How do I calculate the pressure in #"bar"# for #"Cl"_2# if #"1.140 mols"# of it occupies #"5.025 L"# at #"244.5 K"#, using the van der Waals equation of state?

#P = "4.467 bar"#

You'll need the van der Waals constants #a# and #b#, as the van der Waals (vdW) equation of state is:

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

where:

• #P# is the pressure in #"bar"#.
• #R = "0.083145 L"cdot"bar/mol"cdot"K"# is the universal gas constant.
• #T# is the temperature in #"K"#.
• #barV = V/n# is the molar volume in #"L/mol"#.
• #a# is the van der Waals constant that accounts for intermolecular forces. I am going to use units of #"L"^2cdot"bar/mol"^2# but they could be seen elsewhere in other units.
• #b# is the van der Waals constant that accounts for the volume a vdW gas actually takes up (the so-called "excluded volume").

I don't have the van der Waals constants on hand, but there are ways to get them. From McQuarrie Table 16.5, we have the following critical parameters for #"Cl"_2#:

#T_c = "416.9 K"#
#P_c = "79.91 bar"#
#barV_c = "0.1237 L/mol"#

The following relationships are known for #a# and #b#. A full derivation is in the appendix at the bottom of the answer.

#barV_c = 3b#

#P_c = a/(27b^2)#

Therefore, for #"Cl"_2#, we have van der Waals constants of:

#b = "0.04123 L/mol"#

#a = 27P_cb^2 = 27("79.91 bar")("0.04123 L/mol")^2#

#= "3.668 L"^2cdot"bar/mol"^2#

That allows us to obtain the pressure:

#color(blue)(P) = (("0.083145 L"cdot"bar/mol"cdot"K")("244.5 K"))/("5.025 L"/"1.140 mols Cl"_2 - "0.04123 L/mol") - ("3.668 L"^2cdot"bar/mol"^2)/("5.025 L"/"1.140 mols Cl"_2)^2#

#=# #color(blue)("4.467 bar")#

And just to check, we can easily use the ideal gas law to see if we get something similar.

#P = (nRT)/V = ("1.140 mols Cl"_2 cdot "0.083145 L"cdot"bar/mol"cdot"K" cdot "244.5 K")/("5.025 L")#

#=# #"4.612 bar"#

which is pretty similar (#3.24%# difference). The reason why the pressure for #"Cl"_2# is higher in the ideal gas law is that #"Cl"_2# has dominantly attractive intermolecular forces.

That means on average, the #"Cl"_2# molecules slightly attract each other to such an extent that the actual pressure they exert on the walls of their container is smaller than predicted with the ideal gas law.

Indeed, that is the case, since the compressibility factor #Z = (PV)/(nRT)# for #"Cl"_2#, as calculated here, would be #0.9686#, less than #1#, indicating that attractive forces dominate.

#color(white)(.)#APPENDIX: DERIVATION!

DISCLAIMER: Long derivation!

I'll try to go through the derivation that some textbooks (ask you to) go through, to find #a# and #b# in terms of #T_c# and #P_c#.

The van der Waals equation of state (vdW EOS) in the form that separates the pressure term from the rest is:

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

where #barV = V/n# is the molar volume, #a# accounts for the intermolecular forces between real gases, #b# accounts for the excluded volume that real gases (not point masses) take up, and the remaining variables are familiar from the ideal gas law.

Note that equations of state tend to be representable as cubic functions of #barV#.

Now, if you recall, critical points in Calculus are found when the first derivative is equal to #0#. When the second derivative is also #0#, you have an inflection point.

This inflection point flattens out as we approach the critical point, at which the critical parameters are defined. This is shown below:

(in the diagram, #stackrel(~)(v) = barV#.)

To find #a# and #b# in terms of the critical parameters (which are measurable, whereas #a# and #b# are not), we would therefore be taking the first and second partial derivatives of #P# with respect to #V# at constant #T# (#((delP)/(delbarV))_(T)#, #((del^2P)/(delbarV^2))_(T)#), since the critical point occurs at a single temperature and pressure (zero degrees of freedom).

This is a long derivation, so if you want to go through it, you should set aside around an hour.

These two partial derivatives are:

#0 = ((delP)/(delbarV))_(T) = -(RT)/(barV - b)^2 + (2a)/(barV^3)#

#0 = ((del^2P)/(delbarV^2))_(T) = (del)/(delbarV)[((delP)/(delbarV))_(T)]_T#

#= (del)/(delbarV)[-(RT)/(barV - b)^2 + (2a)/(barV^3)]_T#

#= (2RT)/(barV - b)^3 - (6a)/(barV^4)#

Since these are both equal to #0#, we can compare them and divide to get #barV_c#. First, we have these two expressions:

(1) #(RT)/(barV - b)^2 = (2a)/(barV^3)#

(2) #(2RT)/(barV - b)^3 = (6a)/(barV^4)#

Now, divide them as #((1))/((2))# to get:

#(cancel(RT))/(barV_c - b)^2*(barV_c - b)^3/(2cancel(RT)) = (2cancel(a))/(barV_c^3) * (barV_c^4)/(6cancel(a))#

#(barV_c - b)/2 = barV_c/3#

#barV_c - b = 2/3barV_c#

Therefore, the critical volume is #color(green)(barV_c = 3b)#.

Plugging this back into the first partial derivative, (1), gives:

#(RT_c)/(3b - b)^2 = (2a)/((3b)^3)#

#(RT_c)/(4b^2) = (2a)/(27b^3)#

Therefore, with a little rearrangement, the critical temperature is:

#color(green)(T_c = (8a)/(27Rb))#

Similarly, the critical pressure is then found by plugging #T_c# and #barV_c# back into the original vdW equation:

#color(green)(P_c) = (RT_c)/(barV_c - b) - a/(barV_c^2)#

#= ((8acancel(R))/(27cancel(R)b))/(3b - b) - a/((3b)^2)#

#= ((8a)/(27b))/(2b) - a/(9b^2)#

#= (4a)/(27b^2) - (3a)/(27b^2)#

#= color(green)((a)/(27b^2))#

And now that we have all that, we can find #a# and #b# in terms of #P_c# and #T_c#! To do that, try dividing the expressions for #T_c# and #P_c#:

#T_c/P_c = (8cancel(a))/(cancel(27)Rb)*(cancel(27)b^2)/(cancel(a)) = (8b)/R#

As a result, we have:

#color(blue)(b = (RT_c)/(8P_c))#

Finally, we can get #a# by solving the #T_c# equation, now that we know #b#:

#T_c = (8a)/(27R((RT_c)/(8P_c)))#

#= (8a)/(27R)*(8P_c)/(RT_c) = (64aP_c)/(27R^2T_c)#

Solving for #a# gives:

#color(blue)(a = (27R^2T_c^2)/(64P_c))#