## ∆U When ∆T = 0

isochoric/isometric: $\Delta V = 0$
isothermal: $\Delta T = 0$
isobaric: $\Delta P = 0$

Reina Robles 2B
Posts: 71
Joined: Fri Aug 09, 2019 12:16 am

### ∆U When ∆T = 0

Why is ∆U = 0 when ∆T = 0? Thanks!

IScarvie 1E
Posts: 66
Joined: Sat Aug 17, 2019 12:16 am

### Re: ∆U When ∆T = 0

Temperature is a number directly proportional to the energy of a system (at least when you're working with ideal gasses, which we are). So when there is no change in T, there also cannot be any change in U.

Luc Zelissen 1K
Posts: 57
Joined: Mon Jun 17, 2019 7:23 am

### Re: ∆U When ∆T = 0

$\Delta U$ is a function of N and T. If $\Delta T$ does not change, $\Delta U$ will also not change

Amir Bayat
Posts: 115
Joined: Sat Sep 07, 2019 12:16 am

### Re: ∆U When ∆T = 0

Since there is no change in the temperature of the system, there is essentially no change in the internal energy of the system as well.

Ryan Lee 1E
Posts: 50
Joined: Sat Aug 17, 2019 12:16 am

### Re: ∆U When ∆T = 0

I think from the context of class, we are referring to an isothermal expansion of an ideal gas. This means that the gas is expanding reversibly and work is being done to move the piston and increase the volume. In this case, the work being done would be negative. However, in the isothermal reaction that we discuss, temperature of the system is also held constant, which means that any internal energy being lost through work being done is replaced through heat flowing into the system. In other words, as negative work is done to move the piston out, positive heat is flowing into the system, and as a result, deltaU = 0 because all the -w is countered by +q. This in effect, also means that deltaT is 0 since heat is flowing into the system to keep the system at constant T.

Reina Robles 2B
Posts: 71
Joined: Fri Aug 09, 2019 12:16 am

Thank you!