Physics please check repost
posted by Mary .
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.
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