|
|
 |
Reply
| | From:  Albert1145 (Original Message) | Sent: 8/6/2008 9:11 AM |
How are you Steve?
I studying on entropy right now. Its a bit confusing.
Er... I think entropy is sort of the energy of molecules want to be dispersed out into the surrounding rather than staying in a system. Is that right?
My second question is: I know pressure affects entropy, would low pressure have greater entropy than one with higher pressure?
I did my best at wording my own definition of entropy, I'm very sorry if it is extremely confusing. |
|
 First
Previous
2-8 of 8
Next
Last
|
|
Reply
| | | From: ·Steve· | Sent: 8/6/2008 2:03 PM |
Hi Albert, long time no type! I hope chemistry has been treating you well. The entropy of a system is the "degree of randomness" or "degree of disorder" it has. A gas has a higher entropy than a liquid, and a liquid has a higher entropy than a solid. Some people do look at entropy in terms of energy. There is a natural tendency for energy to spread out, to become less concentrated in one location. Entropy can be interpreted as a measure of this tendency. If you have the same number of gas molecules at the same temperature, but in different volumes, then their pressures will be different. The gas in the larger container will exert a lower pressure, and the gas in the smaller container will exert a higher pressure. In the larger volume, the gas molecules are more spread out, so they have more "degrees of freedom." The more spread out the molecules are, the more "random" the system is, and therefore, the higher its entropy is. Steve |
|
Reply
| |
Hi, thanx for your help, I understand entropy a lot better now.
It is long time no type, I've just had a month long holiday, so getting back to entropy isn't really the mood.
I guess my mood right now is a really really huge entropy! :) |
|
Reply
| |
Hi, sorry, I just want to clear a few things up! Could you correct me if I'm wrong?
1. ∆S universe = ∆S system + ∆S surrounding
2. ∆S surrounding = -�?Entropy, H) system / Temperature
3. Positive ∆S surrounding indicates an exothermic reaction
4. Positive ∆S universe indicates a spontaneous reaction
And just to touch a bit on Gibbs free energy:
5. Negative ∆G is spontaneous
Are they right? |
|
Reply
| | | From: ·Steve· | Sent: 8/10/2008 9:17 PM |
1. ∆S universe = ∆S system + ∆S surrounding Correct.  2. ∆S surrounding = –∆(Entropy, H) system / Temperature This should be, ∆S surrounding = –∆(Enthalpy, ∆H) system / Kelvin Temperature 3. Positive ∆S surrounding indicates an exothermic reaction This is correct, according to the equation in (2) above. When ∆H of the system is negative (–∆Hsys), then ∆Ssurr = �?–∆Hsys / T) = positive value. (The temperature T in Kelvins is always a positive number.) 4. Positive ∆S universe indicates a spontaneous reaction Correct. |
|
Reply
| |
| omg i remember that section lol and the enthalpy and all that crazy stuff |
|
Reply
| | | From: ·Steve· | Sent: 8/12/2008 9:17 PM |
Where energy is concerned, that's true. The author of one introductory thermodynamics book said that no textbook on the subject should have the entry "energy, definition of" in the index because there was no such thing as a definition of energy!  |
|
|
First
Previous
2-8 of 8
Next
Last
|
|
|
Free Forum Hosting
Important Announcement
Return to All Message Boards