Change in entropy at constant temperature

AI Thread Summary
The discussion focuses on calculating the entropy change of a perfect gas compressed at constant temperature. The user seeks assistance with determining the heat transfer (Q) necessary for the calculation. Key equations mentioned include the first law of thermodynamics and the work done during isothermal expansion. The contributors clarify that since temperature is constant, there is no change in internal energy, leading to the conclusion that Q equals negative work (Q = -W). Ultimately, the user confirms their understanding and correctness of the solution with the help of the forum members.
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
\Delta E=Q+W
Work for isothermal expansion
W=-nRTln\frac{V_{f}}{V_{i}}

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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