Change in entropy at constant temperature

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Homework Help Overview

The problem involves calculating the change in entropy for a perfect gas undergoing compression at a constant temperature of 273K, with specific initial and final pressures provided.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the First Law of Thermodynamics and the relationship between heat transfer and work done during isothermal processes. Questions arise regarding the calculation of the number of moles and the implications of constant temperature on internal energy.

Discussion Status

The discussion is active, with participants providing guidance on relevant equations and concepts. Some participants have clarified the relationship between heat, work, and internal energy, while others have confirmed the correctness of the original poster's understanding.

Contextual Notes

There is an emphasis on the constraints of the problem, such as the constant temperature condition and the need to apply the ideal gas law for calculations. The original poster expresses uncertainty about calculating heat transfer.

ricof
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Homework Statement



Calculate the entropy change when 0.011m3 of a perfect gas at a constant temperature of 273K is compressed from an initial pressure 1x10^5 N/m/m to a pressure of 1x10^6 N/m/m.

I know the equation but cannot work out Q! Please help!

Homework Equations



dS=Q/T

The Attempt at a Solution

 
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Hi ricof, welcome to PF. Can you combine the First Law with what you know about the energy of an ideal gas at constant temperature?
 
Some more relevant equations.
The first law
[tex]\Delta E=Q+W[/tex]
Work for isothermal expansion
[tex]W=-nRTln\frac{V_{f}}{V_{i}}[/tex]

If you combine this with what you know about change in internal energy at constant temperature, like Mapes said, and that integral of yours, you can figure out the answer.
 
Because T is constant, there is no change in internal energy right?

And also, in the above equation for isothermal expansion, how is n calculated?
 
Last edited:
ricof said:
Because T is constant, there is no change in internal energy right?

And also, in the above equation for isothermal expansion, how is n calculated?

Agreed on the first question. For the second: apply the ideal gas law.
 
Ok, so as dE = 0,

Q = -W

and as PV = nRT,

-W=PV ln (v1/v2) which is Q

Thanks guys, my answer is correct. Thanks for helping!
 

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