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### Delta U

Posted: Sun Feb 09, 2020 6:42 pm
Under what conditions would Delta U = 0 and q = -w ? Thank you

### Re: Delta U

Posted: Sun Feb 09, 2020 6:45 pm
delta U is equal to 0 when temperature is constant. If delta U is 0, and deltaU=q+w, then 0=q+w and when you solve for q then the equation becomes q= -w

### Re: Delta U

Posted: Sun Feb 09, 2020 6:46 pm
delta U can be 0 when a system is doing work as a reversible expansion and when the energy lost is replaced by heat flow into the system (isothermic).

### Re: Delta U

Posted: Sun Feb 09, 2020 8:51 pm
Isothermal systems have delta U = O because the transfer of heat and pressure are both at equilibrium, so all the energy lost by the work of expansion transfers into the system as heat in order to keep the temperature constant. Since work is being done (system losing energy), -w is negative, and since heat is transferring in (gaining energy), q is positive. If these values were not equal, delta U would not equal zero and the system would not be truly isothermal.

### Re: Delta U

Posted: Sun Feb 09, 2020 9:07 pm
delta U = 0 for isothermal expansion or compression of an ideal gas. delta U = 0, when q = 0 and w = 0 (since delta U = 0 + 0 = 0). q = -w when delta U = 0, then 0 = q + w, which is q = -w.

### Re: Delta U

Posted: Sun Feb 09, 2020 9:08 pm
isothermal compression or expansion

### Re: Delta U

Posted: Sun Feb 09, 2020 11:17 pm
DeltaU = 0 when there is an isothermal expansion. Although work is used, and heat is consequently lost, heat flow from the surroundings replaces the heat lost, so temperature remains constant. If we know that deltaU = 0, and that deltaU = w + q, then q = -w. We can confirm that w is a negative value because work is being done by the system (system is losing heat/energy).

### Re: Delta U

Posted: Mon Feb 10, 2020 2:06 pm
Delta U=0 for Isothermal compression or expansion (at constant temperature).