[Needaask]

1)The formula for ΔS is ΔS=Q/T. But I can't really relate this to what I learned. A change in entropy is when the substance becomes more disordered in the final product when compared to its initial product. Why would there be a energy change here?

Watching KhanAcademy's video allowed me to better understand what is meant by becoming more disordered as a macro state point of view. But I don't really understand why there is a energy component here. What does the Q present and how does it apply to the orderliness of the system?

2)The enthalpy or heat content (internal energy) of an element is not 0 right? While the enthalpy change of formation of an element is 0 the actual enthalpy level (same as internal energy) is going to be a positive value right? That being said, we cannot compare the thermal stability of 2 products of different enthalpy of formation reactions with their respective enthalpy changes unless they have the same type and amount of reactants right? Because the enthalpy change gives us the change of energy level from reactant to product only and not the actual enthalpy level right?

So unless they told us the reactants have a total enthalpy level of 100J then the reaction is exothermic as 50J of energy is given out and it expands by 10J. Then we can infer that the products have a total enthalpy level of 40J? Is this correct?

Thanks for the help

[curiouscat]

1)The formula for ΔS is ΔS=Q/T. But I can't really relate this to what I learned. A change in entropy is when the substance becomes more disordered in the final product when compared to its initial product. Why would there be a energy change here?

Heat causes disorder.

So unless they told us the reactants have a total enthalpy level of 100J then the reaction is exothermic as 50J of energy is given out and it expands by 10J. Then we can infer that the products have a total enthalpy level of 40J? Is this correct?

Absolute enthalpies mean nothing. Only differences matter.

[Corribus]

Calling entropy a measure of disorder can lead to some confusion.  Perhaps a better way to think of it is a measure of how energy is concentrated and dispersed.

I have spoken here before about the concept of an "entropic force": http://www.chemicalforums.com/index.php?topic=68287.msg246509#msg246509

Take a simple example of an expanding ideal gas, which is driven purely by entropy.  Actually, take the reverse process - compression of an ideal gas.  It takes a force to do this, which means it takes work, because the force is applied over a distance.  The work it takes to compress an ideal gas at a particular temperature is related to the entropy change of similar expansion at that temperature.  The reason a gas expands is because there's a natural tendency for energy to disperse over a larger volume.  When a gas is compressed, its kinetic energy is concentrated into a small volume.  When the gas expands, that same quantity of energy is now spread over a larger volume.  This is a more favorable situation.  The tendency to expand is higher at higher temperatures because at higher temperatures more kinetic energy is concentrated into the same volume.  This is why the entropy effect on spontaneity is temperature dependent.

You can also go the opposite direction (compress the gas).  However this cannot happen spontaneously, and indeed cannot happen without energy being dispersed elsewhere - in a device which compressed the gas, for instance, some energy is lost as heat, which becomes dispersed throughout the environment.  This is why entropy of the universe must always increase for a process.  It is why also there is no such thing as a perpetual motion machine. No process is 100% efficient.  Some heat (energy) is always dispersed to the universe.

[Needaask]

Calling entropy a measure of disorder can lead to some confusion.  Perhaps a better way to think of it is a measure of how energy is concentrated and dispersed.

I have spoken here before about the concept of an "entropic force": http://www.chemicalforums.com/index.php?topic=68287.msg246509#msg246509

Take a simple example of an expanding ideal gas, which is driven purely by entropy.  Actually, take the reverse process - compression of an ideal gas.  It takes a force to do this, which means it takes work, because the force is applied over a distance.  The work it takes to compress an ideal gas at a particular temperature is related to the entropy change of similar expansion at that temperature.  The reason a gas expands is because there's a natural tendency for energy to disperse over a larger volume.  When a gas is compressed, its kinetic energy is concentrated into a small volume.  When the gas expands, that same quantity of energy is now spread over a larger volume.  This is a more favorable situation.  The tendency to expand is higher at higher temperatures because at higher temperatures more kinetic energy is concentrated into the same volume.  This is why the entropy effect on spontaneity is temperature dependent.

You can also go the opposite direction (compress the gas).  However this cannot happen spontaneously, and indeed cannot happen without energy being dispersed elsewhere - in a device which compressed the gas, for instance, some energy is lost as heat, which becomes dispersed throughout the environment.  This is why entropy of the universe must always increase for a process.  It is why also there is no such thing as a perpetual motion machine. No process is 100% efficient.  Some heat (energy) is always dispersed to the universe.

Oh so when energy is gets dispersed entropy increases. What about a gas heated at constant volume? When a gas is heated it there can be more states of the particles so entropy increases and the Q would be for the heat gained when its heated right? But still the energy isn't dispersed through a larger volume?

