Work done by air in piston with temperature change

Homework Statement

Two pounds of air contained in a cylinder expand without friction against a piston. The pressure on the back side of the piston is constant at 200 psia. The air initially occupies a volume of 0.5 ft3. What is the work done by the air in ft-lbf if the expansion continues until the temperature of the air reaches 100 F?

Homework Equations

Q - W = m(uf - ui), which translates to W = m*cv*(Tf - Ti) [Q is 0 because there's no heat transfer]
PV = mRT

The Attempt at a Solution

W = 2 lbm (0.171 Btu/lbm R)(559.67 R - Ti)

The text says the answer is 45300 ft-lbf

Basically I don't really know how to start this all off. I know I need to find the initial temperature which I think means I need to use PV = mRT but I can't find an agreeable R value in my book (also I have no idea if my cv value is correct). Could someone just point me in the right direction?

Last edited:

verty
Homework Helper
I don't know that Q-W formula, but I think you'll need the initial temperature and final volume, which you can get from the formula PV = mRT, if you think about which of those letters change and which are constant.

I don't know that Q-W formula, but I think you'll need the initial temperature and final volume, which you can get from the formula PV = mRT, if you think about which of those letters change and which are constant.

The thing is I can't really tell what is going to remain constant when I read the question. I think the pressure remains constant and I know mass does which leaves me with Ti/Vi = Tf/Vf. I've already got Vi and Tf but I don't really know what to do with that.

In order to answer this, you must assume that the system comes to rest at state 2 and has achieved equilibrium with its surroundings; ie P2=200psi.
Now we can use the ideal gas equation to solve for v2:
V2=(mRT2)/P2.
Some notes here:
1) watch your units! P is in inches and R is in feet.
2) Use the right R value (85.67)
3) T is absolute.
Once you’ve got v2, then w=P(v2-v1) with P=200 psi.
I know pressure is not constant within the system, but it is outside the system. Think of it as the work done on the environment is the work done by the system.

You need to make an assumption