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!

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1. 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

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posted by kat

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