Reversible vs. Irreversible Expansion

Question

Suppose I have a gas in a piston expanding adiabatically against atmospheric pressure. If I do this reversibly, the final state is uniquely specified by the final volume $V_F$ of the gas.

What happens if I let the same process happen irreversibly? That is, I let the piston expand adiabatically against atmospheric pressure until some maximum volume, or until the pressure in the piston equals 1 atm, whichever comes first. Is there a unique final state in terms of the final volume $V_F$ or final pressure $P_F = 1 \,{\rm atm}$? If not, what else do I need to specify? Maybe the final temperature $T_F$? What information about the expansion is missing? (Neglect gravity and the mass of the piston, friction, etc.)

What is the typical situation, for example, in the case of a (very well insulated) steam engine piston? Is it possible to predict the final state of the gas after the expansion?

Answer 1

Suppose I have a gas in a piston expanding adiabatically against atmospheric pressure. If I do this reversibly, the final state is uniquely specified by the final volume $V_F$ of the gas

You didn’t answer my original comment about this first sentence directly and it wasn’t subsequently edited. The way it is worded it makes it sounds like the gas pressure is initially greater than 1 atm and something is initially holding the piston in place. Then the piston is released allowing it to expand against a constant external pressure of 1 atm. If that’s the case, it is not a reversible adiabatic expansion, but the irreversible process discussed below.

If, on the other hand, you meant in the first sentence that the external pressure and gas pressure are the same and greater than 1 atm due to the application of an external force, and the external force needed to maintain equilibrium is gradually reduced so that the expansion occurs quasi-statically, then we will have

$$PV^{γ}= constant$$

If the expansion is allowed to continue in this manner until the externally applied force is reduced to zero and the gas pressure is in equilibrium with the external pressure of 1 atm, then the final volume $V_f$ will be uniquely specified based on

$$P_{f}V_{f}^{γ}= P_{i}V_{i}^{γ}$$

What happens if I let the same process happen irreversibly? That is, I let the piston expand adiabatically against atmospheric pressure until some maximum volume, or until the pressure in the piston equals 1 atm, whichever comes first.

Now it appears we start with the same initial conditions as the reversible expansion, but instead of gradually reducing the externally applied force that was maintaining equilibrium, we abruptly remove it so that the gas is now expanding non quasi-statically (irreversibly) against 1 atm.

In this case the maximum volume will be achieved if you allow the expansion to reach equilibrium, that is, allow the process to continue until the gas pressure is in equilibrium with the external pressure, 1 atm. If you stop the process before equilibrium is reached the maximum volume will not be reached and the gas pressure will be greater than 1 atm.

ADDENDUM

The following is in response to your follow up question:

More generally, what I'm really asking is this: does the final temperature depend on messy details of the expansion?

Yes it does because the details of the expansion determine how much work is done. We will take the general cases of how to determine the final temperature (and volume) for a reversible and irreversible adiabatic expansion, each ending at the same final pressure, $P_f$. Then we will provide a specific example that shows how the final temperatures and volumes will differ.

REVERSIBLE ADIABATIC EXPANSION:

The process equation is:

$$PV^{k}= constant$$

Therefore

$$P_{i}V_{i}^{k}= P_{f}V_{f}^k$$

If $P_{i}$, $V_{i}$, $P_{f}$ and $k$ are known, you can calculate $V_f$.

Then since $V_f$ and $P_f$ are known, you can calculate $T_f$ using the ideal gas equation, for 1 mole of gas

$$P_{f}V{_f}=RT_{f}$$ Alternatively,

$$\frac{P_{i}V_{i}}{T_{i}}=\frac{P_{f}V_{f}}{T_f}$$

IRREVERSIBLE ADIABATIC EXPANSION.

For any adiabatic process

$$\Delta U=-W$$

For any ideal gas,any process

$$\Delta U=C_{V}(T{_f}-T_{i})$$

Therefore

$$C_{V}(T_{f}-T_{i})=-W$$

For the irreversible adiabatic expansion against constant external pressure and where the external pressure is the final pressure

$$W=P_{f}(V_{f}-V_{i})$$

Putting the last two equations together

$$C_{V}(T_{f}-T_{I})=-( P_{f}(V_{f}-V_{i}))$$

From the ideal gas law

$$\frac{P_{i}V_{i}}{T_i}=\frac{P_{f}V_{f}}{T_f}$$

Now we have two equations and two unknowns, $V_f$ and $T_f$. Solve for the unknowns $V_f$ and $T_f$.

EXAMPLE:

The following is an example that a friend and I worked on recently. We wanted to compare reversible and irreversible adiabatic expansions of an ideal gas where the expansions both terminated at the same pressure. In this case we chose air, which is essentially a mixture of two diatomic gases (oxygen and nitrogen) where $k$=1.4 and $C_V=0.2053\frac{l.atm}{mol.K}$ .

