My equation sheet says 1atm=1E5 Pa, so since i had 2 atm its therefore 2E5. But anyway i figured it out.
W=(2E5)(.002-.01)
W= -1600
Then the internal energy equation
U=-(3.53E3)-(-1600)
U=-1930J
Now i did get this right on my webassign but i kinda have the understanding of why the Q was...
This problem is going to make me go crazy...
4. In the figure below, an ideal gas is slowly compressed at a constant pressure of 2.0 atm from 10.0 L to 2.0 L. This process is represented as the path B to D. In this process, some heat flows out and the temperature drops. If the heat lost from...