textex - 15-12-2013 at 05:11

Well this is more of related to thermodynamics, but I guess that it's some expertise of that here aswell .

Anyways we had a exam before the weekend and there was one task that i feel was pretty self-contradictory. And because I dont want to look like a fool when complaining to the person that made the exam I figured I'd shoot it out here first:

You have Air, which may be assumed as an ideal gas that is at 1Bar and 60 degrees centigrade. It is then adiabaticly compressed to 5 degrees centrigrade.

As I see it the temperature would obviously increase when compressing the gas. I can't think of any situation where such a thing as presented in the task could make sense.

smaerd - 15-12-2013 at 06:21

You're correct it shouldn't be compressed it would need to be expanded for temperature to decrease. As:

T1 = T2 * (V1/V2)^(y-1)

Here's a half-ass proof...

Case: V2 < V1 ie: compression:
1 = (T2/T1) * (V1/V2)^(y-1)
1 = (T2/T1) * (V1^(y-1))/(V2^(y-1))
V2^(y-1) = T2/T1 * V1^(y-1)
V2 = (T2/T1)^(1/y-1) * V1
V2/V1 = (T2/T1)^(1/y-1)
Let V2/V1 be any real number, V, which represents a ratio of the final volume to the initial volume which is less then one but greater then zero because V2 < V1 and is non-negative due to physical law.
V^(y-1) = T2/T1
V^(y-1) * T1 = T2

Where y = 1 + 2/D.O.F. for an ideal monoatomic gas D.O.F. = 3. => y= 1+ 2/3 = 5/3. Which represents the largest value possible for gamma, as the larger the gas molecules the greater degrees of freedom. So for larger molecules y = 1 + 2/D.O.F will be smaller in value. Therefore the exponential (y - 1) = 1 + 2/D.O.F - 1 = 2/D.O.F.. The ratio 2/D.O.F. will always be < 1 because D.O.F. has a minimum value of 3. So therefore, 0 < V^(y-1) < 1 . Let F be any real number which represents V^(y-1).

Edit: Actually V raised to any real number will result in a number 0 < V < 1 . So D.O.F. is kind of irrelevant in this case...

F * T1 = T2

Because 0 < F < 1, T2 must then be less then T1 in the case of compression.

[Edited on 15-12-2013 by smaerd]

Metacelsus - 15-12-2013 at 06:23

Yes, you are right. Maybe they meant 5 bar? (or expanded to 5 C?)

Air can be reasonably assumed to be a diatomic gas, in which case gamma = 7/5.

[Edited on 15-12-2013 by Cheddite Cheese]

textex - 15-12-2013 at 07:36

Thanks for the input guys! I forgot to add that gamma were given in the task as 7/5. I'll take this up with the professor we had that made the exam, which i highly suspect of having very little knowledge in what we are supposed to be taught. Most of the lectures we have had has been consisting deriving equations right from the book, or powerpoint-presentations consisting of content copied right from sites like chemguide.

[Edited on 15-12-2013 by textex]

smaerd - 15-12-2013 at 07:50

Never underestimate a professors breadth of knowledge is one thing I learned. Give them the benefit of the doubt and assume it was a typo. Understandably you are frustrated though. I wonder if there's a way to use LaTeX on here I better go utfse to see.