# Homework Help: Work done by gases

1. Jun 24, 2017

### Timothy Schablin

<Moderator's note: Moved from a technical forum and thus no template used.>

Consider a gas in a closed container with a piston allowed to move. Lets start with a volume of 15 and pressure of 1.5. We add heat to the system, lets say 1200 J. This forces the piston to move increasing the volume to 40. The pressure remains the same, 1.5. When calculating the work done by the gas, how do we go about it?

W = P * change in Volume.

So, since the pressure remains constant, do we just take change in volume, 15 - 40 = -25?
Or 1.5(15-40)?
Or add in the heat to the system? (15-40) + 1200?

Last edited by a moderator: Jun 24, 2017
2. Jun 24, 2017

### olgerm

Volume increases. Change of volume is ΔV=40-15 (you did not say the unit).

W=p*ΔV not W=Δp*ΔV

3. Jun 24, 2017

### Timothy Schablin

apparently, the correct answer is the amount of energy added, or 1200J. This is an isobaric situation so I guess W=PV doesnt work?

All my research on the 'net points to W=PV though....

4. Jun 24, 2017

### Staff: Mentor

What is the exact wording of the question and the analysis that led to this result?
Then I guess you have more research to do. Do you have an actual thermodynamics book?

5. Jun 25, 2017

### olgerm

correct is $W=p \cdot _\Delta V$