# Internal energy

1. Apr 5, 2006

### targa9932001

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 the gas in the process BD is 3.53E3 J, what is the change in internal energy of the gas?

Ok... all i have so far is since the pressure remains constant, it's an isobaric process. Therefore i have to calculate work.
W=(2E5)(.01-.002)
W=1600 J Now is it positive or negative???
Then i know next step is the change in internal energy
U=Q-W
Since the heat is lost from the gas is it negative 3.53E3???
U=(3.53E3)-(-1600)
Now i do not get the right answer... someone please explain why??? thanks.

2. Apr 6, 2006

### Cyrus

1atm = 101.325 kPa,

not 2E5.

3. Apr 6, 2006

### Szeki

HEAT LOST from the GAS is Negative
HEAT SUPPLIED To the gas is POSITIVE
Work Done on the Gas is POSITIVE(Means Compress gas)
Work done by the gas is NEGATIVE(Means Expand Gas)

Internal Energy = Work DOne + Heat
Internal Energy = Kinetic Enery + Potential energy

Hope this helps

4. Apr 6, 2006

### targa9932001

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 negative so was the W. Ah w/e teacher hasnt taught us this... im just trying to get ahead.

5. Apr 6, 2006

### Cyrus

I would redo your caculation, it is incorrect.