Ideal gas through Isobaric process

• BurningUrge
In summary, the problem involves an isobaric process with 12 moles of an ideal gas, where the internal energy decreases by 20 KJ and 4 KJ of heat is added. Using the first law of thermodynamics and the equation for isobaric work, the end volume is calculated to be approximately 0.47m3.

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 truly embarrasing. I skipped my exam in Physics in my first year due to illness, so that's why I am doing this now.

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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?

JBA said:
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.

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BurningUrge said:
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.

1. What is an ideal gas through an isobaric process?

An ideal gas is a hypothetical gas that follows the gas laws perfectly. An isobaric process is a thermodynamic process in which the pressure of the gas remains constant. Therefore, an ideal gas through an isobaric process is a gas that follows the gas laws while maintaining a constant pressure.

2. How does an ideal gas behave during an isobaric process?

An ideal gas behaves according to the gas laws during an isobaric process. This means that the volume of the gas will increase or decrease proportionally to the change in temperature, while the pressure remains constant.

3. What is the equation for an isobaric process in an ideal gas?

The equation for an isobaric process in an ideal gas is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

4. What are some real-life examples of an ideal gas through an isobaric process?

Some examples of an ideal gas through an isobaric process include a gas expanding in a balloon as it is heated, the heating of a gas in a piston-cylinder system, and the heating of air in an air conditioner.

5. What are the limitations of an ideal gas through an isobaric process?

One limitation of an ideal gas through an isobaric process is that it assumes there are no intermolecular forces between gas particles. This is not true in real gases, so the behavior may not be perfectly accurate. Additionally, the process assumes a constant pressure, which may not always be the case in real-life situations.