# The relationship of the osmotic pressure of a polymeric solution and its the molecular mass

Why is the osmotic pressure of a polymeric solution decreases when the molecular mass is increasing?

I am honestly unsure if the above statement is true, at least it seems to me to be true. When I look at the equation of Osmometry it states that:

If $${\pi} =$$ osmotic pressure; $$R =$$ ideal gas constant; $$M =$$ molecular mass; $$T =$$ temperature; $$A_2 =$$ second viral coefficient; $$c =$$ concentration; $$\chi =$$ Flor-Huggins-Interaction parameter; and $$V =$$ volume, following equations are given:

$$\frac{\pi}{RTc} = \frac{1}{M}+ A_2 c \tag1$$

with

$$A_2 = \left(\frac {1}{2}-\chi\right) \left(\frac{V}{M} \right)^2 \tag2$$

2 in 1

$$\frac{\pi}{RTc} = \frac{1}{M} + \left(\frac {1}{2}-\chi \right)\left(\frac{V}{M}\right)^2 c$$

So if $$M$$ will now rise the osmotic pressure will decrease, this is what I understand from the equation. I also created some synthetic data, where I assumed the second viral coefficient to be 5, which seemed like a valid estimate based on a paper I was reading.

So far so good, but what I do not understand is when I now picture a big polymer chain that will occupy space in a solution and a small one, certainly the bigger chain has more chain solvent interactions. Also from my perspective when I think in occupied space as for concentration, and with higher concentration, I get a higher osmotic pressure, then certainly I thought with more occupied space due to higher molecular mass I should see the same effect.

Background - read-only if you are interested: I observed that covalently bound polymers to a surface will detach more easily if they are bigger. Hence I thought this comes from the entropic pressure on the chain and I was hoping that osmotic pressure would back me up here.

The statement is true if the mass concentration (e.g., $$\pu{g//L}$$) of polymers in solution is constant, but false if the number concentration (e.g.,$$\pu{mol//L}$$) of polymers in solution is constant.

Here are some examples corresponding to the above. Let's start with the second case:

Increase molar mass while keeping the number concentration of polymers constant. An example of this would be going from $$N$$ polymers of length $$n$$ to $$N$$ polymers of length $$2n$$. Longer polymers have more effect on osmotic pressure, because they provide more interactions with the solvent. Thus increasing the molar mass at constant number concentration will increase the osmotic pressure.

And now the first case: Increase molar mass while keeping the mass concentration of polymers constant. An example of this would be to go from $$2N$$ polymers of length $$n$$ to $$N$$ polymers of length $$2n$$. Note the mass would be the same in both cases.

The osmotic pressure will decrease because the former has a higher osmotic pressure than the latter. That's because, while the number of possible exchange configurations with the water is higher for a longer polymer than a shorter polymer, the exchange configurations available to the monomers are constrained by the fact that they are connected to each other. If you cleave the polymer into two pieces, you provide more degrees of freedom for the same number of monomers (i.e., for the same mass), and thus increase the osmotic pressure.

The limiting case would be if you divided the polymer into individual monomers. That would give you the maximum possible osmotic pressure for a given mass of polymer (of course, it would no longer be a polymer at that point).

[Note: In my example I described what happens when going from long polymers to short polymers at constant mass concentration: The osmotic pressure increases. The question posits the reverse case (increasing the polymer length at constant mass), so there the osmotic pressure would decrease.]

Going back to the original statement, my guess is that the author was thinking of constant mass concentration, rather than constant number concentration. Since mass concentration is often used for polymer solutions, that would not be surprising.

• Okay, I think I understand it now. Thank you for your explanation! Commented Nov 9, 2021 at 10:01