We say that because there is no external force on the system, the work done on the adiabatic free expansion is $0$(If this is wrong, correct me please). But there is a pressure and volume change in the system and if there is no work how can there be pressure and volume changes?
What causes these changes?
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$\begingroup$ Who says that there is no work done in an adiabatic process or cycle? Do you think that the definition of an adiabatic process is one where no external. force on the system? Where did you see that? $\endgroup$– Chet MillerCommented Jan 20, 2021 at 12:51
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$\begingroup$ It's not just adiabatic process or just cycle. It's the combination of them. When we try to find entropy change in free adiabatic expansion, our teacher used this fact to show that internal energy change is 0 and T is constant $\endgroup$– EcJokCommented Jan 20, 2021 at 13:45
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$\begingroup$ It's not a general fact. In general, an adiabatic process can occur with or without work being done or received by the system. In the specific process of free adiabatic expansion of an ideal gas, it happens that no work is done or received by the system (this is only an approximation, real gases do a small amount of work during free expansion). Since for this specific process we have no heating and no work, then by the first law the change in the internal energy of the system must be zero. $\endgroup$– pglpmCommented Jan 20, 2021 at 14:17
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$\begingroup$ So where does the cycle part come in? $\endgroup$– Chet MillerCommented Jan 20, 2021 at 14:38
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$\begingroup$ Sorry, it's not cycle I typed it wrong. It should be free expansion. $\endgroup$– EcJokCommented Jan 20, 2021 at 14:47
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3 Answers
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Suppose we regard our system as all the matter residing within the confines of the rigid container, and we regard the surroundings as the walls of the rigid container and everything outside. Since the walls of the container is rigid, our system is incapable of doing any work on it; the displacement of the walls is zero. So the work and the heat are zero (insulated container), and the change in internal energy of our system must be zero.
answered Jan 20, 2021 at 17:50
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From your comments it seems that free expansion is the confusing part, so let me address that. There are two points to your question, which must be kept distinct.
First: in general, an adiabatic process can occur with or without work being done or received by the system. So, no, when you have an adiabatic process you can't conclude that there's no force applied to the system and no work exchange.
In the specific process of free adiabatic expansion of an ideal gas, it happens that no work is done or received by the system.
Second: Your intuition is right. During the free expansion of a real gas, a small amount of work is always done. From a macroscopic point of view this happens because of surface tension of the gas even when there's no enclosure. From a microscopic point of view this happens because the particles of the gas do work to overcome their mutual attraction. There can be exceptions, though. For a slightly electrified gas the opposite may happen for example: the expansion is favoured by electrostatic repulsion.
An ideal gas is microscopically modelled as a system of non-interacting particles, and therefore they do not do or receive any work during free expansion. However, this system during such a process cannot be described using the thermodynamics of uniform systems – in other words, using just "$(T,V)$" or "$(p,V)$" state variables. Pressure and temperature gradients appear, and if we tried to force a "$(p,V)$ description" we would encounter mathematical singularities. One needs the full thermodynamics of continuous systems to describe such adiabatic expansion.
However, once the free expansion has taken place and the system has gone back to spatially uniform conditions, we can resume our "$(p,V)$ description" and we have that the internal energy is the same as right before the free expansion. (For some real gases a very approximate $(p,V)$ description of free expansion might be possible, in some circumstances).
A good reference on both of these points is
- Astarita: Thermodynamics: An Advanced Textbook for Chemical Engineers (Springer 1990)
The question of the free expansion of an ideal gas and its mathematical description is discussed in detail in the Appendix to Chapter 2 and in Problem 2.3, and partly in section 5.1 as well.
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how can there be pressure and volume changes?
Although no work may be done, the gas nonetheless expands against a vacuum (increases in volume). When it expands its pressure drops. If it is an ideal gas,
$$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_2}$$
where 1 and 2 denote the initial and final equilibrium states before and after the free expansion.
Since $W=0$ and $Q=0$, from the first law $\Delta U=0$. Also, for an ideal gas internal energy $U$ depends only on temperature so $T_{2}=T_1$. Therefore
$$P_{1}V_{1}=P_{2}V_{2}$$
Hope this helps.
