Work Done By Gases - Calculating Pressure, Volume & Heat

[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. Let's start with a volume of 15 and pressure of 1.5. We add heat to the system, let's 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?

 

[olgerm]

do we just take change in volume, 15 - 40 = -25?

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

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)?

W=p*_ΔV not W=_Δp*_ΔV

 

[Timothy Schablin]

apparently, the correct answer is the amount of energy added, or 1200J. This is an isobaric situation so I guess W=PV doesn't work?

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

 

[Chestermiller]

apparently, the correct answer is the amount of energy added, or 1200J. This is an isobaric situation so I guess W=PV doesn't work?

What is the exact wording of the question and the analysis that led to this result?

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

Then I guess you have more research to do. Do you have an actual thermodynamics book?

 

[olgerm]

apparently, the correct answer is the amount of energy added, or 1200J. This is an isobaric situation so I guess W=PV doesn't work?

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

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