Reading Quiz

Question 1:

Consider eq. (3.39): P = T (∂S/∂V)_U,N. Why is the T there?

Answer:

For units. See the discussion in the paragraph preceeding eq. (3.39).

Because we need units of pressure, and the T helps us get that. This is ok because we've assumed thermal equilibrium.
Partial S/V gives units of pressure/temperature, and multiplication by temperature gets pressure (which we want). This is permitted because the two systems are at the same temperature.
The T is there because the two systems which eq. 3.39 analyzes are in thermal equilibrium. The T is needed to make the answer end up in units of pressure.
The T is there because the units of (dS/dV) are Pa/K, so by dimensional analysis we must multiply by a temperature to get something with units of pressure.
This equation had no proper proof, and was just a postulation. Thus the T was added to make the units come out because entropy has a K term in it. Also T is in the top of the equation because we want the pressure to be high when the temperature is high.
T is there to make the units work out. This is allowed because T_A = T_B.
if you calculate the ds/dv term you get Nk/V. The T makes the ideal gas law pop out.
As far as the text is concerned, it's there to make the units come out right.
We've already assumed that the two systems are in thermal equilibrium, so there is one T associated with them. Without the T, the expression would have a unit of Kelvin in the denominator. Our T is used to take this K out of the units.
The equation assumes the systems in question are in thermal equilibrium. Also, T must be there to get units of Pressure on the left side. Thus, the right side is the same for both systems in thermal equilibrium.

Question 2:

Do you agree or disagree with the statement "The only way to change the entropy of a system is via heat flow"? Briefly explain.

Answer:

I disagree. We've seen several counter-examples to this statement: mixing entropy and free expansion, neither of which involve heat flow.

Disagree! In the example of very fast compression, the change in entropy of the system is done by work, not heat.
I disagree basing my decision on the examples about mechanical work on a piston and free expansion into a vacuum. There's no heat flow here, but extra entropy is created.
I disagree with that statement. You can create new entropy via free expansion. If the hole is tiny enough, no heat flow is detected, and the increase in volume of the gas is small, the entropy of the system is changed.
Disagree. In the case of the free expansion of a gas, no heat flows into or out of the system, but the entropy increases.
Disagree. As we've seen before free expansion of a gas adds entropy to the system without having a heat flow. Here W, Q and U are zero but the change in entropy is not. The example of the fast moving piston also shows entropy gained from processes other than heat flow.
I do not agree. Entropy depends on volume, energy, number of particles,....
Nope. Looking at eq 3.30, if we solve for ds, we get P dv/T, so we can change entropy by changing volume, which has nothing to do with heat flow (well i mean it can but its not NECESSARY.)
No, when a gas expands into a vacuum, no heat is done, but the entropy does increase.
I agree. Even changes in pressure occur do to changes in volume which are caused by work done on/by the gas. This involves heat flow.
No. You can get a dS/dV at constant U by eqn 3.39 by having P/T on the left side of the eqn.

Question 3:

(based on Problem 3.27) Consider eq. (3.46), which gives the thermodynamic identity (holding N fixed). What partial-derivative relation can you derive by considering a process that takes place at constant entropy? Does the resulting equation agree with something you already knew? Briefly explain.

Answer:

Constant entropy means dS = 0, so the thermodynamic identity dU = TdS - PdV becomes dU = -PdV. The physical interpretation of no entropy change must be that no heat can flow, since heat flow is caused by the tendency of the entropy to increase. So this just reproduces the First Law of Thermodynamics for a process involving a "purely mechanical" compression.

If the process is at constant entropy, then dU=-PdV, which is the equation for work done during quasistatic compression.
P = -dU/dV. Looks like an adiabatic expansion of an ideal gas at constant pressure. U = 0 + w = -Pdv
If S is constant, then dS = 0, and dU = -PdV Also, W = -PdV This is another form of the equation for the total energy inside the system Delta_U = Q + W, only there is no heat flow within the system. Work makes up all the energy in the system. This makes sense, considering that no new entropy would be produced within a system without a transfer of heat, unless there was free expansion. I am making an assumption that free expansion does not happen here because the entropy is held constant.
If entropy is constant, dS = 0, so dU = -P dV. This says no heat is flowing between the system and the rest of the universe and the only work done is compression work.
Constant entropy -> dS=0. Therefore dU = -PdV. Here if there is no heat flow into the system then dU is just dW by the first Law. That means that W = -PdV, the work done by an isothermal process.
If dS = 0, we get that dU = -PdV. But -PdV = W, so dU = W, which is consistent with the first law of thermodynamics under adiabatic compression (Q = 0).
dU/dV=-P This makes sence. if you replacs dU with Tds, you can solve the equation to find the ideal gas law.
Constant entropy implies dS = 0, thus dU = -PdV which looks like the energy equation for a quasistatic process in which the heat (in this case TdS) is zero.
du/dv=-P. Yes this result agrees with an equation we learned a while back. delta U = -P*delta V.
if dS=0 then PdV=0 which means work = 0? I think that's what this is getting at but I'm not sure...

Question 4:

Compare eq. (3.17) to eq. (3.48); they look awfully similar. What do we know know about this relationship that we didn't already know after our discussion of Section 3.2?

Answer:

In eq. (3.17), we were limited to constant volume processes with no other kind of work being done. Now, we can use this relation for any quasistatic process.

They sure do look similar. But we only knew that (3.17) worked if volume was held constant, but now we know that (3.48) still works if volume changes quasistatically (and if all other variables are constant and there are no other forms of work done).
It is valid when expansion work is being done if the process takes place quasitatically.
If the change in volume takes place quasistatically, the change in entropy is still Q/T, even if work is being done on it in the process.
We know that this is true even if work is being done on the system.
That this relationship only holds for quasistatic processes. Sometimes you can have dS>Q/T if the processes doesn't happen quasistatically, or slowly enough.
This relationship now applies when the system doesn't have constant volume.
This relation works not for only situations of constant volumes doing no work, but now we can see it works in quasistatic situations where work is done.
The only difference is that now we can generalize the expression to situations when the volume is not constant, and there is work being done.
This new equation allows us to use it even if there is a slight change in volume. Equation 3.17 did not allow for changes in volume.
That this holds true even if there is work being done during the process.

Question 5:

What material from the reading or previous classes would you like me to go over in more detail?

Answer:

Your responses below.

In an isentropic process, are the heat and the temperature both held constant?
I had some problems comleting the hmwk, but I'll talk to you tomorrow morning.
I'm a little hesitant about the new definition of pressure in this section. Sure, it makes sense, but it seems like the factor of T was pulled out of thin air...
This reading was not too complicated. NOthing really stuck out as difficult.
The homework was difficult.