Change in Internal Energy of an Isobaric Process

AI Thread Summary
In an isobaric process involving 1 mole of a monatomic ideal gas, the pressure remains constant while the volume increases, leading to confusion about work and internal energy changes. Calculating work using pressure and volume change yields 10,000 J, while applying the ideal gas law gives a different internal energy change of 415.5 J, highlighting a fundamental misunderstanding. The discussion reveals that the change in internal energy is influenced by temperature change, which must be correctly determined to reconcile these calculations. The confusion stems from mixing concepts of internal energy changes for different materials and ideal gases. Ultimately, understanding the relationships between work, heat, and internal energy in thermodynamics is crucial for accurate calculations.
jeff.berhow
Messages
17
Reaction score
0
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.
 
Physics news on Phys.org
How do you get that second result? What is your ΔT?
 
Ah, sorry, let's say the delta T is 50K.
 
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.
 
jeff.berhow said:
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.
 
Last edited:
  • Like
Likes 1 person
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?
 
Thread 'Gauss' law seems to imply instantaneous electric field'

Thread 'Gauss' law seems to imply instantaneous electric field'

Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
View full post »

Thread 'Do electron density waves accompany EM waves in coaxial cables?'

Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
View full post »

Similar threads

I Question regarding dh, du, cp, cv for ideal gases
Replies
3
Views
2K
B Adiabatic or Isobaric process?
2
Replies
61
Views
6K
Internal energy change in isobaric system
Replies
3
Views
3K
I Internal energy of an ideal gas -- confusion
Replies
4
Views
2K
I Irreversible adiabatic processes & entropy change (clarification needed)
Replies
22
Views
5K
I Adiabatic expansion work far exceeds isobaric of same volume, why?
2
Replies
81
Views
5K
Isothermal, isobaric and adiabatic
Replies
6
Views
2K
I Irreversible Isochoric Process in a Cycle
Replies
8
Views
3K
Is Heat Exchange Equal to Enthalpy Change in an Isobaric Process?
Replies
3
Views
2K
Change in the energy content of an isobaric process
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top