# How can ideal gases be "elastic" and "ideal" at the same time?

The elastic properties of any substance is because of the restoring intermolecular forces operating in it.

And I read that the isothermal elasticity of an ideal gas is given by

$$E_{isothermal} = P$$

And also we can define elasticity of the ideal gas in an adiabatic process.

But I don't know how can we define such quantities for ideal gases too. In the definition of ideal gases, I read that there is no intermolecular forces in it.

So my question is that How do we define Elasticity for an ideal gas ?

If elasticity can be defined in terms of properties other than intermolecular forces then please explain it since I have not heard of other factors.

• Don't you think this property should more closely be observed as an effect of their nature to expand rather than due to intermolecular forces? Jan 16, 2021 at 14:42
• Where did you read this? Does $E_{isothermal}$ mean isothermal elasticity? And what is $P$? Jan 16, 2021 at 14:46
• @Bob D yes it means the same . I read it in my book . I can add a pic of it if you want .. Jan 16, 2021 at 14:48
• @BobD - Kindly take a look here. Jan 16, 2021 at 14:49
• @Bob D the derivation linked by SteelCubes is the one in my book.. Jan 16, 2021 at 14:51

"Elasticity" can have slightly different meanings in different fields. In the thermodynamics (and I mean thermodynamics, not statistical mechanics) of spatially uniform systems, a system or a material (a class of systems) is called elastic if its state does not depend on rates of change of physical quantities. The state of an ideal gas at time $$t$$ is taken to be determined by the instantaneous temperature and volume (or density): $$\bigl(T(t),V(t)\bigr)$$, and is therefore elastic. Another example of elastic material, with the same state variables, is rubber.

The state of a solid or fluid (gases included) at time $$t$$ during a spatially uniform process may depend not only on temperature and volume, but for example also on the rate of change of either or both of them: $$\bigl(T(t), V(t), \dot{T}(t), \dot{V}(t)\bigr)$$, where $$\dot{T} := \frac{\mathrm{d}T}{\mathrm{d}t}$$ etc., or even higher-order derivatives. In this case the material is not elastic.

For example, suppose we keep a real gas at constant temperature $$T$$, and measure its pressure during compression when the instantaneous value of its volume is $$V$$; and then measure it during expansion when the instantaneous value of its volume is again $$V$$. We will find that the pressure measured in the two cases is different: it's higher during compression than during expansion. This happens because the pressure of a real gas is not only a function of $$(T,V)$$, but at the very least of $$(T,V,\dot{V})$$ (and possibly other rates of change), and $$\dot{V}$$ is different during compression ($$\dot{V}<0$$) and during expansion ($$\dot{V}>0$$). Real gases are therefore not elastic.

To make a connection with some more familiar aspects of elasticity, suppose we make a system undergo the following processes at constant temperature:

1. Between time 0 and $$\Delta t$$ we let the volume change according to some given time dependence $$V(t) = f(t)$$. For many systems the work done by the system during this process will be $$W_\text{I} = \int_0^{\Delta t} p[\dotso]\ \mathrm{d}V(t) \equiv \int_0^{\Delta t} p[\dotso]\ \frac{\mathrm{d}V}{\mathrm{d}t}\ \mathrm{d}t = \int_0^{\Delta t} p[\dotso]\ \dot{f}(t)\ \mathrm{d}t$$ where I have left the state-dependence of the pressure undefined as yet. This formula is valid for an ideal or real gas. It also holds for a piece of rubber if we use its elongation $$L$$ instead of the volume $$V$$, and keep in mind that $$p$$ will have opposite sign – it's a tension rather than a pressure – and a different state dependence.

2. Between time $$\Delta t$$ and $$2\Delta t$$ we let the volume change in the reverse way: $$V(t) = f(\Delta t-t)$$. The work done by the system in this process is $$W_\text{II} = \int_{\Delta t}^{2\Delta t} p[\dotso]\ \mathrm{d}V(t) \equiv \int_{\Delta t}^{2\Delta t} p[\dotso]\ \frac{\mathrm{d}V}{\mathrm{d}t}\ \mathrm{d}t = -\int_{\Delta t}^{2\Delta t} p[\dotso]\ \dot{f}(\Delta t -t)\ \mathrm{d}t\ .$$

If the pressure (or tension) only depends on temperature and volume, and not on the rate of change of the latter, we have $$p(t) = p[T, V(t)] = p[T, f(t)]$$ in the first process, and $$p(t) = p[T, V(t)] = p[T, f(\Delta-t)]$$ in the second. By a simple change of variable we see that the integrals in the two processeses are opposite: $$W_\text{I} = -W_\text{II}$$. That is, the work done/received by the system in the first process is equal to the work received/done in the second: it's fully returned. So we can say that the work done on the system is fully stored there. This is characteristic of elastic materials, and you see why an ideal gas is also an elastic material. Indeed, it's elastic because it's ideal.

If pressure depends on the rate of change (or even the whole history) of the volume, we have $$p(t) = p[T, V(t), \dot{V}(t)] = p[T, f(t), \dot{f}(t)]$$ in the first process, and $$p(t) = p[T, V(t), \dot{V}(t)] = p[T, f(\Delta t-t), -\dot{f}(\Delta t- t)]$$ in the second. In this case a change of variable only leads us to $$W_\text{I} = \int_0^{\Delta t} p[T, f(t), \dot{f}(t)]\ \dot{f}(t)\ \mathrm{d}t$$ and $$W_\text{II} = -\int_0^{\Delta t} p[T, f(t), -\dot{f}(t)]\ \dot{f}(t)\ \mathrm{d}t$$, and there is no specific relation between the two integrals (unless $$p$$ has some special symmetric dependence on $$\dot{V}$$). In fact they could be both positive or both negative. For real gases, if the first process is an expansion ($$\dot{f}(t)>0$$) and the second the reversed compression, then the first integral turns out to be less than the opposite of the second: $$W_\text{I} < -W_\text{II}$$, which is an expression of dissipation and irreversibility. For the relation between elasticity, irreversibility, and the second law, see Astarita's text below.

In the more general thermodynamics of spatially non-uniform systems (continuum thermomechanics), a material is called elastic if its stress tensor (the generalization of pressure) is rate-independent and memoryless (that is, the material is not capable of hysteresis).

The microscopic interpretation can be quite different depending on the material. In fact, elasticity depends not only on the microscopic composition but also on the macroscopic level of description (coarseness in space and time) that we're using.

Here are some references:

An extensive discussion of elasticity for spatially uniform systems, and its relation with reversibility, irreversibility, and entropy can be found in

see especially chapter 2.

The general classification of materials as elastic, plastic, viscoelastic, or viscoplastic is given for example in

see section 6.4 there, and also in

Quoting from section 43 there:

A material is called elastic if it is simple [a technical term, roughly referring to the dependence upon rotations] and if the stress at time $$t$$ depends only on the local configuration at time $$t$$, and not on the entire past history of the motion.

From this definition you see again why an ideal gas is an elastic material. This text has also a brief discussion of the historical evolution of the notion of elasticity.

Regarding the micro-macro relationship and its dependence on the chosen level of description, a great text is