Thermodynamics Problem: Reversible and Irreversible Processes

[Pendleton]

Homework Statement

Problem 1) Give one reversible and one irreversible example of a process whereby an ideal gas receives heat dQ and during which dW = -PdV.

Problem 2) If it is possible to go from thermodynamic state A of a given system (not necessarily isolated) to thermodynamic state B with an adiabatic process, but it is not possible to go from B to A adiabatically, then we must have S(B) > S(A).

Relevant Equations

$$PV^ γ = P_0 V_0 ^γ$$

Attempt at A Solution

Problem 1
Reversible Process - A cylinder of ideal gas at pressure P is in mechanical equilibrium with a piston of area A driven by a spring of force F = PA and thermal equilibrium with a reservoir of temperature T. The piston is moved a small distance dx toward the spring and clamped.

By the first law,
$$dU = dQ + dW$$

The process is isothermal
$$dU = 0$$
$$dQ = -dW$$

The work done on the gas is the force of the spring times its displacement
$$dW = Fdx$$

The spring force and gas pressure are in equilibrium.
$$F + PA = 0$$
$$F = -PA$$

Therefore
$$dW = -PAdx$$
$$dQ = PAdx$$

This process is reversible because unclamping the system, letting it return to equilibrium, would allow an opposite heat flow to return it to its original state through a spontaneous reverse process.
$$dW = F(-dx)$$
$$dW = -PA(-dx)$$
$$dW = PAdx$$
$$dQ = -PAdx$$

Irreversible Process - The same system, only with a hot reservoir that rapidly heats and thereby expands the gas.

Problem 2
If state B can be reached adiabatically from state A, then one or more adiabatic processes connect them. If any of those processes was reversible, then state A could be reached from state B adiabatically. However, state A cannot be reached from state B adiabatically. Therefore, every adiabatic process between A and B is irreversible.

Entropy is a state function. Therefore, the difference of entropy between two states is the same regardless of the process between them. Therefore, all possible processes between two states must change the entropy equivalently. Therefore, we may calculate the difference of entropy between two states by identifying a process between them. Here, we have recognized that an irreversible adiabatic process links A to B. Irreversible processes increase entropy. Therefore, the entropy of state B is greater than that of state A.

 

[Chestermiller]

Homework Statement:: Problem 1) Give one reversible and one irreversible example of a process whereby an ideal gas receives heat dQ and during which dW = -PdV.

Problem 2) If it is possible to go from thermodynamic state A of a given system (not necessarily isolated) to thermodynamic state B with an adiabatic process, but it is not possible to go from B to A adiabatically, then we must have S(B) > S(A).
Relevant Equations:: $$PV^ γ = P_0 V_0 ^γ$$

Attempt at A Solution

Problem 1
Reversible Process - A cylinder of ideal gas at pressure P is in mechanical equilibrium with a piston of area A driven by a spring of force F = PA and thermal equilibrium with a reservoir of temperature T. The piston is moved a small distance dx toward the spring and clamped.

By the first law,
$$dU = dQ + dW$$

The process is isothermal
$$dU = 0$$
$$dQ = -dW$$

The work done on the gas is the force of the spring times its displacement
$$dW = Fdx$$

The spring force and gas pressure are in equilibrium.
$$F + PA = 0$$
$$F = -PA$$

Therefore
$$dW = -PAdx$$
$$dQ = PAdx$$

This process is reversible because unclamping the system, letting it return to equilibrium, would allow an opposite heat flow to return it to its original state through a spontaneous reverse process.
$$dW = F(-dx)$$
$$dW = -PA(-dx)$$
$$dW = PAdx$$
$$dQ = -PAdx$$

Irreversible Process - The same system, only with a hot reservoir that rapidly heats and thereby expands the gas.

Problem 2
If state B can be reached adiabatically from state A, then one or more adiabatic processes connect them. If any of those processes was reversible, then state A could be reached from state B adiabatically. However, state A cannot be reached from state B adiabatically. Therefore, every adiabatic process between A and B is irreversible.

Entropy is a state function. Therefore, the difference of entropy between two states is the same regardless of the process between them. Therefore, all possible processes between two states must change the entropy equivalently. Therefore, we may calculate the difference of entropy between two states by identifying a process between them. Here, we have recognized that an irreversible adiabatic process links A to B. Irreversible processes increase entropy. Therefore, the entropy of state B is greater than that of state A.

I am not going to discuss Problem 1 because I am opposed to using differentials for irreversible processes.

For problem 2, I would add that, if the entropy of state B is higher than state A, then, if you could use an adiabatic irreversible path to go from state B to state A, the entropy would increase even more, which would be incompatible with A having a lower entropy than B. If you used a reversible path to go from B to A, you would have to remove heat to decrease the entropy back to that of A, so the process could not be adiabatic.