Change in entropy of an irreversible adiabatic process

  • #1
Homework Statement:
We have 5.32 L of an ideal diatomic gas at 16.3 bar and 371 K. The gas is in an insulated cylinder contained with an insulated piston. We unlock the piston and the gas expands against a constant external pressure of 1.43 bar until the piston is locked again at triple the original volume. Calculate the values of the parameters below for this process. Express all energies in J and entropy in J/K.
Relevant Equations:
W= PexdV
delta U = q+ w
ds= dq/T
Screen Shot 2021-03-02 at 6.56.49 PM.png

I have been able to get everything except entropy. I know it's not zero. I know I have to find a reversible path to calculate it, but keep coming up with strange values so I don't think i'm doing it correctly.
can I do CpdT/T + CvdT/T = ds? Im having trouble calculating my P2 (I know my final pressure is not the constant external pressure) and T2.
 

Answers and Replies

  • #2
21,279
4,728
Please show us how you calculated T2. Also, how did you get the change in enthalpy if you do not know P2?

Please describe for us the alternative reversible process you devised to determine the change in entropy (it may involve two process steps).
 
  • #3
21,279
4,728
I agree with your calculated internal energy change. For an enthalpy change, I get 1.4 times as much, or -2123 J. The next step is to determine the final temperature. You know the internal energy change, the number of moles of gas, and the heat capacity of the gas. From that, you can determine ΔT. What value do you get, and what do you get for the final temperature.

You can get the final pressure P2 knowing the final temperature and employing the ideal gas law, or from the enthalpy change, since you know ΔU, P1, V1, and V2. The values you get from both these methods should agree.
 
  • #4
21,279
4,728
It doesn't look like the OP is going to return to complete this. If anyone else would like to continue for practice, please feel free to do so. I will continue to look on.

Chet
 

Related Threads on Change in entropy of an irreversible adiabatic process

Replies
1
Views
3K
Replies
1
Views
498
Replies
1
Views
2K
Replies
9
Views
65K
  • Last Post
Replies
1
Views
22K
Replies
1
Views
1K
  • Last Post
Replies
12
Views
21K
Replies
0
Views
1K
Replies
0
Views
4K
Replies
5
Views
2K
Top