Change in Internal Energy of an Isobaric Process


by jeff.berhow
Tags: energy, internal, isobaric, process
jeff.berhow
jeff.berhow is offline
#1
Oct1-13, 06:00 PM
P: 16
In an isobaric process of 1 mole of a monatomic ideal gas, the pressure stays the same while the volume and temperature change. Let's take an isobaric expansion where the volume increases by 2m3 and the pressure stays at 5kPa.

If the work done by the gas is the pressure times the change in volume we get 5000J, but if we apply the ideal gas law to that equation, we get nRΔT, which then gives an amount of 415.5J. In the first instance, the change in internal energy decreases which is weird to me. In the second, it acts accordingly and increases.

What is the fundamental issue I am having dealing with this concept, and why do we effectively learn two ways of calculating gas processes?

Thanks in advance. For some reason, Thermodynamics is kicking my butt because I can't conceptualize it correctly.
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nasu
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#2
Oct1-13, 06:10 PM
P: 1,909
How do you get that second result? What is your ΔT?
jeff.berhow
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#3
Oct1-13, 06:24 PM
P: 16
Ah, sorry, let's say the delta T is 50K.

jeff.berhow
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#4
Oct1-13, 06:26 PM
P: 16

Change in Internal Energy of an Isobaric Process


I think I'm starting to understand what I'm getting confused on. Our first chapter discussed change in internal energy in the context of any type of material, and the second chapter discusses ideal gases and their change in internal energy. I think I am getting the two mixed up and mixing and matching them. This is very disheartening.
nasu
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#5
Oct1-13, 06:33 PM
P: 1,909
Quote Quote by jeff.berhow View Post
Ah, sorry, let's say the delta T is 50K.
You cannot just "say" it. It's not arbitrary but determined by the given parameters and process.
You could "say" that the initial temperature is 50 K (for instance). This will give you an initial volume
V1=nRT1/P = 0.0831 m^3
Then V2 should be 2.0831 m^3 and the final temperature
T2=pV2/nR=1253 K.
So ΔT = 1203 K.
With this you will get the same work if you use nRΔT as by using p ΔV directly.
Work which, by the way, is 10,000J and not 5000J.

The nice thing is that no matter what T1 is, the things will "arrange" such that the work will be same, for the given parameters.
Chestermiller
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#6
Oct2-13, 09:55 AM
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Thanks
PF Gold
P: 4,502
Let me guess. You are trying to calculate the temperature change and the heat added Q. If PΔV=10000 J, from the ideal gas law what is nRΔT equal to? If you have n = 1 mole, what is the temperature change? What is the equation for the change in internal energy ΔU of a monotonic ideal gas in terms of n, R, and ΔT? From this equation, what is change in internal equal to? The ideal gas law tells you that ΔU=Q-PΔV? Using this equation and the results so far, what is Q equal to?


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