Calculating q,w,H and U

  • #1

Homework Statement


An ideal gas at 25degree celcius and 100kPa is allowed to expand reversibly and isothermally to 5kPa. Calculate..
1) w per mol (J/mol)
2) q per mol (J/mol)
3) change in molar internal energy U (J/mol)
4) change in molar enthalpy H (J/mol)


Homework Equations



ΔU = q + w
= q - pextΔV
ΔH = ΔU +p ΔV
= ΔU + ΔngasRT
PV=nRT
Um(T)=3/2RT

The Attempt at a Solution



I absolutely have no idea how to do this question. Any direction would be greatly appreciated.
 

Answers and Replies

  • #2
1,198
5
Welcome to the forum.
Hint:
A reversible isothermal process is a polytropic process where PV^n = constant. What is n for this kind of process?
 
  • #3
I am thinking this question should include a molar volume? (molar volume = V/n, which changes the ideal gas law as P(molar volume) = RT?)

n is constant for this kind of process? but that sounds weird.. sorry I am lost..

Also, since it is an isothermal process, would molar U be 0?
 
  • #4
I like Serena
Homework Helper
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176
Welcome to PF, chocolatepie! :smile:

An isothermal process is one in which T is constant.
So molar U will not be zero, but constant.
In other words: ΔU=0.

With constant T=To, you get for one mole: PV=RTo.

To calculate w you need to integrate dw=-PdV.
Do you know how to do that?
 
  • #5
I assuming we need to use Wrev= nRT ln (P2/P1) to solve.
However, we were not given n value...?

PS. And thanks everyone for the warm welcome to PF :D
 
  • #6
I like Serena
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I assuming we need to use Wrev= nRT ln (P2/P1) to solve.
Yep. That will work.


However, we were not given n value...?
n is the number of moles.
The question asks to calculate "per mol".
This means you should calculate using n=1 mol.


PS. And thanks everyone for the warm welcome to PF :D
You're welcome! :wink:
 
  • #7
n is the number of moles.
The question asks to calculate "per mol".
This means you should calculate using n=1 mol.
oh, OK!
And for part (d) where I need to calculate for the change of molar enthalpy, should I use ΔH = qp to solve? or ΔH = ΔU + pΔV?

I just don't know when to use each formula..
 
  • #8
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oh, OK!
And for part (d) where I need to calculate for the change of molar enthalpy, should I use ΔH = qp to solve? or ΔH = ΔU + pΔV?

I just don't know when to use each formula.
Actually, neither formula.
Both have conditions attached.

ΔH = qp only holds in an isobaric process.
ΔH = ΔU + pΔV also only holds in an isobaric process (is this really the formula you have?)

What else do you have for ΔH?
 
  • #9
The note at the bottom of the question says that the expansion does not occur at constant pressure. Thus, iti s not OK to use ΔH = ΔU + pΔV.
I am not sure why the note says "constant pressure" when the question itself says "constant temperature" :-O

I have one that includes heat capacity..? H=CpT (subscript p = const P)
But I don't have the heat capacity given in the question, so I didn't use it.
 
  • #10
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The note at the bottom of the question says that the expansion does not occur at constant pressure. Thus, iti s not OK to use ΔH = ΔU + pΔV.
I am not sure why the note says "constant pressure" when the question itself says "constant temperature" :-O

I have one that includes heat capacity..? H=CpT (subscript p = const P)
But I don't have the heat capacity given in the question, so I didn't use it.
In your relevant equations you have:
ΔH = ΔU + ΔngasRT
(Works only for ideal gasses.)

What about that one?


Btw, Cp=Cv+R for ideal gasses.
And ΔH = ΔU + Δ(PV) is true in general.
 
  • #11
Aha, now I got it. Yes there was a formula on my note.
Thank you so much!!
 
  • #12
I like Serena
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See you! :smile:
 

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