posted by Pierre on .
The temperature of a monatomic ideal gas remains constant during a process in which 4500 J of heat flows out of the gas. How much work (including the proper + or - sign) is done on the gas?
Well, to start - You must compress the gas to get heat out at constant temperature --> work in
The process is isothermal (constant temp)
In an ideal gas, the internal energy U depends only on T, so the change in U is zero so
0 = Q - W
but Q = -4500 J
0 = -4500 + W
W = +4500 = work done BY gas
so -4500 is work done ON the gas
I'm not sure what we are doing wrong but the computer is saying -4500 is wrong
0 = -4500 - W ======= NOT + W
W = -4500 = work done BY gas
so +4500 is work done ON the gas
I prefaced this whole thing by telling you why W had to be positive then did not notice my sign error.