Gas Expansion in insulated cylinder (piston & Diaphragm)

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The discussion centers on analyzing two expansion cases of an insulated cylinder, one with a diaphragm (case A) and the other with a piston (case B). Case A is identified as a free expansion process where the gas expands without heat transfer or work, resulting in no change in temperature and a decrease in pressure. In contrast, case B involves a piston that does work on the surroundings, implying that the expansion is not free and leads to a decrease in temperature and pressure as well. The participants explore the implications of the piston being massless and connected to the outside, ultimately concluding that case B cannot be treated as free expansion and requires additional assumptions to determine final pressure and temperature. The final analysis suggests that the pressure in case B will be lower than in case A due to the work done on the surroundings and the resulting temperature change.
  • #31
voko said:
Why is there no change in temperature in this case?



Consequently, what happens with temperature?


I am having problems with looking at this conceptually. I will try to take another crack at justifying case A. Since there's no heat transfer and work in case A, that means there's no change in internal temperature. For a perfect gas, internal temperature is only a function of temperature. Since there's no change in internal energy, there's no change in temperatre. Temperature and internal energy are directly correlated
 
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  • #32
pyroknife said:
Since there's no change in internal energy, there's no change in temperatre. Temperature and internal energy are directly correlated

And what does that imply for case B?
 
  • #33
voko said:
And what does that imply for case B?
Oh I see. Internal energy and temp would also be directly correlated for case B because the gas is still an ideal gas. There's no heat but there's a work output thus change in internal energy is negative, which implies change in temperature is negative (a decrease in temp from initial to final)!
 
  • #34
Correct. And what does lower final temperature imply for final pressure?
 
  • #35
voko said:
Correct. And what does lower final temperature imply for final pressure?
Sweet. Hmmm, wouldn't we have to know the change in temperature to determine whether the final pressure increase or decrease?

I forgot to mention the final volume is 6 times the original volume.

Let me work this out mathematically.

The left portion (1) has a volume 5x the volume of the right. I have use "f" index to denote final state rather than (2).
For case A:
P1V1=P_f*V_f, V_f=1/5V1+V1=6/5*V1
=>P_f=5/6*P1 (Decrease in pressure for case A)For case B:
P1V1/T1=P_f*V_f/T_f => P_f=5/6*P1*T_f/T1. The ratio T_f/T1 is less <1, thus P_f for case B is less than P_f for case A.
Is there any way to obtain the T_f/T1 ratio though?
 
Last edited:
  • #36
It is generally bad when one symbol means more than one thing, as is the case with your symbols labeled with "f". Your conclusion is correct, however.

I do not think it is possible to obtain final temperature/pressure in case B without making additional assumptions on the process.
 

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