## -w = q

$\Delta U=q+w$

KNguyen_1I
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

### -w = q

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

Brian_Ho_2B
Posts: 221
Joined: Fri Aug 09, 2019 12:16 am

### Re: -w = q

-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
Posts: 105
Joined: Thu Jul 25, 2019 12:17 am

### Re: -w = q

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
Posts: 58
Joined: Thu Jul 11, 2019 12:16 am

### Re: -w = q

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
Posts: 130
Joined: Fri Aug 30, 2019 12:16 am

### Re: -w = q

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.

005384106
Posts: 101
Joined: Sat Aug 24, 2019 12:16 am

### Re: -w = q

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