# Homework Help: How would I calculate the work ?

1. Sep 28, 2009

### hungryhippo

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

Given the following case, how would I calculate work?
Volume initial = 10 cubic metres
Mass = 8 kg of O2
Temperature initial = 300 K
Volume final = 5 cubic metres

a) Work at constant pressure? What is the temperature?

b) Work at constant temperature? What is the pressure?

2. Relevant equations

dw= -Pdv
du = dw +dq
dq = CdT

3. The attempt at a solution

a)
integrate -Pdv
work = P(Vf-Vi) <-- the question doesn't state that it's an ideal gas, but can I still use PV=nrt to find the initial pressure?

As for calculating temperature, I was thinking of dq(constant pressure) = C (constant pressure)(Tf-Ti). Not sure if this is correct?

b) I though work was indpendent of T, thus dw=-Pdv --> P(Vf-Vi) assuming infinitesimal change in P

2. Sep 29, 2009

### Mapes

Hi hungryhippo, welcome to PF. I agree with your work calculation for part (a). Yes, it's a good idea to use the ideal gas law to find the pressure. You could use dq to find the temperature, but it doesn't look like you're given this, so why not use the ideal gas law for temperature too?

For part (b): work is not independent of temperature. If the temperature is kept constant, then the pressure will likely change, which will change the work term from its value in (a).

3. Sep 29, 2009

### hungryhippo

Hey thanks for the warm welcome mapes :)

For part a, even though I wasn't given dq, I thought I could look up the specific heat capacity at constant pressure, to find the change in temperature. As for the ideal gas, I'm still not very sure on that...I guess i'll have to ask for clarification on that question.

As for part b, I'm still unsure of how to approach exactly. Since temperature is held constant, the internal energy would just be equal to work. But then I wasn't sure on how to show the new pressure. Am i still able to use dw=-P(deltaV) since this is for infinitesimal changes in pressure.

4. Sep 29, 2009

### Mapes

Again, you're missing dq. You don't know how much energy has to be added to the system through heating to keep the pressure constant. If you assume ideality, however, the pressure, volume, and temperature are easily related.

No; the internal energy of an ideal gas depends on temperature alone, and temperature is constant for this case.

It's important to realize that in both cases, an undetermined amount of heating has to occur to keep the system at constant pressure or temperature. The internal energy changes of the gas are not due to work alone.

5. Sep 29, 2009

### hungryhippo

Hey mapes, thanks for the clear up. lols I got how to do the question now :) I had to check with the teacher that it was an ideal gas and a reversible process