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Thermodynamics- Isothermal Expansion

  • Thread starter Muntaballs
  • Start date
  • #1

Homework Statement



The temperature of 2 moles of an ideal gas is 366 K. How much work does the gas do in expanding isothermally to 2 times its initial volume?

______ J



Homework Equations



I think they are relevant...
W= nRT ln(Vf/Vi)

PV=nRT

Q constant pressure= 3/2 nRT(Tf-Ti) + nR(Tf-Ti)

Q constant volume = 3/2 nR(Tf-Ti) = 0

constant pressure: Cp= 3/2R=R=5/2R

constant volume: Cv=3/2R


The Attempt at a Solution



Ok, I looked at the problem, and I saw it was isothermal, so according to my textbook I used the formula for it: After plugging in the given values I only got so far...

W=(1.5mol) (8.31 J/mol) (366K) ln (Vf/Vi)

I am pretty much clueless on physics in general, so I my goal isn't to fill in the answer box, but to learn the conceptual ideas behind the process. Like why do you do this or that... Usually responds with just numbers puts me no further in my homework :/

Oh and more on the problem: I remembering my textbook mentioning calculating heat using 2 different formulas based on whether its constant pressure or volume, and then after you get that, you use some different formulas for finding "molar specific heat capacities" to finish solving the problem by pluggin everything in the formula. I dont know if this applies to this current problem or not, but I think its a multi step problem and I dont know what my next move should be because I guess I dont have the basic intuition of thermo.
 
Last edited:

Answers and Replies

  • #2
3,003
2
Hi Muntaballs! :smile: Do you know what the general expression for work done by a system is? I am trying to assess whether you are taking calculus based physics or not, as the approach will differ.

Does the expression [itex]W = \int p\,dV[/itex] look familiar to you (where p= pressure and V = volume)?
 
  • #3
Um, your post seems kind of weird. I dont know what the d stands for. And the symbol before the p...

But if you are referring to Work=Pressure*Volume, thats what I know of it.

Im not taking calculus based physics, I think our class is trying to avoid any calculus at all. I personally havent taken AP calc yet anyways.
 
  • #4
3,003
2
Um, your post seems kind of weird. I dont know what the d stands for. And the symbol before the p...

But if you are referring to Work=Pressure*Volume, thats what I know of it.

Im not taking calculus based physics, I think our class is trying to avoid any calculus at all. I personally havent taken AP calc yet anyways.
Ok then :smile: If you were taking the calculus version, I was going to help you derive the formula for work that you gave (this would help with the "why we use this formula instead of that one" part of your question). Since you are not, you will just have to accept at face value that the formula for work will be different depending on the scenario and that in this particular case (i.e., an ideal gas undergoing an isothermal expansion) the formula is given by your first equation. Though I think you have a typo

This:
W= nRT ln(Vf-Vi)
should be this: W= nRT ln(Vf/Vi) I believe, please check your text or notes.

It is basically plug and chug from here.
 
  • #5
Oh Yes, you're right, i made a mistake in that formula. Ill edit it out.

W= nRT ln(Vf/Vi) is correct, sorry.

but from all the information given from the problem, I plugged in everything, but I dont know how you would get the specific volumes.. There aren't any actual values for volume, except for just stating that it expands by a factor of 2....

so what do i physically type in for the "Vf/Vi" part? 2/1?
 
  • #6
3,003
2
Oh Yes, you're right, i made a mistake in that formula. Ill edit it out.

W= nRT ln(Vf/Vi) is correct, sorry.

but from all the information given from the problem, I plugged in everything, but I dont know how you would get the specific volumes.. There aren't any actual values for volume, except for just stating that it expands by a factor of 2....

so what do i physically type in for the "Vf/Vi" part? 2/1?
Yes! Whenever you think there is not enough information given, just plug in what you know; chances are that the unknowns will cancel. You know that V2= 2*V1, so

[tex]W = nRT\ln\left (\frac{V_2}{V_1}\right ) = nRT\ln\left (\frac{2*V_1}{V_1}\right )=nRT\ln(2)[/tex]

and nRT is known.
 

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