How much work is done by the Adiabatic Compression of Diatomic Gasses

In summary: I'm going to email my professor. I don't see anything in that article that helps me. Like the textbook it just goes over my head. Thanks for...In summary, the gas volume after compression is .249 Liters.
  • #1
grandpa2390
474
14
I apologize if my questions were in the wrong place. I am in "Advanced" Physics Classes. And so I post the questions from those classes in the Advanced Physics forum. And these are questions that weren't covered in Introductory Physics (1st three semesters). But they keep being moved to this forum. I don't understand the criteria. I guess that forum is for doctoral students and above? The "Really Advanced" stuff lol. regular graduate level classes and below go here? Want to make sure my questions are in the right place. 1 liter of ideal diatomic gas (allow for rotation but not vibration) at a pressure of 1 atmosphere and a temperature of 300 K is compressed adiabatically to a pressure of 7 atm. What is the volume of the gas after compression?

so I found the volume to be .249 Liters
the final temperature after compression is 523 K

what I can't find is the work that was done during compression on the gas.
I drew a graph of the compression and found the equation of the line to be ##(-6/.751)##*v+8.989=p
i then tried integrating that over v from 1 Liter to .249 Liters. but the answer I got of 303 J is wrong...

What am I doing wrong?

I apologize about not having the guidelines on here. I accidentally erased them, and then the forum saved the draft. and I can't get them back...
 
Physics news on Phys.org
  • #2
One thing from the homework template that is missing is the "Relevant equations" part. Can you write the relevant equations?

Also, it is very helpful for the helpers to see the details or your previous calculations.
 
  • #3
grandpa2390 said:
what I can't find is the work that was done during compression on the gas.
I drew a graph of the compression and found the equation of the line to be ##(-6/.751)##*v+8.989=p
i then tried integrating that over v from 1 Liter to .249 Liters. but the answer I got of 303 J is wrong...

What am I doing wrong?
What is the relation between pressure and volume during an adiabatic compression/expansion? It is not linear!
 
  • Like
Likes grandpa2390
  • #4
ehild said:
What is the relation between pressure and volume during an adiabatic compression/expansion? It is not linear!
You are right. In class he did an example that was not linear. I'm not sure what the relationship is though because in the example it wasn't necessary. We did get a formula using log(T/T2), maybe that one is it

ln(T2/T1)^3/2 = ln (v1/v2)...or no, this one. W=-NKTln(V2/V1) is probably my best bet

no, that didn't work. I realized that formula is for isothermic because it only takes one T into account.

I found on another post where the helper asked if the person could find ##C_v## I found it using (f/2)NK but I don't understand the rest of his post to see how it is relevant. https://www.physicsforums.com/threads/work-required-for-adiabatic-compression.644227/
 
Last edited:
  • #7
grandpa2390 said:
You are right. In class he did an example that was not linear. I'm not sure what the relationshipy/

Read the problem text carefully.
1 liter of ideal diatomic gas (allow for rotation but not vibration) at a pressure of 1 atmosphere and a temperature of 300 K is compressed adiabatically to a pressure of 7 atm. What is the volume of the gas after compression?
What is the relation between P and V during an adiabatic expansion/compression?
You have to integrate pdV to get the work. Express P with V.
 
  • Like
Likes grandpa2390
  • #8
ehild said:
Read the problem text carefully.

What is the relation between P and V during an adiabatic expansion/compression?

they are inverses (as one rises, the other falls). And they are not linearly related...

I know I have to integrate. I di that initially. the problem is I don't know the relationship. I'm going to try V^gamma = 1/P
 
  • #9
grandpa2390 said:
they are inverses (as one rises, the other falls). And they are not linearly related...

I know I have to integrate. I di that initially. the problem is I don't know the relationship. I'm going to try V^gamma = 1/P
It is close, but not correct. What is the relation exactly between P and V?
 
  • Like
Likes grandpa2390
  • #10
ehild said:
It is close, but not correct. What is the relation exactly between P and V?
I don't know. I have looked through my notes 5 times, and that's all I have got. :(

he just said it was equal to a constant.

I have another formula he said was equal to a constant. VT^3/2

is VT^f/2 = PV^gamma?
 
  • #11
What was equal to a constant?

To read: http://www.chem.zju.edu.cn/~lihaoran/kejian/5_60_l06_s05.pdf
 
  • Like
Likes grandpa2390
  • #12
ehild said:
What was equal to a constant?

