Ideal gas through Isobaric process

  • #1

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 process. The inner energy declines with 20 KJ, at the same time as 4 KJ heat is added. The Pressure at the start was 140 kPa, and the start volume was 0.3m3. What is the end volume?

Homework Equations


The relevant equations we were given was to use the first law of Thermodynamics and then use that in the equation for Isobaric work. They are as follows:

ΔU = Q + W
W = -p1⋅ΔV = p1V1(1-V2/V1) = nRT1(1-V2/V1)

The Attempt at a Solution


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I have made several attempts at finding W to use in the equation for Isobaric work. I tried finding the temperature with pV=nRT to find the immediate temperature, to then use it to find the internal energy, but I struggle finding a way to use this.

I'm soon done with my second year in Mechanical Engineering and have finished with flying colors on my Process Technique class (which builds on exactly this, but a step up), yet I cannot manage to do this and it is truely embarrasing. I skipped my exam in Physics in my first year due to illness, so that's why I am doing this now.
 
Last edited:

Answers and Replies

  • #2
JBA
Science Advisor
Gold Member
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Since the process is isobaric (constant pressure) what does that tell you about the relationship between the internal energy change and the volume change?
 
  • #3
Since the process is isobaric (constant pressure) what does that tell you about the relationship between the internal energy change and the volume change?
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 W = -p1ΔV makes it -W/-p1 = ΔV. Ultimately this makes it that ΔV ≈ 0.17m3. Thus the end resulting volume is 0.47m3. But with my luck it's just coincidential.
 
Last edited:
  • #4
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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 W = -p1ΔV makes it -W/-p1 = ΔV. Ultimately this makes it that ΔV ≈ 0.17m3. Thus the end resulting volume is 0.47m3. But with my luck it's just coincidential.
This looks correct.
 

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