Q: 20 J of heat energy is added to an ideal gas while -40J of work is done BY the same gas. The total change in internal thermal energy of the system is what? Equations I used: [tex]\Delta[/tex]Eth = Heat + Work [tex]\Delta[/tex]Eth = Heat - Ws This was a problem on a recent physics test I took and I said the answer was 60J. However, my instructor insists that the correct answer is -20J. He says that -40J of work is energy leaving the gas irregardless of whether the gas is doing the work or the environment is doing the work. I don't see how this is possible. My reasoning is that (-40J) of work is done by the gas = 40J of work done ON the gas. Thus, [tex]\Delta[/tex]Eth= 20 J + 40J. Can anyone enlighten me on why this is? Thanks a lot!