Work Done By Gases - Calculating Pressure, Volume & Heat

AI Thread Summary
In a closed container with a piston, when heat is added to a gas, the volume increases while the pressure remains constant. The work done by the gas can be calculated using the formula W = P * ΔV, where ΔV is the change in volume. In this scenario, the change in volume is 40 - 15, resulting in a positive value for ΔV. Despite initial confusion, the correct interpretation is that the work done is equivalent to the energy added, which is 1200 J, in this isobaric process. Therefore, W = 1200 J is the accurate conclusion for this situation.
Timothy Schablin
Messages
17
Reaction score
0
<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?
 
Last edited by a moderator:
Physics news on Phys.org
Timothy Schablin said:
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).

Timothy Schablin said:
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
 
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...
 
Timothy Schablin said:
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?
 
Timothy Schablin said:
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##
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top