Work done by irreversible and reversible processes

In summary: Yes. But, if you specify that your system is the gas, the ##P_{ext}## is the pressure on the inside face of the piston. And, if you specify that your system is the gas plus piston, the ##P_{ext}## is the pressure on the outside face of the piston, but the kinetic energy change of the piston needs to be included. Of course, by the time you reach final steady state, the...
  • #1
Hamiltonian
296
190
I am not able to follow the derivation of work done in a reversible and irreversible process as I don't get why the work done should be different in the two processes.
a reversible process is said to be a process that occurs infinitesimally slowly and an irreversible process goes from initial to the final state in a single step in finite time and cannot be reversed.

I don't understand why an irreversible process can't be reversed if you consider the example of a cylinder containing a gas onto which a piston is placed if the piston is pushed down and taken from a height I to a height F infinitesimally slowly it is said to be reversible but if it is taken in one step it is said to be irreversible.
why can't the process be reversed if the piston is taken down in a single step?

either way shouldn't the work done be the same?

my book says work done by an irreversible process is $$P\Delta V$$ and work done by an isothermal reversible process is $$ -2.303nRT\log(Pi/Pf) $$
 
Physics news on Phys.org
  • #2
For the isothermal reversible process, maybe you have a bunch of tiny weights on top of the piston, which you remove one by one to facilitate the reversible expansion of the gas inside. The internal and external forces per unit area on the piston face are always equal and the work done by the gas is$$W = \int_{V_1}^{V_2} P dV = \int_{V_1}^{V_2} \frac{nRT}{V} dV = nRT \ln\frac{V_2}{V_1} = nRT \ln \frac{P_1}{P_2} \quad \left( \approx 2.303nRT\log{\frac{P_1}{P_2}}\right)$$The pressures in the above could be the internal or external pressures to the piston, since they're equal at all times. For the irreversible expansion, however, the force per unit area at the inside piston face is not given by the ideal gas law (since we need to take into account other factors, like viscous stresses in the gas), so instead we deal with the external pressure. If you compress the gas, hold the piston in place and suddenly release the piston, the atmospheric pressure on the external face is considered constant and the work done by the gas during the irreversible expansion, between two equilibrium states, is just$$W = \int_{V_1}^{V_2} P_{\text{ext}} dV = P_{\text{ext}} \Delta V$$(N.B. The force on the inside face is some function ##\vec{F}(t)##, but if you apply the work-energy theorem to the piston between the equilibrium states you'll notice that ##\int \vec{F} \cdot d\vec{r} - P_{\text{ext}} \Delta V = 0##, i.e. the work done by the gas inside is ##P_{\text{ext}} \Delta V##).

It might be easier to visualise if you sketch the two paths on the PV plane, and look at the difference in areas between the two paths :smile:
 
Last edited by a moderator:
  • Like
Likes vanhees71
  • #3
etotheipi said:
(N.B. The force on the inside face is some function ##\vec{F}(t)##, but if you apply the work-energy theorem to the piston between the equilibrium states you'll notice that ##\int \vec{F} \cdot d\vec{r} - P_{\text{ext}} \Delta V = 0##, i.e. the work done by the gas inside is ##P_{\text{ext}} \Delta V##).
For a massless-frictionless piston, the force F the gas exerts on the inside piston face does not only integrate in this way. At all times during the irreversible process, ##F=P_{ext}A##. The viscous stresses and rate of piston movement combine in such a way that this happens.
 
  • Like
Likes vanhees71 and etotheipi
  • #4
Chestermiller said:
For a massless-frictionless piston, the force F the gas exerts on the inside piston face does not only integrate in this way. At all times during the irreversible process, ##F=P_{ext}A##. The viscous stresses and rate of piston movement combine in such a way that this happens.

For the massless piston that has to be true, since the net force on a massless piston must be zero otherwise the acceleration would diverge. But for the massive piston, there must be an imbalance in force per unit area on either side in order to accelerate and then decelerate the piston, so the force per unit area on the inside face would need to vary with time. Is that right?
 
  • #5
etotheipi said:
For the massless piston that has to be true, since the net force on a massless piston must be zero otherwise the acceleration would diverge. But for the massive piston, there must be an imbalance in force per unit area on either side in order to accelerate and then decelerate the piston, so the force per unit area on the inside face would need to vary with time. Is that right?
Yes. But, if you specify that your system is the gas, the ##P_{ext}## is the pressure on the inside face of the piston. And, if you specify that your system is the gas plus piston, the ##P_{ext}## is the pressure on the outside face of the piston, but the kinetic energy change of the piston needs to be included. Of course, by the time you reach final steady state, the viscous stresses will have damped the piston motion, and your equation will apply, but with the upper limit of integration being infinite time.
 
  • Like
Likes vanhees71 and etotheipi
  • #6
Thanks, yes I completely agree with all of that. You said it much better than I did :smile:
 

Related to Work done by irreversible and reversible processes

1. What is the difference between work done by irreversible and reversible processes?

The main difference between work done by irreversible and reversible processes is the efficiency of the process. In an irreversible process, some energy is lost as heat and cannot be recovered, resulting in a lower efficiency. In a reversible process, the system is able to return to its original state and no energy is lost, resulting in a higher efficiency.

2. How is work done by irreversible and reversible processes calculated?

The work done by a reversible process can be calculated using the formula W = -PΔV, where P is the pressure and ΔV is the change in volume. For an irreversible process, the work done is calculated using the formula W = -∫PdV, which takes into account the varying pressure and volume throughout the process.

3. Can work done by an irreversible process be reversed?

No, the work done by an irreversible process cannot be reversed. This is because some energy is lost as heat during the process, making it impossible to return the system to its original state. This is why irreversible processes are considered less efficient than reversible processes.

4. What are some examples of irreversible processes?

Some examples of irreversible processes include friction, combustion, and diffusion. These processes involve a transfer of energy from one form to another, resulting in a loss of energy as heat and a decrease in efficiency.

5. How can reversible processes be achieved in real-world situations?

Reversible processes are idealized and cannot be achieved in real-world situations. However, by minimizing energy losses and using efficient systems, we can approach reversible processes. This is why many industrial and engineering processes are constantly striving to increase efficiency and reduce energy losses.

Similar threads

Replies
4
Views
2K
Replies
27
Views
8K
Replies
56
Views
3K
  • Mechanics
Replies
5
Views
2K
  • Mechanics
Replies
3
Views
7K
  • Introductory Physics Homework Help
Replies
14
Views
791
  • Other Physics Topics
Replies
8
Views
2K
Replies
22
Views
2K
  • Thermodynamics
Replies
3
Views
821
Replies
12
Views
1K
Back
Top