Entropy Calculation

[Scienc]

Hi everyone,

I am studying thermodynamics, and I'm confused about item b. I understand why the professor used W = 0 since Pexternal = 0; however, I'm unsure why Delta T also equals 0. Should I consider that process B is also isothermal, or is the phrase "an isothermal reversible expansion" only applicable to item c?

From what I recall, an expansion for Pexternal = 0 does not necessarily result in a Delta T = 0, or it does.

 

Edited December 5, 2024 by Scienc

[swansont]

6 hours ago, Scienc said:

From what I recall, an expansion for Pexternal = 0 does not necessarily result in a Delta T = 0, or it does.

No temperature change for an ideal gas.

[Scienc]

25 minutes ago, swansont said:

No temperature change for an ideal gas.

For an ideal gas that expands against P = 0, is Delta T = 0?

[swansont]

6 hours ago, Scienc said:

For an ideal gas that expands against P = 0, is Delta T = 0?

Yes. The temperature is just a measure of the kinetic energy of the particles comprising the gas, and that doesn’t change since there’s no work being done. There’s nowhere else for that energy to go.

[sethoflagos]

12 hours ago, Scienc said:

I'm confused about item b. I understand why the professor used W = 0 since Pexternal = 0; however, I'm unsure why Delta T also equals 0.

Consider case b) being physically realised by flow through a perfectly insulated porous plug. The pore resistance of the plug is sufficiently high for kinetic energy terms to be very small.

For an ideal gas, the operating equation is d(PV) = VdP + PdV = nRdT

The VdP term represents a differential loss of internal energy converted to kinetic acceleration of the flow through the plug.

The PdV term represents a differential increase of internal energy due to heating from the frictional resistance opposing that acceleration.

For low flow rates these two terms become equal in magnitude, opposite in sign. Hence dT = 0

I've attached a copy of my backpocket cribsheet for this sort of system. 

13 hours ago, Scienc said:

Should I consider that process B is also isothermal, or is the phrase "an isothermal reversible expansion" only applicable to item c?

'Isothermal' generally implies heat exchange between the system and surroundings. This case is an adiabatic one that just happens to maintain a constant temperature. The process is far from reversible due to the large increase in total entropy.  

Porous Plug.pdf

[studiot]

You might like to compare with the Joule Thomson effect where cooling does take place

 

this is a good pdf

 

https://byjusexamprep.com/liveData/f/2021/12/joule_thomson_effect_76.pdf