0
$\begingroup$

An option in my test says:

"The osmotic pressure of a dilute solution is the same as it would exert if it exists as a gas in the same volume of the solution and at same temperature."

I'm not able to think how should I relate between a solution and a gaseous mixture. Am I missing something?

$\endgroup$
  • 1
    $\begingroup$ The equations are analogous but in my opinion that is a misstatement. The entire solution, if evaporated, would not exert the same pressure. It is more subtle than that. It is the solute that would exert an equivalent pressure if a gas. $\endgroup$ – Buck Thorn May 16 at 16:57
  • $\begingroup$ thank you ill check the derivation of the osmotic pressure $\endgroup$ – charan sahith May 16 at 17:15
  • $\begingroup$ Dah.. deleted my earlier comment. // Still believe... The sentence doesn't seem to be readily comprehensible to me. It might make more sense if the full context was known. // I think what the sentence is trying to point out the association of gas pressure and osmotic pressure. Think of a setup to measure osmotic pressure. Initially the opposite sides start off with the same water level. Think of solute side as a gas with $V_0$ and $P_0$. If the solute side liquid had been a gas and confined to its original volume, then the pressure would go up to $P_1$, where $P_1$ is the osmotic pressure. $\endgroup$ – MaxW May 16 at 17:47
  • $\begingroup$ my teacher had said that in calculation of osmotic pressure we consider the solution to be ideal so when it is ideal the state , that is gas or liquid doesnt matter. Does this seem right? I think you are also trying to tell the same thing $\endgroup$ – charan sahith May 17 at 5:36
  • $\begingroup$ @charansahith The condition of ideal behavior (essentially ignoring interactions between has molecules or solute molecules) is a key condition during derivation of an expression for the osmotic pressure. The underlying reason for the similarity has presumably to do with entropy associated with a volume expansion (rather than enthalpy changes) playing a central role in both gas and osmotic expansion work. $\endgroup$ – Buck Thorn May 17 at 18:24
1
$\begingroup$

In both the case of the osmotic pressure of a dilute solution and the case of the pressure exerted by an ideal gas, the solute or gas may be described as composed of non-interacting (ideal) particles, and the mathematical expressions (equations of state) describing the two situations are very similar (in one case $p=cRT$, in the other $\pi = cRT$)$^\dagger$. However it might be less confusing if equivalent to state that in both cases the equations describe similar relationships between the work required to change the volume of the system and the accompanying change in the concentration of gas or solute. In both cases work can be done by the system through an expansion, but in one case the expansion results from pressure exerted by the gas, while in the other it results from pressure exerted by the solvent. In the case of osmotic pressure, since the chemical potential of the solvent is coupled to that of the solute (as described by the Gibbs-Duhem relation) it is possible to relate the osmotic pressure to the solute concentration (in the limit of an ideal solution as described by Henry's Law).

$^\dagger$As I commented, the equations are analogous but in my opinion, "The osmotic pressure of a dilute solution is the same as it would exert if it exists as a gas in the same volume of the solution and at same temperature." is a misstatement. The entire solution, if evaporated, would not exert the same pressure. It is more subtle than that. It is the solute that would exert an equivalent pressure if a gas.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.