# Work Done On Ideal Gases

1. Jan 29, 2012

### Rapier

1. The problem statement, all variables and given/known data
I want to start by saying that my instructor is a particle physics guy. He loves to talk about particles seems to want to rush through fluids and thermo so he can get to particles and the 'real modern physics.' He is skipping, skimming and not really covering a lot of this material and I'm having to search it out on my own. Keeping that in mind, if you could dumb any answers down and explain your reasoning/thinking it would greatly help me. Now, on to the problem! :)

There are three problems in this week's homework that deal with the work performed by changing pressure/temperature on ideal gases.

I'm copying the whole problem so that you have all the information. The first bits are easy-peasy.

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A quantity of ideal gas at 40 oC and 150 kPa in a sealed balloon occupies a volume of 2.7 Liters (L).

(a) How many moles of gas are present?
n = mol *
.1555 OK

(b) If the temperature is raised to 60 oC and the pressure is raised to 350 kPa, what volume does the gas now occupy (assuming no leaks)?
V = L *
1.23066 OK

(c) The gas now has its pressure lowered from 350 kPa to 150 kPa isothermally at temperature 60 oC. Calculate the final volume of the gas.
V = L *
2.8715 OK

(d) Calculate the work done by the gas during the isothermal process described in part (c).
W = J

(e) The gas is now compressed from the volume found in part (c) back to its original volume of 2.7 L at a constant pressure of 150 kPa. Calculate the work is done by the gas during this isobaric process?

W = J

2. Relevant equations
PV= nRT
W = ΔKE
KE = 1/2(mv^2)
KE = 3/2(kT)

3. The attempt at a solution
Work is the change in kinetic energy. So I need to be able to calculate the kinetic energy before and after the lowering of the pressure. I realise intellectually that changing the pressure requires some kind of work, but I can't seem to find any equations that take pressure into account in relation to kinetic energy. In this case, the temperature is not changing so the 3/2(kT) would be zero.

I guess my question boils down to, how does a change in pressure equate to a change in kinetic energy?

Sorry for the wordiness, but I'm really trying very hard to understand all this. Thanks, you guys are great!

2. Jan 29, 2012

### Rapier

Oh!

W is in Joules. 1 J = 1 Nm

W = ΔP * ΔV?

The units work out. Which will help me solve these problems, but could anyone (if I'm correct) kind of walk me through how we get to W = ΔP * ΔV?

Ok, I tried that:

W = (350e3 Pa - 150e3 Pa)*(.0028715 m^3 - .00123066 m^3)
W = 328.168 J

And it tells me NO! :(

Last edited: Jan 29, 2012
3. Feb 1, 2012

### Rapier

I haven't gotten any responses, but I was able to get some assistance and figured it out. I thought, for posterity, I would toss this up so that others can use it, if needed.

Part D)
W = ∫ F*dx
Multiply both sides by Area: WA = ∫(F/A)*dx
Solve for W: W = ∫(F/A)*(Adx) <-- grouped nicely
P = F/A and (Adx) = V: W = ∫(Vi→Vf)Pdv
PV=nRT: W = nRT∫(Vi→Vf)dv/v
Notice one of our 'favourite' integrals?: W = nRT ln |v| (Vi→Vf)
W = nRT (ln |vf/vi|)

So that works out pretty good. This is good for isothermal (no temp change) volume changes.

Part E, talks about isobaric (no pressure change)
W = ∫(Vi→Vf)Pdv
W = P∫(Vi→Vf)dv <--Since the pressure is a constant so we can pull it through the integral
W = Pv (Vi→Vf) <-- Easy integral
W = P(Vf - Vi)

It makes quite a bit of sense now, but the really difficult part is the 'screwey math stuff' in Part D where you need to multiply both sides by the Area and then isolate the Work. I might have never seen that if someone hadn't pointed that out.

You are welcome, posterity! :)