Posted by Mary on .
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!

Physics repost please check 
kat,
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