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This may not a good quesiton I guess, but it is very confusing while I was studying the classical states of matter :- SOLIDS, LIQUIDS, GASES.(Also please correct me wherever my concepts are wrong, and use the usual meanings of the symbols)

  1. In solids, we have modulus of elasticity(young's modulus (Y), shear modulus(η) etc.), which are used to find the expansion in them, when a Force is applied, given by the equation F/A=y*x/l where x is linear expansion.
  2. But for fluids we do not consider these modulus. But for them a property called Surface Tension exists.

This is my first question, that how are the molecular interactions different and are their different interactions result to different properties in solids, liquids and gas. If yes, How ??(please explain with actual forces acting on the particles)

  1. Now, in comparison of liquids with gases both have different thermodynamics. In Gases the pressure due to gravity (specifically I am talking about P = ρgh, which are considered in liquids but not gases) are not considered. Similarly in liquids the Pressure due to root_mean_squre velocity is not considered (that is P=1/3ρvrms2).
  2. Also for liquids we consider ΔQ=msΔT, but for gases, we useΔQ = ΔU + ΔW. Also ΔU = nRfΔT/2, but for fluids the ΔU = SΔA (where S, is surface tension, A is area).

This my second question that are the two relations different? If yes/no why ?

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    $\begingroup$ Gravitational pressure gradients are considered in gasses....that's why the atmosphere gets thinner the higher you go. Gasses act a lot more like fluids than you indicate. $\endgroup$
    – Triatticus
    Commented Jan 23 at 7:32
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    $\begingroup$ For solids, there is also surface tension, just that it is not usually as interesting as the many other moduluses. If you want to consider the pressure due to gravity in gases, that is also doable; it is just usually not a big contribution. $\endgroup$ Commented Jan 23 at 7:34
  • $\begingroup$ I have seen stress tensor constructed from constitutive equations based on matter properties and velocity and gradient of velocity through modeling things like gas laws, hydrostatic pressure, strain tensor (elastic media), strain rate tensor (viscous media). I tried adding them all at once in a program but I had some bug so could not tell if it looked real. $\endgroup$
    – Emil
    Commented Jan 23 at 9:06
  • $\begingroup$ You are asking quite different questions. It would be better to have separate questions. $\endgroup$ Commented Jan 23 at 11:25
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 And that is only my question how they can be related? At the atomic level they are made of particles which have just different motions, and hence their property is different and that is question, how different motions of molecular particles are resulting to different properties and how are all those properties relatable? $\endgroup$ Commented Jan 23 at 15:15

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There are multiple misconceptions in the question that should be addressed, as you invited. Looking at them provides an opportunity to review relations between the basic states of matter.

But for fluids we do not consider these [elastic moduli].

That's not correct. Ideal fluids have a shear modulus of zero, but they have a nonzero bulk modulus, which is one of the elastic moduli. Specifically, the bulk modulus mediates changes in size, whereas the shear modulus mediates changes in shape. (Young's modulus mediates length changes for uniaxial loading of a rod or bar.)

Ideal fluids have a shear modulus of zero because they're not quite well bonded enough to resist unbalanced loads over the time scale of interest. A basic analogy is an increasingly crowded room of people: a few people walking around with minimal contact is broadly similar to a gas; a crowd of people in physical contact but still able to squeeze by each other is broadly similar to a liquid; and a very dense crowd jammed in place is broadly similar to a solid.

Take the latent heat of water as a measure of the degree bond breaking from state to state:

We have to break only 1 in 2265/334 ≈ 7 bonds on average to melt ice to liquid water, essentially. In this way, the density remains similar, and substantial bonding persists in the material, but the liquid state is now "loose" enough to rearrange over relevant time scales.

But for [fluids] a property called [s]urface [t]ension exists.

Gases are fluids, and gases are arguably defined by a lack of surface tension; this is what allows them to expand to fill an arbitrarily large volume having an arbitrarily large surface area. In gases, we assume that intermolecular bonding is completely lost; the energy required to break these bonds is supplied by the latent heat of vaporization.