Also, during expansion there is no energy supplied even though it becomes more disperse. So what would be placed under the Q here?

Thanks Corribus

[curiouscat]

What about a gas heated at constant volume?

S increases.

Also, during expansion there is no energy supplied even though it becomes more disperse. So what would be placed under the Q here?

You can have an isentropic expansion. ΔS=0

[Needaask]

What about a gas heated at constant volume?

S increases.

Also, during expansion there is no energy supplied even though it becomes more disperse. So what would be placed under the Q here?

You can have an isentropic expansion. ΔS=0

But if a gas expands doesn't S increase? And if so, ΔS=Q/T what would Q be in this case?

Also, may i check with you about the enthalpy level of an element? According to wikipedia http://en.wikipedia.org/wiki/Enthalpy H=Q+w and since any element like hydrogen has some internal energy and also it occupies volume and exerts a pressure, it has to be non-zero right?

[curiouscat]

You can have an isentropic expansion. ΔS=0

But if a gas expands doesn't S increase? And if so, ΔS=Q/T what would Q be in this case?

Zero, I think. Assuming processes is reversible.

[curiouscat]

Also, may i check with you about the enthalpy level of an element? According to wikipedia http://en.wikipedia.org/wiki/Enthalpy H=Q+w and since any element like hydrogen has some internal energy and also it occupies volume and exerts a pressure, it has to be non-zero right?

Absolute enthalpy is meaningless. Only differences matter.

[Needaask]

You can have an isentropic expansion. ΔS=0

But if a gas expands doesn't S increase? And if so, ΔS=Q/T what would Q be in this case?

Zero, I think. Assuming processes is reversible.

But why would it be zero? As Corribus mentioned isn't the energy more dispersed now?

[Corribus]

What I wrote was for your conceptual help, maybe not to be generalized too much.  The entropy change of a true expansion depends on the expansion conditions.  If the expansion is reversible and adiabatic (no heat energy gained or lost from the system), the entropy change is also zero.  In fact, adiabatic and isentropic are often interchangeably used terms. 

Remember, lots of things are going on at one time in the expansion of a gas, and how some parameters change depend a lot on what other parameters are kept constant.  Gas expansion can happen with and without a change in temperature, with and without a change in pressure, with and without heat exchange.  These will all impact the degree to which energy is dispersed, and hence the entropy of the system.  Also bear in mind that even if the entropy of the system is unchanged during the process, more than likely the entropy somewhere else (in the universe) is rising. 

I think this is a good website that may help you understand entropy in physical processes:

http://www.av8n.com/physics/gas-laws.htm

See esp. section 3

In the end, thermodynamics is a very difficult subject.  Far more difficult and bloody impossible to make sense of, if you want to know my opinion, than quantum mechanics.  Do not lose too much sleep over it if something doesn't make sense.  After studying thermodynamics in one form or another for over 15 years (depending on when I choose the starting point), I still often have situations where I think I understand something perfectly, and then someone says "But why....?" and then I'm like, "Hell, that doesn't agree with what I thought was the case!"

[Needaask]

What I wrote was for your conceptual help, maybe not to be generalized too much.  The entropy change of a true expansion depends on the expansion conditions.  If the expansion is reversible and adiabatic (no heat energy gained or lost from the system), the entropy change is also zero.  In fact, adiabatic and isentropic are often interchangeably used terms. 

Remember, lots of things are going on at one time in the expansion of a gas, and how some parameters change depend a lot on what other parameters are kept constant.  Gas expansion can happen with and without a change in temperature, with and without a change in pressure, with and without heat exchange.  These will all impact the degree to which energy is dispersed, and hence the entropy of the system.  Also bear in mind that even if the entropy of the system is unchanged during the process, more than likely the entropy somewhere else (in the universe) is rising. 

I think this is a good website that may help you understand entropy in physical processes:

http://www.av8n.com/physics/gas-laws.htm

See esp. section 3

In the end, thermodynamics is a very difficult subject.  Far more difficult and bloody impossible to make sense of, if you want to know my opinion, than quantum mechanics.  Do not lose too much sleep over it if something doesn't make sense.  After studying thermodynamics in one form or another for over 15 years (depending on when I choose the starting point), I still often have situations where I think I understand something perfectly, and then someone says "But why....?" and then I'm like, "Hell, that doesn't agree with what I thought was the case!"

Ohh so heat must be supplied for S to increase? I thought even if no heat was supplied, now that the gas expands it is more disorderly so the entropy must increase?

I get what you mean by it being difficult. I feel like it is very abstract especially on the entropy part.

