-w = q

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-w = q

Postby KNguyen_1I » Mon Feb 03, 2020 3:36 pm

Why does -w = q at isothermal, reversible conditions? What does this conceptually imply?

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Re: -w = q

Postby Brian_Ho_2B » Mon Feb 03, 2020 3:40 pm

-w = q for isothermal reversible expansions because temperature is constant. This is because when we apply heat to such a system, all of that heat is converted into the work used for the expansion. This relation also allows us to derive the second law of thermodynamics: delta S = q/T, where temperature is constant.
-w = q = nRTln(V2/V1) --> q/T = nRln(V2/V1) = delta S.

Paul Hage 2G
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Re: -w = q

Postby Paul Hage 2G » Mon Feb 03, 2020 3:50 pm

We arrive at the "-w=q" relationship by knowing that for an ideal gas, U(total)=3/2(nRT). Therefore ΔU=3/2(nRΔT). When ΔT=0, ΔU=0. Since ΔU=q+w=0, you can rearrange the expression to arrive at -w=q.

Hannah Romeo 1J
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Re: -w = q

Postby Hannah Romeo 1J » Mon Feb 03, 2020 5:05 pm

For isothermal expansion, no net heat is lost or absorbed from the system. Because the pressure and volume of the system change very slowly over time, the energy released by work is pushed back into the system as heat. As a result, the net energy of the system is 0 resulting in -w = q.

Jainam Shah 4I
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Re: -w = q

Postby Jainam Shah 4I » Mon Feb 03, 2020 7:07 pm

That expression holds true for isothermal reversible expansion. As Dr. Lavelle explain today in lecture the net change in temperature is zero. The work does means energy does exit the system, but at the same time energy released by the system renters the system as heat. Since these changes are minute and nothing drastically changes the net change in temperature is zero.

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Re: -w = q

Postby 005384106 » Tue Feb 04, 2020 1:21 pm

What is the units for temperature that we use in these problems?

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