# Ideally, how to achieve isothermal expansion of an ideal gas?

In an isothermal expansion of an ideal gas, there cannot be change in internal energy otherwise the temperature would change. So hence we know that all the heat go into work. Knowing that PV=nRT and T being constant, we see that P is inversely proportional to V.

Does that mean in order to feasibly realize an isothermal expansion we would need to do the following:

1. Put the ideal gas in a cylinder with a piston and some amount of weight on it to reach some pressure.
2. Find a heat reservoir equal to temperature of the gas.
3. Slowly lift weight off of the piston according to $P \propto 1/V$.

The last part is kind of the strange part. We have to lift some weight off otherwise the volume won't change. But why can't we quickly lift the weight off? So the external pressure changes from some large number to a very smaller number? Would that be approximately an adiabatic expansion?

The isothermal expansion is a theoretical ideal. An isothermal process requires the system is in perfect equilibrium with its surroundings at all times so it would have to be done infinitely slowly. As you say in your question, any process done at a finite speed is necessarily out of equilibrium.

However in real life provided heat flow is fast enough processes can be so close to isothermal that we can treat them as perfectly isothermal. That is, the error involved in assuming they are isothermal is negligibly small.

• If you allow the expansion (or compression) to take place slowly and continuously, there's enough time at every point for the temperature to equilibrate: Since all the gas will be at same temperature as the reservoir. This is an iso-thermal process.
• If you do the expansion (or compression) rapidly, so fast there's no time for thermal energy to flow and the gas to equilibrate, then the pV work being done will change the temperature of the gas. That's not an iso-thermal process, because the temperature changes. If you do it fast enough, so no thermal energy is exchanged, then it's adiabatic.

So the same sequence of operations if done fast enough is adiabatic and done slow enough is isothermal. What's "enough"? Compare the energy that can flow during the ideal amount, and see how close it is.

Two ways: 1) Expand the ideal gas into a vacuum, and there is no change in the temperature. However, this is not the type of expansion I surmise you are asking about.

2) Expand the gas against a piston so that it does work. Now, since the work is accomplished through the individual collisions of gas atoms, each collision transfers a small amount of momentum and energy to the piston. So the gas cools. Or think about it as change in internal energy = heat + work. The temperature reflects the kinetic energy of the gas, which is also the total energy of the gas since it is ideal.

Since the gas is cooling during expansion as described so far, the only way to maintain the same temperature is to heat the gas in sync with the transfer of collisional energy to the piston. This heat could transfer from some other part of the container enclosing the gas.

If you lift the weight off quickly (so that the external force on the gas is much lower), the expansion will be irreversible. Under these circumstances, even though the cylinder is in contact with a constant temperature reservoir at its surface, only the surface of the gas will be at the specified temperature, and the interior of the gas will drop to a lower (non-spatially-uniform) temperature throughout the expansion (until final equilibrium is reached, and the gas is again at the reservoir temperature). In the interim, heat conduction will be occurring within the gas.

In addition, the force that the gas imposes on the piston will be a combination of the effect of viscous (tensile) stresses within the gas plus what we ordinarily would consider the gas pressure under static conditions. Initially, for the irreversible case, all the change in force will be supported by viscous stresses, which are proportional, not to the change in volume but, to the rate of change of volume. So, during the irreversible expansion, the gas ceases to behave like an ideal gas and ceases to satisfy the ideal gas law (which applies only at thermodynamic equilibrium) within the cylinder. As equilibrium is approached, the rate of change of volume decreases, and the viscous stresses make less of a contribution to the overall force of the gas on the piston. At final equilibrium, the viscous stresses are again zero, and the force on the piston is entirely described by the ideal gas law.

To get a better feel for all this, imagine that the gas behaves like a combination of a spring (preloaded in compression) in parallel with a viscous damper, with the combination joined mechanically to the rear end of the cylinder and to the piston. The spring mimics the ideal gas behavior of the gas and the viscous damper mimics the viscous behavior of the gas. If the piston moves very slowly, the force on the piston is dominated by the spring (ideal gas behavior). But, if the piston moves very rapidly, the change in the force on the piston is dominated by the viscous behavior. In line with this conceptualization, a crude approximation to the behavior of a gas experiencing either a slow- or a rapid deformation can be expressed as: $$\frac{F}{A}=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}$$where F is the force of the gas on the piston and k is a constant proportional to the gas viscosity. The first term captures the thermodynamic equilibrium behavior of the ideal gas, and the second term captures the viscous contribution.

To get a better understanding of the viscous Newtonian behavior of fluids (including gases), I highly recommend Transport Phenomena by Bird, et al.