Initial conditions for both processes:

$P_i$=10 atm

$V_i$=2.46 $l$

$T_i$=300K

Final conditions for both processes:

$P_f$=5 atm

$V_f$=?

$T_f$=?

For the reversible adiabatic process, the external pressure starts at 10 atm and is very slowly reduced to a final pressure of 5 atm, so that the gas pressure essentially equals the external pressure during the expansion.

For the irreversible adiabatic process, the external pressure was abruptly reduced from 10 atm to 5 atm and the expansion allowed to proceed irreversibly until the gas pressure reached equilibrium with the external pressure of 5 atm.

We then applied the general equations above to both processes.

Curve C2 is for the reversible adiabatic expansion. Curve C1 (red curve) is for the irreversible adiabatic expansion.

RESULTS:

Reversible process-

$V_f$=3.12 $l$

$T_{f}$=190 K

Irreversible process-

$V_f$= 4.22 $l$

$T_f$=257 K

So we see that for the same final gas pressure, the final volume and temperature will be different for the two processes.

Now we can also compare the work done by the gas for the two processes. Remembering that for any adiabatic process, $\Delta U=-W$, and for any ideal gas, any process, $\Delta U=C_{V}(T_{f}-T_{i})$, plugging in the values for the final temperatures, the work done for the two processes is

Reversible process: 22.6 $l.atm$

Irreversible process: 8.83 $l.atm$

More work is done for the reversible process than the irreversible process. The difference between the work done is referred to as the “lost work” due to the generation of entropy for the irreversible process.

This was an example in which the final pressure was 5 atmospheres. You can do the same calculations for your example of a monatomic gas expanding against a constant pressure of 1 atmosphere using the same equations.

Hope this helps.

Answer 2

Since the expansion is adiabatic ($Q=0$), the final temperature depends on a single quantity associated with the expansion: the amount of work done by the gas.

The gas does not need to change its temperature at all as it expands. If the piston is massless and is released against zero external pressure, the gas simply expands irreversibly and the final temperature is equal to the initial temperature. That is,

$$\Delta U = -W = C_V(T_f-T_i) = 0$$

anf thus

$$ T_f = T_i. $$

If, however, work is done by the gas as it expands, we will have

$$E_f = \frac32 Nk_BT_f = E_i - W = \frac32 Nk_BT_i - W$$

where $W$ is the work done by the gas as it expands. Thus,

$$ k_BT_f = k_BT_i - \frac23 \frac W N .$$

The above equation was for a monoatomic ideal gas, but in general we will have

$$T_f = T_i - \frac{W}{C_V}.$$

The maximum work that can be extracted from the gas is when the external pressure is infinitesimally smaller than that of the gas. This what is known as a reversible process. In that case, we know that, for an ideal gas, the work done in an adiabatic expansion is just

$$W = \frac{P_f V_f - P_i V_i}{1-\gamma}.$$

Using

$$P_i V_v^\gamma = P_f V_f^\gamma,$$

we find the lowest temperature reached by any expansion:

$$T_f = T_i \left(\frac{V_i}{V_f}\right)^{\gamma -1}.$$

The final temperature will therefore be in the range:

$$\boxed{T_i \left(\frac{V_i}{V_f}\right)^{\gamma-1} \leq T_f \leq T_i}$$

In the specific case you are referring to, you say the external pressure is atmospheric. Presumably, you also mean that the internal pressure is greater than atmospheric. In that case, the work done is

$$W = P_{\rm atm} (V_f - V_i).$$

How do we determine $V_f$? Let's do this again for a monoatomic ideal gas. We know that the final volume will satisfy:

$$ E_f = \frac32 P_{\rm atm} V_f = \frac32 E_i - P_{\rm atm} (V_f - V_i).$$

Solving gives

$$V_f = \frac{3 P_i + 2 P_{\rm atm}}{5 P_{\rm atm}}V_i$$

and thus

$$ W = \frac35(P_i-P_{\rm atm})V_i$$

and

$$\boxed{ k_B T_f = k_B T_i - \frac25(P_i-P_{\rm atm})V_i .}$$

Answer 3

The answer you gave to your own question involves the free adiabatic expansion of a gas and not an adiabatic expansion against atmospheric pressure which was your original question. That’s OK, but I think the answer for a free expansion can be simpler than you put forth.

In the example of your answer, $W=0$ and $Q=0$. By the first law

$$\Delta U=Q-W=0$$

For an ideal gas the change in internal energy, for any process, depends only on temperature change according to

$$\Delta U=C_{v}\Delta T=0$$

Therefore there is no change in temperature.

Hope this helps