Recent content by BurningUrge

This isn't really a hard question, but it's been a while since I did this and I would just like to make absolute sure. This is an exceptionally simplistic draft of a wireblock (open at one side to allow for wire to be threaded over while each side of the wire is attached to the ballast). It's...

It might be due to my exhausted mind being slow, but from what I understand now is this; They tell me that internal energy is -20 KJ, which means that ΔU = -20KJ. And then they tell me that Q is +4KJ. This means that ΔU = Q + W is actually -20KJ = 4 KJ - 24 KJ. Slamming this into the equation...

Homework Statement This is probably a real easy task for most, but I simply CANNOT manage to calculate it, even given the correct answer. I will translate it as best I can and hope I don't phrase it in a way that causes misunderstandings: 12 Moles of an ideal gas go through an Isobaric...

Okay, so I did that. And I came up with the right answers. V2=0.148m3 And using that I could find the pressure as well with p1V12=p2V22 Ending at p2=4.6 bar. Greatly appreciate your time and help!

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...

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.

I've read back and forth in my chapter and tried consulting with my formula 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...

Then I have my answer and I am greatly appreciative of your help. Thank you very much!

Yes there is. The 1 -> 2 and 3 ->4 both link the process from the initial volume to the unknown. Does it have anything to do with the pv2 = constant that is the exponent for the revesible process happening in 3 -> 4? Or put in this way, I can figure out the volume using p1V12 = p3V32, where V3 = V2

Okay, so I rechecked the values, and the Spesific volume is meant to be 0.5, NOT 0.05. One slight tap too much on the 0 there. Also, I did sketch the p-v diagram, yes. I'm pretty sure I got the understanding of how the process looks. And when you say that I need to use Spesific volume, it...

This is quite possibly very basic, but I have a horrible habit of forgetting basics as I move up the ladder in difficulty during the semester. This is a question regarding Thermodynamics, taken from Chapter 1; Introduction and the First Law of Thermodynamics of the book Applied Thermodynamics...