Thermochemistry: Irreversible Expansion of Ideal Gas

AI Thread Summary
When one mole of an ideal gas expands irreversibly against a constant external pressure of zero, the work done (w) is zero, as there is no resistance. The heat transfer (q) is also likely to be negligible due to the rapid expansion. The change in internal energy (ΔU) remains uncertain without additional information on heat exchange, while the entropy (ΔS) increases due to the irreversible nature of the expansion. This scenario illustrates the inefficiency of energy transfer in thermodynamic processes. Understanding these concepts is crucial for grasping the fundamentals of thermochemistry.
kido
Messages
6
Reaction score
0

Homework Statement



State whether the thermodynamic quantities q (heat), w (work), ΔU (internal energy) and ΔS (entropy) are greater than, equal to, or less than zero when one mole of an ideal gas expands irreversibly and against a constant external pressure of zero.

Homework Equations



Well, I know that ΔU = w + q, and w = PΔV.

The Attempt at a Solution



As P = 0, w = 0. If the system was insulated, q would be equal to zero, but it doesn't say anything about that. I have no idea about ΔU or ΔS. I'm really struggling to understand thermochemistry!
 
Physics news on Phys.org
Hi kido, welcome to PF. If a gas is expanding against an external pressure of zero, there's no resistance and it's therefore probably expanding pretty fast. Too fast for any significant heat transfer. Does this help you move forward?
 
Imagine a long cylinder with a removable barrier half way and imagine an ideal gas in the left half of that cylinder. I'll try to draw it:
.__________________________________|__________________________
(____________gas here_______________|__________________________)

Now the molecules of that gas are randomly moving at a given speed in the left half of the cylinder. The temperature of that gas is proportional to (1/2)mv2 where v is the velocity of the gas particles (assume they are all the same.)

Assume there is no gravitational attraction between the gas particles and that they are bouncing back and forth with perfectly elastic collisions between themselves or the walls of the cylinder.

If you remove the barrier and let the ideal gas molecules flow to the right half of the cylinder there will be no loss of speed in the particles and the temperature of the gas will remain the same as it is only dependent on the squared velocity of each molecule.

The pressure in the cylinder will drop but the temperature will not. No work has been done by this gas with this maneuver. In order to return the gas to its original state (to the left half of the cylinder) you will have to push the gas with a piston to the left which will take work. This will create heat and will require work - hence, you have created a situation in which no work was done yet work has to be done to restore the original state of affairs. This is irreversible loss of energy and an increase in entropy and you "got nothing for it."

Poor Sadi Carnot who in 1828 described the perfect inefficiency of a heat engine. Must have been frustrating:

Efficieny = (T2 - T1)/T2
 
Last edited:
H-E-L-P

How do I draw diagrams on PF?
 
You don't. But you can always draw something on your computer and upload picture to attach it to your post. See "additional options" frame below edit field in advanced editor, look for a "Manage attachments" button.

Some simple diagrams can be drawn with LaTeX, but I have no idea how - I have just seen some simple diagrams done by others, probably in this thread. Or check here. Or search forums.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top