What is the major factor that accounts for most of the difference in these two values of pressure (ideal gas law vs. van der Waals equation)?

me

I'm not sure what the difference between B and D is? Why wouldn't they both affect it?

1 Answer
Truong-Son N.
Nov 4, 2017

It's a matter of semantics. While #B# does change the size of the particles relative to atomic #"Cl"# (which causes #D#), it changes the nature of the problem (consequently changing the van der Waals constants #a# and #b#, and the pressures calculated!).

#D# does not require you to change the nature of the problem; the forces of attraction of #"Cl"_2# molecules to each other change the pressure that is needed to compress them to that molar volume of #"0.33 L/mol"#.

The above pressures say that since a lower pressure is actually needed to reach #"0.33 L/mol"# when using the vdW equation of state, the #"Cl"_2# molecules are very compressible, and their compressibility factor

#Z = (PV)/(nRT)#

is less than #1# (i.e. forces of attraction dominate forces of repulsion for #"Cl"_2#). In fact, for #"Cl"_2# at this temperature,

#Z = (("22.5 atm")("2.00 L"))/(("6.00 mols")("0.082057 L"cdot"atm/mol"cdot"K")("273 K"))#

#~~ 0.3348 < 1#

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