Lastly, I understand that absolute enthalpy isn't important but i can't say that pure elements have 0 enthalpy right? My teacher was explaining enthalpy of formation in class and he mentioned that when comparing 2 reactions, the more exothermic enthalpy of formation reaction would indicate that the products is more thermally stable than the other reaction's products. He based it upon the idea that pure elements have 0 enthalpy (as he mentioned that enthalpy of formation for elements is 0) so comparing 2 different reaction's products applicable.

However, I disagree with that notion as H=Q+PV so pure substances definitely have a Q as they have some internal energy unless they are at 0K. While some pure elements don't have a PV like graphite, gases such as hydrogen definitely have some pressure exerted over a fixed volume. So how can an element be at 0J?

Thanks for the help

[Corribus]

Needaask,

I had some time to think further about your question.  I can't claim to never make errors in discussing such topics, but here is my current understanding.

I believe a lot depends on how you define the system.  Entropy, as has been remarked earlier, is really only practically useful as a relative concept.  Thus we usually want to know how the entropy of an entire system has changed during a process.  Which means we want to know how energy has been dispersed not within a system, but between the system and its surroundings.  This is what really defines the entropy change of interest.  In an adiabatic process, no heat is exchanged between the system and surroundings.  Therefore the entropy change is zero, because energy has not been redistributed between the system and the surroundings in a statistically meaningful way.  (That is, the energy of the system is the same with respect to the energy of the surroundings.)  But again, this is all based on how we define the system.  The system, I need not remind you, is always an artificial construct.

We might ask again why expansion of an ideal gas is spontaneous.  We know that in an adiabatic expansion, the entropy change for the system is zero.  Can we ask what this means?  I admit to some speculation here, but let us speculate anyway.  In a perfectly adiabatic system, there absolutely is no heat exchange between the system and the surroundings.  This would be akin to a perfectly insulated container.  When an ideal gas expands, the temperature of the gas decreases because the system does work (this is independent of Joule Thompson heating/cooling of a real gas).  An interesting question might be - if the amount of available expansion volume is infinitely large, can the gas expand far enough to fill it?  That may be a trick question (can anything occupy an infinite space?), but ignoring that, we are left with the fact that the farther the gas expands, the cooler it gets, and the pressure goes down.  I don't really have the answer here, but if the temperature drops, gas particles get slower on average.  Eventually, and in theory, I suppose they may stop moving altogether, in which case no more change is possible.  So I might conclude that for a truly reversible, adiabatic proceses, where entropy change is zero, the end "product" can never be reached.

Anyway, that's a fun thought exercise but let's be realistic here.  In fact, gasses do expand spontaneously, so why?  Ok, here's the important thing.  It's not possible to have a truly adiabatic process.  No insulator is perfect, no vaccuum is complete, so no matter how well we isolate a container from the external environment, there will be some heat exchange.  As an ideal gas expands, as we've said, the temperature will drop.  This will create a temperature differential between the system and the external environment.  Heat will then flow between the external environment to the system to compensate, because (you probably see what's coming) this is thermodynamically, and indeed entropically, favorable.  Assuming the environment is large enough to be considered an "infinite" reservoir of heat, and assuming the transfer is fast (though it ultimately makes no difference), the temperature of the system will remain equal to the temperature of the environment, which means the process is "isothermal" (something else that's an idealization).  Because heat must transfer to an expanding gas in any realistic system, the entropy change of an expanding ideal gas is always positive, at least as far as I can tell.  The "adiabatic expansion" is really an idealized limit, which shows that even for a perfect system, the smallest the entropy change can be for an expansion is zero.  For any real expansion, this limit is unreachable, which means an expansion of an ideal gas is always accompanied by a positive change in entropy.

The only reason adiabatic processes are really worth even mentioning (aside from their conceptual usefulness) is this: because internal energy is a state function, we can use any path we want to calculate changes in internal energy (or other state function properties) for a process by separating it into idealized components.  If we want to know how the properties of a gas system change from going from one pressure/volume to another, we can split it up into two sequential changes (an idealized isothermal expansion, followed by an idealized adiabatic expansion), both of which are easy to calculate, and we expect the changes will be equivalent to a single process which is neither isothermal nor adiabatic.  To see a practical explanation of how this works, I encourage you to read a bit about the Carnot cycle, if you have not done so already.

As a final remark, expansion of a real gas is somewhat different because you have additional enthalpic factors to consider, which boil down to interactions between molecules and distribution of energy within internal vibrational/rotational modes which impact the amount of work that is done during an expansion.  Expansion of a real gas, for example, may no longer be spontaneous if it takes more Free Energy to overcome the intermolecular forces than is gained by the entropic effects of heat redistribution between the environment and the system.

