How Do State Functions Apply to a Multi-Step Thermodynamic Process?

In summary, the conversation is discussing the calculation of change in U, S_sys, and H values for a monatomic, ideal gas undergoing four reversible steps: freely expanding into a vacuum, being heated at constant volume, expanding adiabatically, and being warmed up isobarically. The question is whether to add the values for each step or just take the difference between step 4 and step 1, and if the temperature is the same, the change in U should be zero. However, there is some confusion about the line of reasoning and the concept of a state function, as well as a contradiction in describing one of the steps as reversible when it involves free expansion into a vacuum.
  • #1
jaejoon89
195
0
Thermodynamics question -- please help

A monatomic, ideal gas undergoes 4 steps, all reversible

1) At initially 293 K and 1 atm, it is let freely expand into a vacuum, tripling the volume.
2) It is then heated to 393 K, keeping V constant.
3) It is then let expand adiabatically, doubling the volume.
4) It is then warmed up to 293 K in an isobaric manner.

-----

To calculate the total change in U, S_sys, and H values do I add the values for each step or just take step 4 minus step 1? I realize they are state functions, but this isn't a chemical reaction per se, just a series of physical processes. Plus, the change in U for #1 is zero, and for #4 it is nonzero. I assume that even though U is a state function, I don't take change in U for #4 minus change in U for #1 since that would not be zero. Since the temperature is the same, it is isothermic and ideal and therefore the change in U must be zero. Thus, adding the state function values for each step results in what I imagine is the correct answer. Is this correct? I'm not sure about the line of reasoning so I'd appreciate some help...
 
Physics news on Phys.org
  • #2


A state function is a state function; you can look at the end value only (compare #4 to #1) or move from #1 to #4 step by step. What would be the problem with a nonzero change in U?

By the way, it's a contradiction to describe #1 as reversible. Reversible means, among other things, that the pressure on the other side of an expanding boundary is equal. Free expansion into a vacuum is irreversible.
 
Back
Top