In adiabatic expansion does the internal energy of an ideal gas decrease?

Question

By First Law of thermodynamic, for an ideal gas, if there isn’t heat transfer, work done by the gas is equal to decrease in internal energy of the gas.

Suppose that I have a perfectly-insulated syringe closed at one end and a frictionless piston on the other. The syringe initially contain ideal gas of volume $V$. If I pulled the piston outward, the volume of gas would increase. Since I am the one applying force, work is done by me instead of by the contained gas. So, in this case, does the internal energy of gas remain constant?

Answer 1

The ideal gas inside the syringe is always applying a force on the piston, because of its pressure. Therefore, for the piston to remain motionless, there must be another force from the outside that contrasts the push of the gas inside. From the text of your question it seems that in the initial situation, e.g. when the piston is still, you are not applying any force on it (correct me if I'm wrong). This means that there must be something else doing it, otherwise the gas would expand without need of your intervention.

I suppose that your hypothetical experiment takes place on Earth in ordinary conditions, so that there is the air of the atmosphere outside of the syringe. If this is the case that you mean (again, correct me if I am wrong), then the piston is still at first because there's the atmospheric pressure outside that equilibrates it. In this case, your confusion may derive from forgetting the atmospheric pressure.

When you pull the piston you are obviously doing work, but not on the gas inside the syringe. Indeed, its pressure is pushing in the same direction in which you're pulling, so both you and the internal gas are doing work. But against what? Against the atmospheric pressure, which is opposing the expansion. Indeed, as the volume of the ideal gas increases, the volume of the atmosphere decreases. Therefore, both you and the gas are losing energy, in favor of the atmosphere.

Answer 2

Microscopically, gas molecules impacting on the outward-moving surface get reflected with a lower velocity. Of course, the average molecular velocity is orders of magnitude larger than the velocity of the piston and the difference in energy for each collision is very small. But it all adds up, and the gas is cooling.

Answer 3

Foundations

The first law has two forms.

$$\mathrm{IUPAC:}\ \Delta U = q + w\ \ \ \mathrm{Clausius:}\ \Delta U = q - w$$

In the former, work done by the system is negative. In the latter, work done by the system is positive. We can safely stay with the IUPAC form and recognize that all we need to do is take the negative of any work term that we derive to obtain the Clausius expression.

Work comes in various forms, including mechanical, shaft, chemical, electrical, and magnetic among others. For a system that undergoes only mechanical work, the vector expression for (IUPAC) work is

$$w = \int\ \vec{p}_{ext} \bullet d\vec{V} $$

By convention, we take expansion as the positive direction of $d\vec{V}$. When $\vec{p}_{ext}$ points inward, expansion goes against the pressure, work is done by the system, and the result is negative $w$. Alternatively, when $\vec{p}_{ext}$ points inward, compression goes with the pressure, the system has work done on it, and $w$ is positive.

The four types of mechanical expansion/compression work are as follows:

• Free expansion/compression - In this case, $p_{ext}$ is constant and $p_{ext} \equiv 0$. No work is done on or by the system.

• Constant pressure (isobaric) expansion/compression - In this case, $p_{ext}$ is constant and $w = \vec{p}_{ext} \bullet \Delta\vec{V}$. By example, work is positive (done on the system) when the system is compressed by the external pressure.

• Expansion/Compression under a varying pressure - In this case, $p_{ext}$ may have a time dependence or may depend on the position of $dV$.

• Reversible expansion/compression - In this case, $p_{ext} = p_{int}$ at any and all steps during the process.

Application to the Problem at Hand

A syringe is pulled with a force $F$. Assume that the syringe has an end cap area $A$ and that it is also subject to an external pressure $p_a$ (i.e. from the atmosphere). The net external pressure seen by the system (the inside of the syringe) is

$$ \vec{p}_{ext} = \vec{p}_a + \frac{\vec{F}}{A} $$

Allow that $p_a$ and $F$ are constant. Realize that $\vec{F}$ points in the opposite direction to $\vec{p}_a$, and $\vec{p}_a$ points inward (opposite $d\vec{V}$). Finally, recognize that the piston expands to give $\Delta V > 0$ (positive). We can now write the IUPAC work as

$$w = \left( -p_a + \frac{F}{A} \right) \Delta V$$

Insights

When you would apply enough force to generate $F_{FE} = p_a A$, the system mimics the case of free expansion. In an adiabatic process, since $w = 0$ and $q = 0$, we find $\Delta U = 0$. When the system is an ideal gas, the end temperature will be the same as the starting temperature.

For any force less than $F_{FE}$, the system expands against a positive external pressure and does work. We find the term $w < 0$. In an adiabatic process with $q =0$, we find $\Delta U < 0$. When the system is an ideal gas, the end temperature will be lower than the starting temperature.

For any force greater than $F_{FE}$, the system expands against a net negative (outward pointing) pressure. The system has work done on it. We find the term $w > 0$. In an adiabatic process we $q = 0$, we find $\Delta U > 0$. When the system is an ideal gas, the end temperature will be higher than the starting temperature.

Other Thoughts

A classic example given in introductory thermodynamics is to propose to add a weight to push on top of a piston and to calculate the work and change in internal energy when the piston compresses. The case of pulling on a syringe is the same as inverting the piston and hanging a weight to pull on the piston rather than to put a weight on top to push on it.

All considerations above are irreversible. You cannot immediately model the example of pulling on a syringe as a reversible process. This will at first blush violate the principles that $p_{ext} \equiv p_{int}$ at all steps during a reversible process. You could attempt to force this case to a reversible example by proposing to pull on the syringe only with enough force to allow this

$$-p_a + \frac{F}{A} = p_{int} $$

at any and all stages during the expansion of the syringe. When the system contains an ideal gas and the process is adiabatic, this gives

$$p_{ext} = \left(-p_a + \frac{F}{A}\right) = p_o\left(\frac{V_o}{V}\right)^\gamma $$

where $p_o, V_o$ are the initial pressure and volume inside the syringe and $\gamma$ is the ratio of specific heat molar capacities of the gas $\bar{C}_p/\bar{C}_V$.

In those cases where the syringe has anything other than an ideal gas (for example a liquid), you are going to face terrible difficulties to find a viable mechanical equation of state for the fluid $p_{int}(V, T)$ to allow the above to be true. In other words, pretending that the process must be reversible to understand it may be far harder to justify than just analyzing the thermodynamics robustly for their own right.

You may also find this discussion interesting as well.

Answer 4

By definition, work done by a force $\vec F$ on a body over an incremental displacement is $$\delta W=\vec F\cdot \text d\vec r.$$ The value of this work depends only on the force $\vec F$ and the displacement $\text d\vec r$. It is independent of whether there are other external forces acting on the body (in which case each force will have its own work).

Thus, the gas in the syringe is performing work, due to the fact that it is applying pressure (i.e a force) on the piston, and the piston is moving (a displacement). Therefore, the internal energy will decrease by an amount $$\text dU=-P\text dV.$$