Question on Work and Pressure/Volume

[tf2medic]

okay doing some physics review and...the kaplan book is bringing up a work equation as W = Pressure * change in volume. it says you can find the work done by calculating the area under a curve of a PvsV graph. that's all kaplan says...does anyone mind explaining where that relationship comes from mathematically? if you can do that, then that means some W=integral of some P(V) function. i've never heard of that so...i tried the opposite approach by taking the derivative of that W=PchangeinV with respect to time so dW/t = something related to P and V but dW/t actually = Power. so
that means pressure and volume are also related to power through some equation? I'm a little confused.

also, that equation kaplan mentioned said holds true for an isobaric condition where pressure is constant and there's a change in volume. BUT it still said you can always take the area under the curve of a PvsV graph to give you work. so i don't understand where that comes from. kaplan doesn't give any of the calc-explanation behind it and just wants you to blindly memorize it. i took the engineering physics and i don't recall any integrals/derivatives relating work to P and V. if you really can use the area of PvsV to find work, then an integral has to be involved. i just don't understand the details. so...help?? thanks

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[tncekm]

that equation kaplan mentioned said holds true for an isobaric condition where pressure is constant and there's a change in volume. BUT it still said you can always take the area under the curve of a PvsV graph to give you work.

I'll do my best to answer your question.

Work is always the area under the curve of a P vs. V graph.

For an isobaric situation you can use the equation W=-P*deltaV and you don't need to calculate the integral because you can get the area under the curve simply by multiplying P*deltaV--in short, the integral isn't needed, and that's why its not used.

If you don't have an isobaric situation, you must calculate the integral to find the area under the curve, so you can't use the simplified formula.

Edit: this might help: http://hypertextbook.com/physics/thermal/pressure-volume/

 

[tf2medic]

yeah...your explanation doesn't really explain why you're able to take the area under the curve for the work, so initially your post didn't help. BUT the link you provided does explain it!

W = ∫ F · ds = ∫ P dV = − area on PV graph


that's why! the 3rd part of the equation tells you.

W=∫ P dV= − area on PV graph


Makes perfect sense. that's what i assumed it was based on what the Kaplan book said, but since I had never seen that equation before in terms of P and V I was confused and thought I was just making it up. apparently it exists tho...haha.
the - sign in the front makes perfect sense too because Kaplan says you can only see the above relationship to get work done ON a gas, which is negative work so it needs that - sign.
in the case of isobaric, pressure is constant so you can just move the P in front of the integral giving you:
W=∫ P dV= P∫ dV = P(changeinV)
awesome. thanks.

for other situations, you can still use W=∫ P dV
but you would need the actual equation for P and then integrate with respect to P and V. so in other words you'd have to solve the double integral to get work in that case. so yea that might be more of a pain, but taking the area under the curve, if the curve is given and the area's shapes are nice, is a faster way. the instance you couldn't really just take the area under the curve for the work would be if PvsV is actually more a curve and not straight lines. the Kaplan book actually shows a drawing of a curved PvT graph as its part D picture. however, kaplan is stupid and says "in part D you would also find the work using the area of the curve." except that's not accurately possible by just looking at it! it doesn't say you'd have to take the double integral to get an accurate #. hell, kaplan doesn't give any of the simple calc. derivations on where the equations come from. they just want you to blindly memorize the equation.

i thought most important to the physics section is knowing how to use the equations AND understanding what they mean. how can you understand what the equations are actually doing if you don't know where they come from??

 

[tf2medic]

one last question, where does the - sign come from?? the Kaplan book says the area under the curve would give you the work. it says nothing about a - sign anywhere or in the isobaric equation. the only time you'd mathematically use the - sign is if the area under the curve happens to fall under the x-axis. in that case you need the - sign since you are only considered with adding the magnitudes.

do they simply, by principle, add the - sign in the front to tell you it's negative work and thus is work being done on the gas/system. that would make sense since they are always adding a sign in the end in physics in order to indicate the meaning of the answer even tho mathmetically the *-1 has no derivation. is this right?? thanks

 

[tncekm]

Its just a sign convention.

If work is greater than zero, this means work was done ON the system BY the environment, or volume decreases.

i.e. W=-P&#916;V; if &#916;V < 0, then W > 0.; increases in volume means work done BY the system on the environment.
i.e. W=-P&#916;V; if &#916;V > 0, then W < 0.; decreases in volume means work done by the environment ON the system.

I'm with you; I really like knowing how, where,why,etc, with regards to all of the concepts.