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CarolinaGirl

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So I just can't seem to get these down. I have studied and studied and I am still missing them in the 1001 chem book. Is there an easy way to understand these? Do I just need to memorize the info? I get confused when the question starts throwing out changes in systems, surroundings, universe and says as one inc. one dec. and this is the outcome, etc.

How are you getting this info? Is this big on the mcat?
 

DiverDoc

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IMO, yes it is big on the MCAT. You should hammer these topics until you do understand, along with delta G and their relationships. There are numerous topics/concepts more difficult than this, are you having trouble with anything else?

Also view the sticky posted by Q in the study Q&A, there should be a good explanation there.
 

xucardsfan08

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Entropy = randomness
Enthalpy = amount of work possible from a constant pressure system

The best way to learn this is to first understand the concepts of open/closed systems, surroundings, adiabatic systems, etc.
Next, begin with U (total energy) and work your way to H, then to G.
 

Pansit

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So I just can't seem to get these down. I have studied and studied and I am still missing them in the 1001 chem book. Is there an easy way to understand these? Do I just need to memorize the info? I get confused when the question starts throwing out changes in systems, surroundings, universe and says as one inc. one dec. and this is the outcome, etc.

How are you getting this info? Is this big on the mcat?

Yes this is important...Basic chemistry often relates to a lot of biological systems as well...especially with energy...etc..you need to master these concepts down before the test...and I do mean CONCEPTS...although you need to memorize formulas etc...rarely will you asked about it directly...
 
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RPedigo

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Entropy = randomness
Enthalpy = amount of work possible from a constant pressure system

The best way to learn this is to first understand the concepts of open/closed systems, surroundings, adiabatic systems, etc.
Next, begin with U (total energy) and work your way to H, then to G.

H is enthalpy, and H = U + PV

So the change in enthalpy is:

dH = dU + d(PV)

Using multiplication rule of d(XY) = YdX + XdY

dH = dU + VdP + PdV

And U = q + W

dH = dq + dW + VdP + PdV

And W = -PdV

dH = dq - PdV + VdP + PdV

Simplifying:

dH = dq + VdP

You said constant pressure system, so dP = 0 (change in pressure is zero)

dH = dq
Commonly written as dH = q(p) to denote constant pressure


...so how is that related to the work? If you assume that q = -w, then that's absolutely correct, but that only happens when dU = q + w = 0, which is only in an isothermal process. Or is this just an assumption we make for the MCAT?
 

LeafNinja

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Just pick up a freshman chemistry book and read the chapters (around 3-4) about therodynamics. Entropy and enthalpy are all covered in the 1-3 laws of thermodynamics.

Rpedigo:

For an isothermal system (constant temperature): ∆U = W, since q=C∆T

note* the C in the equation is molar heat capacity, hence no m before it
-----
For a constant volume system: ∆U = q, since W=P∆V
 

LeafNinja

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If you assume that q = -w, then that's absolutely correct, but that only happens when dU = q + w = 0, which is only in an isothermal process. Or is this just an assumption we make for the MCAT?

For an isothermal process, dU = q + w = 0 is wrong.

Isothermal I thought meant T is constant, and thus, q is 0?

That would make dU = w, not dU=0

someone correct me if this is wrong please.

Also can you explain why q=-w?

only when something has constant volume AND temperature, can dU = 0, and q=-w
 

xucardsfan08

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If you really really want to understand the relationships, brush up on your partial derivatives, pick up a physical chemistry book (or thermodynamics) and then you can learn in detail, the relationships between U, H, T, P, V, S, and A.
It is real quite interesting.
Here is where it all comes together:
17356d2c22f43f6fa8f8df912e15e180.png

8449af4a1a6d427aec80440e5128f433.png

6dbe1450676cdda25243794fbe1095a5.png

1636052ecfb2e9fac13e204f19ec3daa.png
 

RPedigo

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Rpedigo:

For an isothermal system (constant temperature): ∆U = W, since q=C∆T

For an isothermal process, dU = q + w = 0 is wrong.

Isothermal I thought meant T is constant, and thus, q is 0?

That would make dU = w, not dU=0

someone correct me if this is wrong please.

Also can you explain why q=-w?

only when something has constant volume AND temperature, can dU = 0, and q=-w

These are both incorrect.



For an isothermal process, dU = 0, and thus q = -w because U = q + w

This is correct.


Adiabatic means no heat transfer, and thus dq would be equal to 0. Isothermal simply means that the temperature is constant-- that does not mean that heat cannot flow in or out of the system, just that it is offset by work done.
 

MrDocNYU

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So I just can't seem to get these down. I have studied and studied and I am still missing them in the 1001 chem book. Is there an easy way to understand these? Do I just need to memorize the info? I get confused when the question starts throwing out changes in systems, surroundings, universe and says as one inc. one dec. and this is the outcome, etc.

How are you getting this info? Is this big on the mcat?

Dear CarolinaGirl,
I apologize on behalf of my really smart peers for throwing on you details usually taken into consideration only in Pchem and Thermo classes. There are actually just a few things that you will need to know for the MCAT, they are all covered in your local Freshman Chem textbook, very uninterestingly, that is. The following should serve you as an introduction.

Systems are part of the universe that we are interested in. You (or MCAT) choses the system. Nothing is special about it.
Isolated systems cannot exchange matter or energy with their environments. (an insulator)
Open systems are the opposite.
Closed systems can exchange heat but not matter.


