Clarification on osmotic pressure

In today's lecture, we came used the following equation:

Pi = kB T x_solute / v

Then, we substituted in to get

Pi = kB T x_solute * N_solvent / V_solvent

Here, N_solvent and V_solvent apply to the right hand side of the osmotic pressure experiment; that is, the solution (non pure) phase. Since v is for the pure phase, here we are making the approximation that the molar volume of the solvent is the same on the left and right hand side of the experiment. This is almost certainly the case for small, dilute amounts of solvent.

MSS

Saturation pressure

In case the term "saturation pressure" is causing you confusion, here is something to think about. The saturation pressure at a given temperature is the pressure at which a gas will turn into a liquid upon compression; alternatively, it is the pressure at which a liquid will vaporize upon expansion. The "saturation temperature" is similar: at a given pressure, it is the temperature at which a gas will condense upon cooling, or a liquid will boil upon heating. Thus, there is nothing more complex about a saturation pressure than simply the pressure of liquid-vapor phase equilibrium.

MSS

PS4 5b

In the PS4 5b, showing dCp/dV=0, some used

d/dV [ (dS/dT)_P ]_T = d/dT[ (dS/dV)_T]_P,

However, this converstion does not give 0.

You can't switch the order of the derivatives because they are not compatible (you have different independent variables). d/dV [ (dS/dT)_P ]_T not equal to d/dT[ (dS/dV)_T]_P

To do the switch, you have to have the same set of independent variables. You could do this, for example:

d/dV [(dS/dT)_P]_T
= (dP/dV)_T * d/dP[(dS/dT)_P]_T
= (dP/dV)_T * d/dT[(dS/dP)_T]_P

And then you will find that the right hand term is zero.

Hope that helps,
Scott

Test yourself

Just with a quick look, without consulting any textbooks or resources, see if you are able to fill in the following in your head:

dS = ? ; S = ? ; maximized at constant ?,?,?
dE = ? ; E = ? ; minimized at constant ?,?,?
dA = ? ; A = ? ; minimized at constant ?,?,?
dH = ? ; H = ? ; minimized at constant ?,?,?
dG = ? ; G = ? ; minimized at constant ?,?,?

If you weren't able to rattle off the correct answers without hesitation, I would strongly suggest you invest a short amount of time committing these equations to memory. The reason that I want you to have these equations so down pat is that having them at the forefront of your memory will enable you to analyze and tackle problems so much more efficiently: for example, you will be able to quickly access Maxwell relations, integrate out thermodynamic properties, and connect derivatives to measurables, without having to flip through notes and handouts. The ability to call upon such basic analyses quickly and intuitively is not only important to this course, but a key part of your success as a careful researcher, no matter what field.

As we progress into new material in the next few weeks, it's critical that you don't get lost in the new concepts as a result of difficulties with these fundamental equations. Just a little advice,

Cheers,
MSS

Notes on Maxwell relations

If you're confused about Maxwell relations, here is something of an additional explanation. Consider the derivative:

(dN/dT)_(V,mu)

The first thing you want to do is identify from which thermodynamic potential this derivative stems. To do that, we look at the independent variables. These are the derivative variable (T) as well as the constant condition variables (V,mu). Thus, we want a function whose natural variables are T,V,mu.

We don't actually know offhand such a function, but we can construct one. Look at the Helmholtz free energy,

dA = -S dT -P dV + mu dN

Here we have independent variables T,V, and N. We need to swap mu for N. To do this, we just perform a Legendre transform to get some new potential Y:

Y = A - mu N
dY = -S dT - P dV - N dmu

Now we can use Y to develop our Maxwell relation. We see that our original function is just

-d2Y/dT dmu = -d(dY/dmu)/dT = dN/dT

If we switch the order of the derivatives, we get

-d2Y/dmu dT = -d(dY/dT)/dmu = dS/dmu

Therefore, we find that:

(dN/dT)_(V,mu) = (dS/dmu)_(T,V)

So the general rule of thumb is:

1. Determine which thermodynamic potential underlies the derivative of interest.
2. Construct that potential via Legendre transforms, if necessary.
3. Find the second derivative giving rise to your derivative of interest, and evaluate it in two different cases where the order of the derivatives is switched.

There is one point on which you must be careful. Not every derivative of a thermodynamic property can be related by a Maxwell relation. For example, consider:

(dT/dS)_(V,N)

This derivative comes from the second derivative of the energy:

d2E/dS2

Note that this is not a cross-derivative, between two independent variables. Therefore (dT/dS)_(V,N) cannot be related to another derivative.

MSS

Solutions to practice problems today

Hi class-

You can find solutions to the practice problems you worked today in teams at:

www.engr.ucsb.edu/~shell/che210a/practice4.pdf

Cheers,
MSS

PS3, problem 5b/c

Here's a slightly clarified explanation of this part of the problem. Here, by constant pressure process, we mean that the entire system is at the same pressure throughout the process. That is, even though it may not be at equilibrium, the pressure that it exerts on the container walls remains constant. This should be all that you need to simplify the first law (differential form) to show that constant pressure processes are also constant enthalpy processes. Note that we don't need to assume quasi-static or reversible behavior; we only need to know that the pressure is constant.
MSS