# What is the difference between reversible and irreversible work done in adiabatic process?

The work done in an adiabatic reversible process is given by: $$W = R\frac{\Delta T}{1-\gamma}$$

The work done in an irreversible process is given by:

$$W = P_{ext}\Delta V$$

By the 1st law of thermodynamics, the change in internal energy must be equal to the work done. The change in internal energy is given by:

$$\Delta U = nC_v\Delta T$$

where $C_v$ is the molar heat capacity at constant volume.

Can't this equation be applied to both the processes? Why are there different formulae for the two types of adiabatic processes?

• Even the formula for reversible expansion, is pooison's ratio is expanded, we get the same result as heat capacity*change in temperature. Mar 17, 2016 at 17:14
• – user7951
Mar 17, 2016 at 17:19
• Forget about the book. Show us how YOU would solve the entire adiabatic expansion problem of an ideal gas (a) for the reversible case and (b) for the irreversible case in which the external pressure is held constant. Let's see what you can do. Mar 17, 2016 at 17:56
• Lets say 1 mol of gas in a piston expanded from v1 litre to v2 litre. Now we want to calculate the work done. a) work done= -p(external)*dv but p(ext)=p(gas) for reversible process. PV^gamma(poisson's ratio)=k(some constant). P=k/v^gamma. Putting in work equation and integration. Rest answer in next comment. Mar 17, 2016 at 18:11
• Work=-(k/1-gamma)*|v/v^gamma|. Using the PV^gamma=k relation. Work=-(k/1-gamma)*|pv/k| = -|pv|/1-gamma. Now using pv=rt and putting limits work= R(T1-T2)/1-gamma. Its the same as -molar heat capacity at constant volume*change in temperature. B) work =-P(ext)dv. Integrating from v1 to v2. Work=-P ext (v2-v1). Mar 17, 2016 at 18:27

In the case of both a reversible expansion and an irreversible expansion, the gas pressure at the interface with the piston matches the externally applied pressure during the deformation. However, in the irreversible expansion, the gas within the cylinder is not uniform with respect to pressure or with respect to temperature.

So the interface pressure (which determines the amount of work done) does not match the average pressure of the gas within the cylinder. In addition, the pressure at the interface depends not only on the gas volume but also on the rate at which the gas volume is changing. So for an irreversible process, we have very little control over what is happening inside the cylinder.

The only thing we can control is the pressure at the interface $p_{ext}$ where the work is being done. We do this by whatever means necessary, including feedback control systems. In the typical scenario, we drop the pressure at the interface to a constant value, less than the initial pressure in the cylinder, and then hold it at that value until the gas volume has increased to the point where the gas in the cylinder is again at equilibrium. At this point, the gas pressure throughout the cylinder is described by the ideal gas law again.

If we follow this game plan for the irreversible process, even if the final pressure matches the value be obtain in the reversible process, the final temperatures and the final volumes will not match.

If you would like me to provide a detailed analysis of the irreversible case, illustrating in detail what transpires, I will be glad to do so. But right now, I'll give you a chance to ask some questions.

Irreversible and Reversible Expansions

Initial conditions: $T_i, P_i, V_i$

Irreversible expansion:

$$\Delta U=-\int_{V_i}^{V_f}{P_{ext}dV}$$ In the irreversible expansion we are considering, $P_{ext}$ is controlled to be constant at the final pressure $P_f$ with $P_f<P_i$. The expansion is allowed to continue until the gas re-equilibrates at the final pressure and volume. Therefore, we have: $$nC_v(T_f-T_i)=-P_f(V_f-V_i)\tag{1}$$

As noted in my discussion above, the ideal gas law cannot be applied to the intermediate states during the irreversible expansion, but it can be applied to the two equilibrium end states. Therefore, $$V_f=\frac{nRT_f}{P_f} \tag{2}$$ and $$V_i=\frac{nRT_i}{P_i}\tag{3}$$If we combine equation $(1)$, $(2)$ and $(3)$, we obtain:$$C_v(T_f-T_i)=-R\left(T_f-\frac{P_f}{P_i}T_i\right)\tag{4}$$

We can solve this equation for $T_f/T_i$ and a function of $P_f/P_i$ to obtain: $$\frac{T_f}{T_i}=\frac{1+(\gamma-1)(P_f/P_i)}{\gamma}\tag{5}$$ Note that this equation differs from the relationship one obtains if the expansion is carried out reversibly.

Reversible expansion: For a reversible expansion, we have $$dU=nC_vdT=-PdV=-\frac{nRT}{V}dV$$ or $$C_vd\ln T=-Rd\ln V=-R(d\ln T-d\ln P)$$This integrates to: $$\frac{T_f}{T_i}=\left(\frac{P_f}{P_i}\right)^{\frac{(\gamma-1)}{\gamma}}\tag{6}$$ Note from equations 5 and 6, that if the temperature ratios are the same in both cases (equal work), the pressure ratios are not, and if the pressure ratios are the same in both cases, then the temperature ratios are not (unequal work).

• In both cases change in internal energy would equal the magnitude of work? Or not. If yes then change in internal energy= specific heat at constant V*dT. Both would have the same work done if I knew the limits for the integral. Mar 17, 2016 at 21:04
• Yes. In both cases, the change in internal energy would equal the magnitude of work. The two amounts of work could turn out to be different and the two changes in internal energy could turn out to be different in the reversible and irreversible cases. Do you want me to do a reversible problem and an irreversible problem so that you can compare? If somehow you could force the work and internal energy changes in the two cases to be the same, the final pressures and the final volumes will not be the same. It is much easier just to set up two focus problems for comparison. Mar 17, 2016 at 21:14
• If you could do separate problems that would be really awesome. Mar 17, 2016 at 21:40
• I'll be glad to. It shouldn't take much. I'll be back tomorrow morning. Bed time now. Mar 18, 2016 at 3:36
• See edited version of my answer. Mar 18, 2016 at 13:49