Entropy of Isothermal Processes

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Jar-Yee 3L
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Entropy of Isothermal Processes

Postby Jar-Yee 3L » Fri Jan 15, 2016 2:31 pm

For an ideal gas, ΔU=0 when ΔT=0, evidently because an ideal gas only has kinetic energy, so without heat there would be no change in internal energy.

ΔU = 0 = q+w,
q = -w.
With ΔT = 0, then q = 0. As a result, w = -q = 0 as well.

Eventually, a proof with this concept would give us ΔS=nRln(Vf/Vi), the equation for the change in entropy of an isothermal process.

Since Vf and Vi both exist, how come work is equal to 0? w= -PΔV, and ΔV exists.

Ronald Yang 2F
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Re: Entropy of Isothermal Processes

Postby Ronald Yang 2F » Fri Jan 15, 2016 3:58 pm

When ΔT = 0, that doesn't mean q is also zero. In the example given today in lecture, for a reversible, isothermal expansion, heat must constantly be put into the system to keep the temperature constant, hence to keep ΔT = 0. The heat put in by the heat reservoir is used as work in expansion, so ΔV is not equal to zero, hence work is not zero.

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Re: Entropy of Isothermal Processes

Postby Chem_Mod » Sun Jan 17, 2016 12:20 am

The above discussion is correct! It is noteworthy that when =0, it DOES NOT mean q=0, only =0

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