To read: http://www.chem.zju.edu.cn/~lihaoran/kejian/5_60_l06_s05.pdf

I'm going to email my professor. I don't see anything in that article that helps me. Like the textbook it just goes over my head. Thanks for trying.
 
  • #13
ehild said:
What was equal to a constant?

To read: http://www.chem.zju.edu.cn/~lihaoran/kejian/5_60_l06_s05.pdf
double post
 
  • #14
grandpa2390 said:
I apologize if my questions were in the wrong place. I am in "Advanced" Physics Classes. And so I post the questions from those classes in the Advanced Physics forum. And these are questions that weren't covered in Introductory Physics (1st three semesters). But they keep being moved to this forum. I don't understand the criteria. I guess that forum is for doctoral students and above? The "Really Advanced" stuff lol. regular graduate level classes and below go here? Want to make sure my questions are in the right place.
This is the kind of problem you do in a freshman/sophomore physics class. The advanced physics forum is for upper division and beyond. A good indicator is if the answer you're looking for is a numerical one, it's probably an intro physics question.

grandpa2390 said:
Like the textbook it just goes over my head.
You need to learn how to read the book. It's a skill like any other, and you have to develop and practice it. Don't expect to be able to simply read the book and have the material be clear to you on the first or second pass. If it were that easy, you wouldn't need to take a class.
 
  • #15
vela said:
Don't expect to be able to simply read the book and have the material be clear to you... If it were that easy, you wouldn't need to take a class.

Exactly! That is why I am here.
I have read the chapter several times and the linked articles. If I say it went over my head, then I am not lacking the skill, I am lacking the background.
 
Last edited by a moderator:
  • #16
You still haven't answered my question about the relevant equations. What equations to you have involving work?
 
  • #17
DrClaude said:
You still haven't answered my question about the relevant equations. What equations to you have involving work?
I did answer it. W= (integral of) -pdv
You take the equation of the line and find the area beneath it. Then multiply that by 101 and that gives you work. My problem is I don't know how to find the equation of the line for adiabatic compression/expansion. I emailed my professor though. @ehild did the best he could for me. It sounds like it is a very simple formula or algorithm that I am missing. If I weren't 9 credit hours from graduating, I'd switch majors... :(
once again, thanks for trying in a respectful manner.
 
  • #18
DrClaude said:
You still haven't answered my question about the relevant equations. What equations to you have involving work?
My professor explained to me what it is. W=-Pi*Vi^(7/5)*integral(Vi, Vf, V^(-7/5)*dV). I told you I was an idiot. Vi and Vf are constants. We know them. Vf is the variable. I was expecting, for some reason, a line that could be graphed without initial conditions. I'm an idiot. :)

it's not an inability to read. I proposed that solution earlier, but a dumb flaw in my reasoning prevented me from pursuing it.
 
  • #19
Remember:

pVγ is constant along a reversible adiabat. (γ=Cp/Cv)

which means that P2V2γ=P1V1γ. You know P1, V1 and P2. It is diatomic gas, from that you get γ, and you can calculate V2.
From the equation for an adiabat, ##P=P_1\frac{V_1^\gamma}{V^\gamma}##, so
##W=-\int_{V_1}^{V_2}{P_1\frac{V_1^\gamma}{V^\gamma}dV}##
 
  • Like
Likes grandpa2390

FAQ: How much work is done by the Adiabatic Compression of Diatomic Gasses

1. What is adiabatic compression?

Adiabatic compression is a process in which a gas is compressed without any heat exchange with its surroundings. This means that the temperature of the gas remains constant during the compression.

2. How is work defined in the context of adiabatic compression?

In the context of adiabatic compression, work is defined as the energy required to compress a gas. It is calculated by multiplying the change in volume by the external pressure.

3. What are diatomic gases?

Diatomic gases are molecules composed of two atoms, such as oxygen (O2) and nitrogen (N2). These gases are commonly found in the Earth's atmosphere and have unique properties that make them ideal for studying adiabatic compression.

4. How is the amount of work done by adiabatic compression calculated?

The amount of work done by adiabatic compression can be calculated using the formula W = PΔV, where W is work, P is external pressure, and ΔV is the change in volume of the gas. This formula assumes that the compression is reversible and adiabatic.

5. What are some real-world applications of adiabatic compression of diatomic gases?

Adiabatic compression of diatomic gases has many practical applications, such as in internal combustion engines and refrigeration systems. It is also used in the production of compressed air for various industrial processes and in the compression of gases for storage and transportation. Additionally, the study of adiabatic compression is crucial for understanding the behavior of gases in various thermodynamic processes.

Back
Top