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

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

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

To determine the amount of work done by the gas, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added or removed (Q) minus the work done by the system (W).

ΔU = Q - W

Since the temperature remains constant, ΔU is equal to zero (ΔU = 0). Therefore, we can rearrange the equation as follows:

W = Q

Given that Q is -4250 J (since heat is flowing out of the gas), the work done by the gas is also -4250 J.

So, the work done by the gas is -4250 J (negative because work is done on the gas).