# Calculating q,w,H and U

• chocolatepie
In summary, at 25 degrees Celsius and 100 kPa, a reversible isothermal process expands reversibly and isothermally to 5 kPa. The energy change is ΔU=q+w-pextΔV and the enthalpy change is ΔH=ΔU+pΔV.

## 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)

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.

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?

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?

Welcome to PF, chocolatepie! 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?

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

chocolatepie said:
I assuming we need to use Wrev= nRT ln (P2/P1) to solve.

Yep. That will work.

chocolatepie said:
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.

chocolatepie said:
PS. And thanks everyone for the warm welcome to PF :D

You're welcome! 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..

chocolatepie said:
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?

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.

chocolatepie said:
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.)

See you! 