Calculating Heat Transfer in a Reversible Isothermal Process with a Slow Leak

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SUMMARY

The discussion focuses on calculating heat transfer in a reversible isothermal process involving a rigid tank with a slow leak. The initial conditions include a tank volume of 0.5 m3, with air seeping in at 1 bar and 21 degrees Celsius. Participants emphasize treating the air as an ideal gas and applying the first law of thermodynamics, specifically the equation Q = T[S(2) - S(1)], where S represents entropy. The conversation highlights the importance of understanding the ideal gas law, PV = nRT, and the assumptions regarding negligible kinetic and potential energy.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with ideal gas behavior and the ideal gas law (PV = nRT)
  • Knowledge of entropy and its role in thermodynamic processes
  • Basic concepts of control volumes in thermodynamics
NEXT STEPS
  • Study the application of the first law of thermodynamics in isothermal processes
  • Learn about the properties of ideal gases and how to use the ideal gas law effectively
  • Explore entropy calculations and their significance in reversible processes
  • Research control volume analysis in thermodynamics and its applications
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Students and professionals in thermodynamics, particularly those studying heat transfer, ideal gas behavior, and the application of the first law in practical scenarios.

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Homework Statement


Rigid tank volume .5m^3 initially evacuated. Tiny hole develops, air seeps in @ 1 bar and 21 degrees C. Pressure in tank reaches 1 bar, slow leak so temp remains 21 deg C inside tank. What's amount of heat transfer?


Homework Equations



Q=T[S(2)-S(1)]
Pv=nRT

The Attempt at a Solution



Ok, I understand this is a reversible isothermal process but it seems that I am not given enough information to solve the problem. Please help to send me in the right direction! Thanks!
 
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Have you taken a look at the the first law? You know of the assumptions for the problem, I would also add to treat the air as an ideal gas.
 
I'm very new to thermo but I haven't seen this formula Q=T[S(2)-S(1)], unless the S's means Volumes, I write it as capital V's

but anyways.. not the point... for rigid tanks, isn't the volume constant so you can solve for Q then use Q to find the missing variables in Pv=nRT

the temperature will give you the pressure, and you would have to treat this as an ideal gas, you're given little v, and I think you can get R from Q, or by the chartes... so then you just have P?

I havn't covered rigid tanks yet, so I'm probably very wrong...
 
The S probably stands for entropy. He probably got the first equation from the 2nd law of themo, S2 - S1 = Q/Tb + sigma, where sigma will equal zero since it's a reversible process.

PV =nRT works, but i think PV = mRT is a better variation of the ideal gas for this problem.

I'm guessing the others assumption that KE and PE are negligble. Take a control volume and you can see what you need to do from there.
 
Last edited:
thank you...

Thanks for the help - I appreciate the responses...have returned to school after a LONG break and some of the basics are pretty hazy

I thought that a control volume would only be applied to fluids?

By the S's I was meaning entropy, didn't know how to show using subscripts.

Side question - I know that when you are given two properties of a fluid (such as temperature and pressure) that you should know what state they are in (superheated etc)...but how exactly is that determined?
And, iff you are given both temp and press which table should you be using?
 
if I do need to do a control volume - how exactly do i do that?
 

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