What's the difference between perfect and ideal gas?

Question

There are all kinds of different dinstinctions in the internet, and I'd like to see what you guys thought.

Actually, I've been asked if the heat capacity of an ideal gas es independent of temperature. I've said no, even though it's practically invariant under small ranges.

Namely the result $C_V/n = \frac{3}{2}R$ is derived from a perfect gas and not an ideal gas and is only an approximation to the latter.

Is this true?

What really is an ideal gas? What's its difference with a perfect gas? Is my answer to the question I was posed correct?

Thanks.

Answer 1

An ideal gas is the same as a perfect gas. Just different naming. The usual name for such gases (for which is assumed that the particles that make up the gas have no interaction with each other) is ideal gas, perfect gas is what such a gas is named in Atkins physical chemistry book. Personally I like the perfect gas naming better as it illustrates the perfect nature of the assumptions made about it.

For the simple systems (like monoatomic gases) when we can assume a perfect/ideal gas, $C_{V,m}$ is independent of temperature. For real gases this is certainly not the case.

Note that $C_{V,m}=\frac{3}{2}R$ only holds for monoatomic gases.

Answer 2

As I recall P. W. Atkins etal point out that the interaction between particles in both ideal and perfect gases are constant (i.e. they don't vary with T or P). The difference is that for a percent gas the interactions are not only constant but they are equal to zero. For an ideal gas they just are constant.

Also, in regard to the comment made just above (The second point) should be that "The size of the particles is negligible when compared to the volume they occupy". In fact the distances are large in comparison and that is why the forces of attraction between particles are negligible.

Answer 3

Namely the result $C_V/n=\frac{3}{2}R$ is derived from a perfect gas and not an ideal gas and is only an approximation to the latter. Is this true?

So, let's look first at where $C_p-C_V=R$ comes from and then look at $C_V=\frac{3}{2}R$ to see what we find.

We start with the definition of heat capacity as being a change in energy per change in unit temperature,$$\Delta H=\int_{T_1}^{T_2}n\cdot C_p\,\mathrm dT$$ Now, I assume heat capacity to be independent of temperature

Then,$$\Delta H=n\cdot C_p(T_2-T_1)$$

Since $H=U+PV$ and pressure is held constant here, we rewrite the expression as $$\Delta U + P\Delta V=n\cdot C_p(T_2-T_1)$$ By the same integration performed above (but with $C_V$) we find that $\Delta U=n\cdot C_v(T_2-T_1)$ Combining those expressions and simplifying,

$$C_p-C_V=P\frac{\Delta V}{n\cdot\Delta T}$$

Using the ideal gas law, with constant pressure, we find,$$\frac{\Delta V}{\Delta T}=\frac{nR}{P}$$ Plugging that in, $$C_p-C_V=R$$

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Now, for a monoatomic ideal gas, energy can only be stored in translation. Invoking the equipartition theorem to avoid having to do math and a little physics, we see that the energy of a monoatomic gas will be,$$U=\frac{3}{2}Nk_\mathrm bT$$ For $N=N_\mathrm A$ particles, we have, $$U=\frac{3}{2}RT$$

So, because $$C_V\equiv\left(\frac{\partial U}{\partial T}\right)_{P,n}$$ We see that, $$C_V=\frac{3}{2}R$$

Conclusions:

We see that in our derivation of the relationship $$C_p-C_V=R$$ we both used the ideal gas law and assumed heat capacity to be independent of temperature.

So, to answer the question quoted at the top of this answer, $\frac{C_V}{n}=\frac{3}{2}R$ is derived from the ideal gas, not the perfect gas.

And, in answer to the other question, our derivation required that we assume heat capacity to be constant with a change in temperature, so it was incorrect to say that heat is dependent on temperature for an ideal gas. It is true, however, that heat capacity varies with temperature for a real gas.

As to whether or not there is a difference between an ideal and perfect gas, I would look at that Wikipedia page posted in a comment above, but it seems superfluous to define something as a perfect gas when an ideal gas is already well understood and the perfect gas essentially behaves the same.

Hope that helps explain some of the math behind this.

Answer 4

Simply put, an ideal gas follows the ideal gas law. In thermodynamics and statistical mechanics, many physical relations are derived using the ideal gas law, sometimes accounting for Van der Waals forces or other non-negligible effects, depending on the conditions. The way I understand it, a perfect gas is an ideal gas, but is never treated as having non-interacting particles. I don't think the distinction is terribly important unless you plan on doing some thesis or dissertation in the fields of physical chemistry or condensed matter physics...

Answer 5

P. W. Atkins says that it is preferable to use the term perfect gas because the particles show no interactions. On the opposite, ideal solutions are constituted by different particles that show the same kind of interactions among them (the toluene/benzene mixture is a good approximation).

Answer 6

In terms of macroscopic thermodynamics (no molecular-based arguments are necessary), the underlying reason for distinguishing between an "ideal gas" and a "perfect gas" is that a mixture (solution) of gases may be an "ideal solution", but each of its components need not be an ideal gas.

In the case of a single component gas phase, it is more precise to use the term "perfect gas", which refers to a gas obeying the equation of state (commonly referred to as the "ideal gas equation of state"): $$PV = nRT \tag{1}$$ where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the absolute temperature.

From the thermodynamic relation

$$(\mathrm d \mu/\mathrm dP)_T = V/n \tag{2}$$

we obtain the chemical potential of a single perfect gas as

$$\mu_i(T,P) = \mu_i^0(T;P^0) + RT\ln(P/P^0) \tag{3}$$

where $P^0$ is set by convention as $\pu{1 bar}$ (but pre-1990 tables used $P^0 = \pu{1 atm}$.)

So far, we could have used the term "ideal gas". However, a problem arises in the case of solutions. The generalization of Eq. (3) to a mixture (solution) of perfect gases is

$$\mu_i(T,P) = \mu_i^0(T;P^0) + RT\ln(x_iP/P^0) \tag{4}$$

where $x_i$ is the mole fraction of species $i$. (Note that thermodynamic manipulations on Eq. (4) readily give the ideal-gas EOS for the mixture

$$PV = n_tRT \tag{5}$$

where $n_t$ is the total number of moles.)

Eq. (4) may be written as

$$\mu_i(T,P) = \mu_i^*(T,P) + RT\ln(x_i) \tag{6}$$

where

$$\mu_i^*(T,P) = \mu_i(T;P^0) + RT\ln(P/P^0) \tag{7}$$

It is then useful to define the general concept of an "ideal solution" (for any phase) as one for which the chemical potential of each species is given by Eq. (6), in which case $\mu_i^*(T,P)$ is the chemical potential of pure species i at the T and P of the mixture. Note that $\mu_i^*(T,P)$ need not be given by Eq. (3). A solution of perfect gases (whose individual chemical potentials are given by Eq. (4)) is thus a special case of an ideal solution.

Answer 7

What I have learned at high school is that ideal gas and perfect gas are synonyms. That is, the names basically refer to the same type of gas which:

1. Has negligibly small forces of attraction between its molecules;
2. Has negligibly small distances between the particles;
3. Has particles randomly roaming about at high speeds;
4. Has molecules whose collision with walls of its container are perfectly elastic.