So to answer your questions directly:

Question: "so heat must be supplied for S to increase?"

Answer: Yes, supplied by or lost to the environment (or exchanged outside the local system in some way). 

Question: "I thought even if no heat was supplied, now that the gas expands it is more disorderly so the entropy must increase?"

Answer: Disorder within the system is not what primary concern here.  It's disorder in the way is heat is distributed between the system and the environment.  Do note, however, that while adiabatic expansion of a gas has zero entropy regardless of how the expansion takes place (fast, slow), if the system or the surroundings form hot spots during the expansion (which can happen if the expansion is too fast to "iron out" nonisotropic statistical distributions of particles, heat transfer will occur between these micro-regions, which would create a local, positive entropy change.  This is why I am careful to point out that what is defined as the system is quite important.  It also speaks again to the "approximation" or "idealization" of an adiabatic change.

Question: "I get what you mean by it being difficult. I feel like it is very abstract especially on the entropy part."

Answer:  Ok, not really a question, but my own chief problem is that thermodynamic properties can be presented in so many different ways.  They are all ultimately equivalent, but figuring out how the statistical interpretation of entropy is equivalent to the classical interpretation of entropy, is equivalent to the molecular interpretation of entropy... it's all very hard to keep straight.

Question: "Lastly, I understand that absolute enthalpy isn't important but i can't say that pure elements have 0 enthalpy right?"

Answer: You can define anything you want to have 0 enthalpy, if all we're interested in is relative values.  Why, for example, do we call sea level an "altitude of zero"?  Or the freezing point of water as "zero degrees Celcius".  We could pick anything we want as the reference point.  It's not about having "zero enthalpy" just as it's not about having "zero altitude".  Taken alone, this has no physical meaning.  What is meaningful is what the enthalpy of a substance (or the height of a mountain, or the temperature of a pot of water) is to the accepted reference point.  As long as you use a consistent reference point that doesn't change according to conditions, and that other people know and accept, then it doesn't matter.

[Needaask]

Needaask,

I had some time to think further about your question.  I can't claim to never make errors in discussing such topics, but here is my current understanding.

I believe a lot depends on how you define the system.  Entropy, as has been remarked earlier, is really only practically useful as a relative concept.  Thus we usually want to know how the entropy of an entire system has changed during a process.  Which means we want to know how energy has been dispersed not within a system, but between the system and its surroundings.  This is what really defines the entropy change of interest.  In an adiabatic process, no heat is exchanged between the system and surroundings.  Therefore the entropy change is zero, because energy has not been redistributed between the system and the surroundings in a statistically meaningful way.  (That is, the energy of the system is the same with respect to the energy of the surroundings.)  But again, this is all based on how we define the system.  The system, I need not remind you, is always an artificial construct.

We might ask again why expansion of an ideal gas is spontaneous.  We know that in an adiabatic expansion, the entropy change for the system is zero.  Can we ask what this means?  I admit to some speculation here, but let us speculate anyway.  In a perfectly adiabatic system, there absolutely is no heat exchange between the system and the surroundings.  This would be akin to a perfectly insulated container.  When an ideal gas expands, the temperature of the gas decreases because the system does work (this is independent of Joule Thompson heating/cooling of a real gas).  An interesting question might be - if the amount of available expansion volume is infinitely large, can the gas expand far enough to fill it?  That may be a trick question (can anything occupy an infinite space?), but ignoring that, we are left with the fact that the farther the gas expands, the cooler it gets, and the pressure goes down.  I don't really have the answer here, but if the temperature drops, gas particles get slower on average.  Eventually, and in theory, I suppose they may stop moving altogether, in which case no more change is possible.  So I might conclude that for a truly reversible, adiabatic proceses, where entropy change is zero, the end "product" can never be reached.

Anyway, that's a fun thought exercise but let's be realistic here.  In fact, gasses do expand spontaneously, so why?  Ok, here's the important thing.  It's not possible to have a truly adiabatic process.  No insulator is perfect, no vaccuum is complete, so no matter how well we isolate a container from the external environment, there will be some heat exchange.  As an ideal gas expands, as we've said, the temperature will drop.  This will create a temperature differential between the system and the external environment.  Heat will then flow between the external environment to the system to compensate, because (you probably see what's coming) this is thermodynamically, and indeed entropically, favorable.  Assuming the environment is large enough to be considered an "infinite" reservoir of heat, and assuming the transfer is fast (though it ultimately makes no difference), the temperature of the system will remain equal to the temperature of the environment, which means the process is "isothermal" (something else that's an idealization).  Because heat must transfer to an expanding gas in any realistic system, the entropy change of an expanding ideal gas is always positive, at least as far as I can tell.  The "adiabatic expansion" is really an idealized limit, which shows that even for a perfect system, the smallest the entropy change can be for an expansion is zero.  For any real expansion, this limit is unreachable, which means an expansion of an ideal gas is always accompanied by a positive change in entropy.

