Calculating q,w,H and U

[chocolatepie]

1. The problem statement, all variables and given/known data
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)


2. Relevant equations

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

3. The attempt at a solution

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

[LawrenceC]

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?

[chocolatepie]

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?

[I like Serena]

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?

[chocolatepie]

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

[I like Serena]

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:

[chocolatepie]

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..

[I like Serena]

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?

[chocolatepie]

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.

[I like Serena]

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.

[chocolatepie]

Aha, now I got it. Yes there was a formula on my note.
Thank you so much!!

[I like Serena]

See you! :smile: