Why is there less loss of energy in reversible process?

I found this:

The reversible expansion does the maximum amount of work because the gas is pushing against the maximum possible external pressure.

Why does pushing against maximum possible external pressure result in less loss of energy (or maximum work)?

  • $\begingroup$ Does this assume that the volume change in the reversible case is the same as in the irreversible case? $\endgroup$ – Chet Miller Jan 2 at 3:56
  • $\begingroup$ Who says energy is being lost? $\endgroup$ – Chet Miller Jan 2 at 17:01
  • $\begingroup$ In general, energy released by a reversible process can do the maximum amount of work because less of the energy is lost as heat. A process that is done quickly (irreversibly) tends to generate turbulence and friction resulting in heat loss to the surroundings. Source:usna.edu/Users/chemistry/morse/_files/documents/SC112-Chapter18/… $\endgroup$ – Ardent Jan 3 at 7:16
  • $\begingroup$ I don't think it is a matter of heat being lost to the surroundings. I think it is a matter of less heat being gained from the surroundings in the irreversible process. I will add an answer looking specifically at this. $\endgroup$ – Chet Miller Jan 3 at 13:29

Let's consider as an example to case of an ideal gas in a cylinder receiving heat from a reservoir that is held throughout the process at the same temperature T as the initial temperature of the gas (so that there is no temperature change for the gas between its initial state and its final state). In the irreversible process, we will drop the external pressure suddenly from the initial pressure $P_1$ to the final pressure $P_2<P_1$, and will hold the external pressure constant at this value throughout the process. In the reversible process, we will carry out the expansion in such a way that the external pressure is lowered very gradually from its initial value to its final value. Since the initial and final temperatures of the gas are the same in the reversible case and the irreversible case, we have for both cases that the change in internal energy of our ideal gas is zero ($\Delta U=0$) and the heat added from the reservoir is equal to the work done by the gas.


In the reversible case, we have that $$Q_{rev}=\int_{V_1}^{V_2}{P_{ext}dV}=\int_{V_1}^{V_2}{\frac{nRT}{V}dV}=nRT\ln{(V_2/V_1)}=P_1V_1\ln{(V_2/V_1)}=P_1V_1\ln{(P_1/P_2)}$$


In the irreversible case, we have that $$Q_{irrev}=P_2(V_2-V_1)=\frac{P_1V_1}{V_2}(V_2-V_1)=P_1V_1\left(1-\frac{V_1}{V_2}\right)=P_1V_1\left(1-\frac{P_2}{P_1}\right)$$


$$\frac{Q_{irrev}}{Q_{rev}}=\frac{\left(1-\frac{P_2}{P_1}\right)}{\ln{(P_1/P_2)}}$$ If one evaluates the right hand side of this equation for any value of the pressure ratio $P_2/P_1<1$, one finds that the ratio of the irreversible heat absorbed to the reversible heat absorbed is always less than unity. This means that the irreversible process is able to absorb less heat during the process (and does less work) than the reversible process. The physical mechanism for why this happens is that viscous friction (enhanced by turbulence) converts some of the mechanical energy available from the pressurized gas into internal energy, so less heat absorption is required to maintain the gas temperature (and internal energy) at its initial value in the irreversible process.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.