This gives a reasonable basis of understanding to distinguish between states of matter based on the degree of bonding: Fluids rearrange easily relative to solids and correspondingly have a negligible shear modulus, and gases—in contrast to condensed matter, meaning liquids and solids—lack a positive surface energy provided by intermolecular bonding.

In [g]ases the pressure due to gravity...[is] not considered

Yes it is, for a region of gas for any appreciable height. The static fluid pressure equation $P=\rho g h$ (density $\rho$, gravitational acceleration $g$, height $h$) still applies, where the pressure is a gauge pressure relative to the free surface. This pressure variation is often ignored, however, for small/short containers.

In gases, the pressure is solely entropic: Unbonded molecules move freely in relation to the thermal energy of the gas, and the measured pressure arises from momentum transfer when molecules bounce off container walls. The mean-squared velocity is a key parameter in this analysis. In condensed matter, however, the pressure arises from the chemical bond and its tendency for attraction at large distances and repulsion at close distances. The equilibrium distance at which these counteracting responses balance determines the molecular spacing and thus the density, which is nearly fixed relative to the gas case.

The rest of the question continues along the line of models that you may have seen in only one context or circumstance but apply more broadly; gaining physics understanding includes filling in the gaps of when and how these models can be applied in conjunction across states of matter.

For example, the equation $Q=mc\Delta T$ (heating $Q$, mass $m$, specific heat capacity $c$, temperature change $\Delta T$) can always be applied to sensible heating—that is, heating (when nothing else is occurring) of a single state. You may have seen it only in the context of liquids at this point, but it's not limited to liquids.

The equation $Q=\Delta U+W$ (heating of a system $Q$, energy change $\Delta U$, work done by the system $W$) is the First Law; it can be applied to any closed system. You may have seen it only in the context of gases at this point, but it's not limited to gases.

The equation $\Delta U=S\Delta A$ (surface tension $S$, surface area $A$) links the change in energy for a change in area alone, when nothing else is happening. It applies to all materials, but note again that $S$ is not well defined or can be taken as zero for gases.

The equation $\Delta U = nRf\Delta T/2$ (amount of gas $n$, gas constant $R$, degrees of freedom $f$) describes how the energy of a gas depends on its temperature. (More generally, and as a more advanced topic, the equation $\Delta U=C_P\Delta T+(\alpha TK-P)\Delta V$, with constant-pressure heat capacity $C_P$, thermal expansion $\alpha$, bulk modulus $K$, and pressure $P$, applies to all states. This reflects the complications of intermolecular bonding. In the ideal gas, $\alpha=1/T$, $K=P$, and the heat capacity can be linked simply to the degrees of freedom of free molecular motion, recovering the first equation.)

In summary, I'd suggest linking each new equation you learn to a context and a scope. In several cases here, for instance, the implicit context of the equation is that only one specific process is occurring, and not other processes. That's why we see the internal energy change $\Delta U$ being equated to a variety of different things. (In some cases, the effects can be combined and the energy changes summed. As an example, if heating $Q$, pressure–volume work $W$, and surface area changes are all relevant, we could write $\Delta U=Q-W+S\Delta A$.) Finally, the scope is important to distinguish situations where a model doesn't apply from situations where you simply haven't yet seen the model being applied.

Edit: As requested, derivation of $\Delta U=C_P\Delta T+(\alpha TK-P)\Delta V$. Expand $dU$ in $T$ and $V$ (two variables because we can either heat the system or do pressure–volume work on it) as

$$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV;$$

From the fundamental relation,

$$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial (T\,dS-P\,dV)}{\partial V}\right)_TdV;$$

From a Maxwell relation,

$$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right] dV;$$

From the triple product rule,

$$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left[T\left(\frac{\partial V}{\partial T}\right)_P\left(-\frac{\partial P}{\partial V}\right)_T-P\right] dV.$$

Now just replace the definitions of the material properties with their symbols, assume constant material properties and $P$ and $T$, and integrate to turn the differentials into finite changes.

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  • $\begingroup$ My confusion is now clear. Actually I was never taught the equation ΔU = CpΔT+(aTK-P)ΔV. By the way can you please derive it? $\endgroup$ Commented Jan 24 at 14:07
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    $\begingroup$ Sure; edited to add. $\endgroup$ Commented Jan 24 at 19:45

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