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# PV diagrams - part 1: Work and isobaric processes

PV diagrams - part 1: Work and isobaric processes. Created by David SantoPietro.

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• At , the equation is written as (-PdeltaV) but the commentator said this reflected as work done ON the gas. Wouldn't this be work done BY the gas as this is negative work and it is going to the right (increase in volume)?
• This issue is very simple if you simply remember that work = -PDV. Why the negative sign in front of PDV?Well it is there to make sure that we are in line with the sign convention. First remember that work done by the gas (expansion) is negative work...this makes sense because DV is positive but the work is negative because of the negative sign in front of PDV.....Now if the gas is compressed (work done on the gas) the work is positive....because DV is negative and thus, -P(-DV)= +PDV= positive work.....
• Assuming the gas is the system here.
At work done by the gas (positive delta V) it was marked in green that it's -W (i.e. negative work), but then at it was stated that "work done by the gas is P delta V" and it's +W? I've watched this many times and I'm still confused. Isn't work done by the system negative and work done on the system positive? Halp plz
• () Does the formula W=PdV only work for isobaric processes? How does using integration to solve for areas under curve (other processes) differ from this formula?
(1 vote)
• At , you say that if you're going to the left, then work will be positive. Shouldn't the work be negative because volume is decreases therefore, gas is being compressed? And gas compression is when the surrounding does work on the system?
(1 vote)
• At he explains that the area under the curve is "Work done BY the gas" [Wby]. In the previous video, he showed us "Work done ON the gas = Negative Work done BY the gas"[Won = - Wby]. Now, if the volume is increasing on the isobaric graph, that would result in a positive deltaV and using the equation Wby = PdV, Work done BY the gas [Wby] would be postive. When you plug in this +Wby into the First Law of Thermodynamics equation (dU=Q+Won), the Work would become negative since Won = - Wby giving us dU=Q + (-Wby). So if the volume is decreasing (going to the left on the graph) Vinitial = lets say 4 and Vfinal = lets say 2, would result in a negative volume change and therefore, a negative Wby using P(-V). When plugging this negative Wby into the First Law of Thermodynamics equation, remember Won = - Wby, so a negative of a negative Wby would result in positive work: dU=Q + (-(-Wby)).
• At , why is pressure always positive?
(1 vote)
• I love how David was on the edge of inventing calculus at , but he didn't even mention integrals. Its like wanting very badly to sneeze but it never happens.
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• Why is it that when the bar goes to the right the work is seen as positive and when the bar is going to the left the work is seen as negative? should be around
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• Is it possible to calculate the work of an evolution if the pressure, temperature, and volume are all changing? I often encounter problems where a piston is moving (meaning the volume is not constant), but then so is everything else. And I'm not sure which type of evolution it counts as.
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• Why isn't the work done by gas equal to the heat income in an isobaric process? Don't we have the same energy used by the gas to expand the voume that we got from the heat coming in?
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• At , wouldn't work done by the gas mean negative area since the energy is lost? So wouldn't the graph actually be showing work done on the gas?
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