Question

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.

Answer 1

Delta U would be 2531J + 1101J instead of 2531J - 1101J, because the system is doing work instead of having work done to it.

(2531J + 1101J)= 3632J

3632J = 3/2(3.0mol)(8.31)(T final - 345K)

3632J = 37.395(T final - 345K)

3632J/ 37.395 = T final - 345K

97.12 + 345K = T final

442.125K = T final