Delta U = Qin - Wout = 0
Since Qin is negative, so is Wout.
What is happening is this:
Gas with a piston on top is being compressed slowly while heat is allowed to flow out. Work is done ON (not by) the gas to maintain the constant temperature.
It is an isothermal compression. The entropy decreases.
It makes no difference that the gas is monatomic
A few homework problems.
1) The temp of a monotomic ideal gas remains constant (isothermic) during a process in which 4250 J of heat flows out of gas. How much work is done? Pos/
delta T = 0 and delta U = 0 by inference
Q = 4250
U = 3/2 nRT
Thanks for help/hints
2 answers
Thank you drwls! Explaining what the question was asking for helped me realize the problem was actually much simpler than what I was trying to do.
W = -4250
W = -4250
2 answers