Question

Critical temperature for CO2 and CH4 are 31.1º and 81.9º C respectively. Which of these has stronger intermolecular forces and why ?

EDIT: It should be a negative sign on `T_c` for `"CH"_4`...
- Truong-Son

Answer 1

The one with stronger intermolecular forces of attraction would have a smaller average distance between particles due to greater interaction, and that would give rise to a higher critical temperature, i.e. it is more difficult to achieve that state of liquid-gas coexistence.

[Note that the critical molar volumes are nearly the same, so that is not a good way to check.]

It then means that `"CO"_2` has the stronger intermolecular forces. Further discussion is found here on the `a` and `b` vdW constants.

---

Quantitatively, we can use the van der Waals equation of state (vdW EOS) for simplicity,

`P = (RT)/(barV - b) - a/(barV^2)`

where:

• `P,V,n,R,T` are known from the ideal gas law.
• `a` is a constant accounting for intermolecular attractions.
• `b` is a constant accounting for excluded volume due to the amount of intermolecular repulsions.

Thus, `a` and `b` inversely correlate with the amount of intermolecular forces of attraction going on.

I had derived the following equations already from the vdW EOS:

`barV_c = 3b`, the critical volume

`T_c = (8a)/(27bR)`, the critical temperature

`P_c = a/(27b^2)`, the critical pressure

`a = (27R^2T_c^2)/(64P_c)" "" "" "b = (RT_c)/(8P_c)`

The typical size of `a` is between `1` and `60` `"bar"cdot"L"^2"/mol"^2`, and often closer to `1` than `60`. The typical size of `b` is between `0.01` and `0.50` `"L/mol"`, much smaller than `a` usually.

From the equations above, a difference in `bbula` is the primary cause for a different `bbul(T_c)` due to the first-order dependence of both and `a` being larger than `b`.

For comparison, fixing the sign on the numbers you quoted, we get...

`(T_c("CH"_4))/(T_c("CO"_2)) = (color(red)(-)81.9 + "273.15 K")/(31.1 + "273.15 K") = 0.629`

So it is reasonable to expect `a` for `"CH"_4` to be smaller, not larger. This is why it is important to watch your signs!

Here are the values of `a` and `b`:

`a_("CH"_4) = "2.300 bar"cdot"L"^2"/mol"^2`

`a_("CO"_2) = "3.658 bar"cdot"L"^2"/mol"^2`

`b_("CH"_4) = "0.04301 L/mol"`

`b_("CO"_2) = "0.04286 L/mol"`

Checking your numbers, I get that

`T_c("CH"_4) = (8a)/(27bR) = (8 cdot 2.300)/(27 cdot 0.04301 cdot 0.083145) "K"`

`=` `"190.57 K"`, while NIST shows `"190.6 K"`.

`= -82.6^@ "C"`, far off from `+81.9^@ "C"`!

while

`T_c("CO"_2) = (8a)/(27bR) = (8 cdot 3.658)/(27 cdot 0.04286 cdot 0.083145) "K"`

`=` `"304.18 K"`, while NIST shows around `"304.2 K"`.

`= 31.03^@ "C"`, which is close to what you have.

And indeed, `a` is larger for `"CO"_2`, while `b` is slightly smaller for `"CO"_2`, so the intermolecular forces of attraction are stronger on `"CO"_2`. Further discussion is found here.