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
EDIT: It should be a negative sign on
- Truong-Son
1 Answer
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
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,
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
From the equations above, a difference in
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
Here are the values of
#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,