Three moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2531 J of heat are added to the gas, and 1101 J of work are done on it. What is the final temperature of the gas?

delta U= 3/2nR(T final -T initial)

(2531J - 1101J) = 3/2(3.0mol)(8.31)(T final - 345K)

1430J = 37.395(T final - 345K)

1430J/ 37.395 = T final - 345K

38.24 + 345K = T final

383.2404K = T final

This answer is incorrect. Please explain to me where I went wrong.

It still seems to me that if heat is added to the gas, the U increases, and if work is done on the gas, the U increases also. I don't understand why you subtracted.

Thanks!

your formula is correct. however since it says the work is done on the system work has to be negative so you would do 2531-(-1101) or 2531 + 1101. That should give u the right answer

The equation you used for the change in internal energy is correct, but the sign of the heat and work terms should be switched in this case. When heat is added to a system, the internal energy increases, so the sign of the heat term should be positive. On the other hand, when work is done on a system, the internal energy also increases, so the sign of the work term should be positive as well.

Let's rewrite the equation using the correct signs:

∆U = 3/2nR(Tfinal - Tinitial)
(2531J - 1101J) = 3/2(3.0mol)(8.31)(Tfinal - 345K)
1430J = 37.395(Tfinal - 345K)
1430J / 37.395 = Tfinal - 345K
38.24 + 345K = Tfinal
383.2404K = Tfinal

So the final temperature of the gas is indeed 383.2404K. Apologies for the previous confusion caused by incorrect sign usage.