} + \frac {-P_2 Aδ}{T} + \frac {P_2 - P_1}{T} Aδ$$
$$ = \frac{P_1 - P_2}{T} Aδ + \frac {P_2 - P_1}{T} Aδ$$
$$dS_U = 0$$