# Thermodynamics -- Internal Energy

1. Nov 30, 2015

### BurningUrge

I've read back and forth in my chapter and tried consulting with my forumla sheet, but I cannot seem to find the correct formula. I bet I am looking at this question wrong, but I am fairly sure I need to use the -100 kJ for something to find the final volume and pressure , along with the given pv2=constant.

Basically, given the information below, how would you go about starting to solve this?

1. The problem statement, all variables and given/known data
A mass of gas at an initial pressure of 28 bar, and with an internal energy og 1500 kJ, is contained in a well-insulated cylinder of volume 0.06m3. The gas is allowed to expand behind a piston until it's internal energy is 1400 kJ; the law of expansion is pv2=constant. Calculate:

1) the work done (Already calculated, -100kJ)
2) the final volume
3) the final pressure

2. Relevant equations

3. The attempt at a solution

2. Nov 30, 2015

### Staff: Mentor

Please show your calculation of the work done, specifying whether it is the work done by the gas on the surroundings, or the work done by the surroundings on the gas.

Chet

3. Nov 30, 2015

### BurningUrge

Okay, so the work done is by the internal energy as it expands.

So Q + W = U1-U2. Since there is no heat transfer, only work done, it's basically only W = U1-U2.

W = 1500 kJ - 1400 kJ => W = 100 kJ. But since it's an expansion done BY the internal energy as it is allowed to expand, it's -100kJ.

4. Nov 30, 2015

### Staff: Mentor

The equation should be Q + W = U2-U1. That's why W is negative (using the sign convention that W represents work done on the system).

Now you need to develop another equation for expressing the work in terms of the pressure and volumes. What is the equation you learned for determining the work as an integral involving pressure and volume?

Chet

5. Nov 30, 2015

### BurningUrge

That would quite probably be this nifty little thing;

W = -∫pdV , from position 1 to 2. (Couldn't find the icon for definite integral)

Since pv2 = Constant, I rewrite the equation to -> -∫ c/V2 dV, => -c ∫ 1/V2 dV => -p1V12 ∫ 1/V dV.

When I can't write the correct definite integral it looks abit messy, but the finished integration should look like this;

-p1V12 (-1/V2-(-1/V1)

Finishing this off with :

-p1V12 (-1/V2-(-1/V1) = -100kJ

Am I currently on the right path?

6. Nov 30, 2015

### Staff: Mentor

Yes. I haven't checked your "arithmetic," but you're on the right path.

Chet

7. Nov 30, 2015

### BurningUrge

Okay, so I did that. And I came up with the right answers.

V2=0.148m3

And using that I could find the pressure aswell with p1V12=p2V22

Ending at p2=4.6 bar.

Greatly appreciate your time and help!