# Thermodynamics Problem: Reversible and Irreversible Processes

• Pendleton
In summary, for Problem 1, there is a reversible process where an ideal gas receives heat and the work done is equal to the negative of the pressure times the change in volume. For Problem 2, if it is possible to go from state A to state B adiabatically but not vice versa, then the entropy of state B must be higher than that of state A. This is because any process between two states must change the entropy equivalently, and an irreversible adiabatic process from A to B would increase the entropy, making it impossible to reach state A adiabatically.
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.

Last edited:
Pendleton said:
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.

## 1. What is the difference between a reversible and irreversible process in thermodynamics?

A reversible process is one in which the system and its surroundings can be returned to their original states by reversing the direction of the process. This means that at every step, the system is in equilibrium with its surroundings. On the other hand, an irreversible process is one in which the system and its surroundings cannot be returned to their original states by reversing the direction of the process. This is because irreversible processes involve energy loss, such as heat transfer to the surroundings.

## 2. How does entropy play a role in reversible and irreversible processes?

Entropy is a measure of the disorder or randomness in a system. In a reversible process, the entropy of the system and its surroundings remains constant, as the process can be reversed and the system returns to its original state. In an irreversible process, the entropy of the system and its surroundings increases, as energy is lost and the system becomes more disordered.

## 3. Can all thermodynamic processes be reversed?

No, not all thermodynamic processes can be reversed. Reversibility depends on the conditions under which the process occurs. For example, a process that involves friction or heat transfer to the surroundings is irreversible, as it cannot be reversed without external influence. However, under ideal conditions, some processes can be reversible.

## 4. How do reversible and irreversible processes affect the efficiency of a system?

Reversible processes are more efficient than irreversible processes, as they involve minimal energy loss. This is because reversible processes occur slowly and at every step, the system is in equilibrium with its surroundings. On the other hand, irreversible processes involve energy loss, leading to a decrease in efficiency.

## 5. Can a reversible process occur in real life?

In theory, yes, a reversible process can occur in real life. However, in practice, it is difficult to achieve perfect reversibility due to factors such as friction, heat transfer, and other forms of energy loss. Engineers and scientists strive to design systems and processes that are as close to reversible as possible, in order to increase efficiency and reduce energy loss.

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