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?

J

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
so
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

so

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.

To calculate the work done on the gas, we can use the First Law of Thermodynamics, which states that the change in internal energy (∆U) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):

∆U = Q - W

In this case, we know that the temperature of the gas remains constant, which means there is no change in internal energy (∆U = 0).

Since the temperature remains constant, the heat added to or removed from the system only accounts for work. Therefore, we can rewrite the equation as:

0 = -W

Simplifying the equation, we find that the work done on the gas is:

W = 0

Therefore, the work done on the gas in this process is zero (0 J).