1. Thermodynamic questions are common in the MCAT because they really affect all science. You should know that the First Law is that Energy is conserved. No surprise here. If someone proves this wrong, scientists are out of business. The first law also says that in most situations, energy (U) comes in the form of either work (w) or heat (q). U = W + Q. So if you got a system that has a certain amount of internal energy, you won't just lose this energy since it is conserved. The only way that the energy of that system can be decreased is by heat or work. You may know in physics that work is w= F*d = P*V. Or you know now. Heat is still not defined, we think that we know what it is, but we don't.

2. The Second Law introduces Entropy. What it says is that all processes in this world that happen spontaneously will result an increase in entropy, or at the very least, the entropy should remain the same. You can think of entropy as a measure of disorder, but it is defined mathematically as dS = dq/T, where T is the absolute temperature. Entropy is a very powerful concept.

3. The third law of thermodynamic makes the bold assumption that we cannot reach absolute zero. At the limit of that temperature, entropy of all system is zero. However, it does not mean that all molecules would become static (no type of motion whatsoever) at 0 K.

Yes, I knew that all you asked me for was enthalpy and entropy, but you need to understand the above concepts before going futher. Good.

Now that we have talked somehow about entropy, let's talk about enthalpy.

Enthalpy is just a mathematical expression defined as H = U + PV. The only reason that this ENTHALPY thing was invented is because the change in enthalpy turns out to be equal to the energy of a system when pressure is constant. The advatage is that the CHANGE in enthalpy is easily measurable. For experimental scientists, it means that it is worth going through the headache of confusing generations of science students.

If you have an enthalpy problem, nine out of ten times, all you will have to do is take the change in enthalpy of the products and substract from it the change in enthalpy of the reactants. You should multiply everything by their stoichiometric coefficients, that means that you usually need a balanced equation. Know that a reaction with negative DH is exothermic and gives off heat, while positive DH is endothermic and absorbs heat. This might not be intuitive. Make sure you know it.

Seriously, I just answered all your questions. But there are related questions that MCAT wants you to know about.

There is a concept called Gibb's Free Energy, (DG) it supposed to mean the amount of useful work that can be obtained from a system at constant temperature and constant pressure. The famous MCAT equation is DG = DH - TDS. DG has to be negative for a reaction to be spontaneous. A -DG is exerGONIC, a +DG is endergonic. This equation implies that just because a system is exoTHERMIC, does not mean that it will be spontaneous.

I think that you are set.

Now some demystifications from previous posts:
- You need to learn definitions for certain questions
Adiabatic: no heat in or out
Isothermal: temperature is constant
Isobaric: pressure is constant
Isochoric:constant volume

- And a few facts
- Heat taken by no system while there is no phase change: q=mcDT (mcat!)
- For gases, U is only dependent on temperature, so DU is 0 for isothermal processes. Then from first law, 0= Q+W or W=-Q. (and its not wrong for ideal gases. Don't know about other things.)
- For an adiabatic process, no heat is transfered so, U = 0 + W.

Anyone needs a science tutor?
 

HippocratesX

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MrNYUDoc, could u explain that last statement about "heat taken by no system while there is no phase change" a little more? I'm not quite sure I understood that part?
 

RPedigo

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MrNYUDoc, could u explain that last statement about "heat taken by no system while there is no phase change" a little more? I'm not quite sure I understood that part?

http://pages.physics.cornell.edu/courses/p101-102/p101/9/st/gif/q01p01.gif

You can see here that there are times where adding heat increases the temperature, and then BAM! a plateau is hit where no temperature increase occurs... yet we're adding heat. This seems like strange phenomenon, but it's really not too out of this world.

Basically, when we hit that plateau, we're at a phase change (solid -> liquid, liquid -> gas), and we need to overcome the energy required to do that specific phase change. If we're heating water from -10 celcius to 50 celcius, we have to do more calculations than just q = mc∆T because we have a phase change (solid -> liquid). So we use q = mc∆T to go from -10 celcius to 0 celcius-- now we're at the point where solid ice turns into liquid water, so we have to overcome the ∆H of fusion (or melting, same thing, different nomenclature). Then once we've overcome that phase change energy, we're in a liquid state and can continue heating. Use q = mc∆T to heat up to 50 celcius from 0 celcius now and we're done!

Just be cognizant that phase changes have an energy change associated with them.
 

MrDocNYU

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http://pages.physics.cornell.edu/courses/p101-102/p101/9/st/gif/q01p01.gif

You can see here that there are times where adding heat increases the temperature, and then BAM! a plateau is hit where no temperature increase occurs... yet we're adding heat. This seems like strange phenomenon, but it's really not too out of this world.

Basically, when we hit that plateau, we're at a phase change (solid -> liquid, liquid -> gas), and we need to overcome the energy required to do that specific phase change. If we're heating water from -10 celcius to 50 celcius, we have to do more calculations than just q = mc∆T because we have a phase change (solid -> liquid). So we use q = mc∆T to go from -10 celcius to 0 celcius-- now we're at the point where solid ice turns into liquid water, so we have to overcome the ∆H of fusion (or melting, same thing, different nomenclature). Then once we've overcome that phase change energy, we're in a liquid state and can continue heating. Use q = mc∆T to heat up to 50 celcius from 0 celcius now and we're done!

Just be cognizant that phase changes have an energy change associated with them.

Yes!!
Isn't it cool that you can add heat to a system without raising the temperature by a single degree?
 
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