Change in entropy of an irreversible adiabatic process

In summary: They tried using the equation CpdT/T + CvdT/T = ds, but are having trouble with calculating their final pressure and temperature. They asked for help in determining T2 and in finding the change in enthalpy without knowing P2. They also asked for an alternative reversible process to determine the entropy. The calculated internal energy change and enthalpy change were agreed upon, but they still need to find the final temperature and pressure. In summary, Chetan is seeking assistance in calculating the entropy for a reversible process and determining the final temperature and pressure using alternative methods.
  • #1
gjb24mrspotts
1
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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? I am having trouble calculating my P2 (I know my final pressure is not the constant external pressure) and T2.
 
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  • #2
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
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
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
 

FAQ: Change in entropy of an irreversible adiabatic process

What is entropy and how does it relate to irreversible adiabatic processes?

Entropy is a measure of the disorder or randomness in a system. In an irreversible adiabatic process, there is no exchange of heat with the surroundings, meaning the system is isolated. The change in entropy in this process is a measure of how much the disorder or randomness increases or decreases.

How is the change in entropy calculated for an irreversible adiabatic process?

The change in entropy for an irreversible adiabatic process can be calculated using the formula ΔS = Q/T, where Q is the heat absorbed or released and T is the temperature of the system. This formula is based on the second law of thermodynamics, which states that the change in entropy is equal to the heat transferred divided by the temperature.

Can the change in entropy be negative in an irreversible adiabatic process?

Yes, the change in entropy can be negative in an irreversible adiabatic process. This would occur if the system becomes more ordered or less random. However, this is not a common scenario as most irreversible processes tend to increase the disorder or randomness in a system, resulting in a positive change in entropy.

How does the change in entropy in an irreversible adiabatic process affect the efficiency of a system?

The change in entropy in an irreversible adiabatic process has a direct impact on the efficiency of a system. The second law of thermodynamics states that the efficiency of any real-world process is always less than 100%, and this is due to the increase in entropy. In an irreversible adiabatic process, the change in entropy results in a decrease in the efficiency of the system.

What factors can influence the change in entropy in an irreversible adiabatic process?

The change in entropy in an irreversible adiabatic process can be influenced by several factors, including the temperature difference between the system and its surroundings, the amount of heat transferred, and the nature of the process itself. Other factors such as the pressure and volume of the system can also play a role in determining the change in entropy.

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