11
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ According to Wikipedia, a perfect gas is not the same thing as an ideal gas. Specifically, it seems that all perfect gasses are ideal, but not all ideal gasses are perfect. Both types obey the ideal gas law, but they differ with respect to the temperature and pressure dependence of their heat capacities. None of the current answers seem to have this information. $\endgroup$ – Nicolau Saker Neto May 3 '15 at 1:00
  • $\begingroup$ An interesting discussion on PhysicsForum suggests that what we'd call an 'ideal gas' is actually a 'perfect gas' to engineers. What strains my imagination is how you'd get $C_p$ or C_v$ dependence without intermolecular interactions. $\endgroup$ – chipbuster Sep 1 '15 at 16:18
  • 1
    $\begingroup$ @chipbuster You can define the molar internal energy of the interactionless gas (monoatomic) as $\frac{3}{2}R + U(T=0K)$ you can get Cv from there and Cp as Cv+R $\endgroup$ – DLV Sep 1 '15 at 18:52
  • 1
    $\begingroup$ @David Thanks, shows you how weak my thermo is :P $\endgroup$ – chipbuster Sep 1 '15 at 19:08
  • $\begingroup$ I forgot to put a 'T' in front of $\frac{3}{2}R$ to get $U(T)=\frac{3}{2}RT+U(0)$ btw. Can't edit my original comment... $\endgroup$ – DLV Sep 1 '15 at 23:15
6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ Well I was asked if Cv has not a function of temperature for the ideal gases. Perfect gases' Cv is not a function of T, but Ideal ones do seem to depend.. Hmm.. $\endgroup$ – DLV Jan 28 '15 at 2:57
  • $\begingroup$ Yes, I agree with this. Physicists define a perfect gas as one that obeys the ideal gas law plus exhibits a heat capacity that is independent of temperature. An ideal gas, as we engineers call it, is one that obeys the ideal gas law plus exhibits a heat capacity that does depend on temperature. Real gases actually approach ideal gas behavior at low pressures, and do exhibit temperature-dependent heat capacities. $\endgroup$ – Chet Miller Nov 18 '15 at 1:59
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You are right, that is exactly what Atkins says. Just bear in mind that different people define the terms differently.. which is confusing, but such is chemistry.... $\endgroup$ – orthocresol Sep 19 '16 at 4:52
3
$\begingroup$

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, $$C_p-C_V=R$$


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.

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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.
$\endgroup$
  • $\begingroup$ Welcome to Chemistry SE! Next time please format your questions. $\endgroup$ – Asker123 May 3 '15 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.