The only reason adiabatic processes are really worth even mentioning (aside from their conceptual usefulness) is this: because internal energy is a state function, we can use any path we want to calculate changes in internal energy (or other state function properties) for a process by separating it into idealized components.  If we want to know how the properties of a gas system change from going from one pressure/volume to another, we can split it up into two sequential changes (an idealized isothermal expansion, followed by an idealized adiabatic expansion), both of which are easy to calculate, and we expect the changes will be equivalent to a single process which is neither isothermal nor adiabatic.  To see a practical explanation of how this works, I encourage you to read a bit about the Carnot cycle, if you have not done so already.

As a final remark, expansion of a real gas is somewhat different because you have additional enthalpic factors to consider, which boil down to interactions between molecules and distribution of energy within internal vibrational/rotational modes which impact the amount of work that is done during an expansion.  Expansion of a real gas, for example, may no longer be spontaneous if it takes more Free Energy to overcome the intermolecular forces than is gained by the entropic effects of heat redistribution between the environment and the system.

So to answer your questions directly:

Question: "so heat must be supplied for S to increase?"

Answer: Yes, supplied by or lost to the environment (or exchanged outside the local system in some way). 

Question: "I thought even if no heat was supplied, now that the gas expands it is more disorderly so the entropy must increase?"

Answer: Disorder within the system is not what primary concern here.  It's disorder in the way is heat is distributed between the system and the environment.  Do note, however, that while adiabatic expansion of a gas has zero entropy regardless of how the expansion takes place (fast, slow), if the system or the surroundings form hot spots during the expansion (which can happen if the expansion is too fast to "iron out" nonisotropic statistical distributions of particles, heat transfer will occur between these micro-regions, which would create a local, positive entropy change.  This is why I am careful to point out that what is defined as the system is quite important.  It also speaks again to the "approximation" or "idealization" of an adiabatic change.

Question: "I get what you mean by it being difficult. I feel like it is very abstract especially on the entropy part."

Answer:  Ok, not really a question, but my own chief problem is that thermodynamic properties can be presented in so many different ways.  They are all ultimately equivalent, but figuring out how the statistical interpretation of entropy is equivalent to the classical interpretation of entropy, is equivalent to the molecular interpretation of entropy... it's all very hard to keep straight.

Question: "Lastly, I understand that absolute enthalpy isn't important but i can't say that pure elements have 0 enthalpy right?"

Answer: You can define anything you want to have 0 enthalpy, if all we're interested in is relative values.  Why, for example, do we call sea level an "altitude of zero"?  Or the freezing point of water as "zero degrees Celcius".  We could pick anything we want as the reference point.  It's not about having "zero enthalpy" just as it's not about having "zero altitude".  Taken alone, this has no physical meaning.  What is meaningful is what the enthalpy of a substance (or the height of a mountain, or the temperature of a pot of water) is to the accepted reference point.  As long as you use a consistent reference point that doesn't change according to conditions, and that other people know and accept, then it doesn't matter.

Thanks Corribus for the great post. Thanks for taking the time to think about this.

I think I better understand it now. Would the temperature drop as ΔU is 0 and since now work is done by system, w becomes negative and in order for the ΔU to remain constant, Q has to decrease reducing the temperature? Then when this happens due to temperature differences, heat gets absorbed into the system causing the entropy to increase?

If so, then what would the T be? Because I'm thinking this cooling effect during expansion occurs simultaneously as the heating effect so I'm not too sure what would the temperature of the system be. If they occur together, I'd think the temperature remains constant? But according to other sources online the temperature decreases during expansion even though heat is absorbed into the system while this occurs.

Also, according to Wikipedia internal energy is the energy needed to create the system but excludes the energy to displace the system's surroundings;only KE+PE. So why does internal energy if work is done onto the system only? If only work is done to the system the kinetic and potential energy of the gas remains the same so shouldn't the internal energy remain the same?

For enthalpy, but in any case we shouldn't be able to tell what is the stability of the products of two different enthalpy of formations right? Because stability is the level of the enthalpy and since we only know the difference of enthalpy we cannot determine the actual enthalpy level right?

Thanks again for the help