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Calculating work done by gas?

 
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Jun13-12, 07:44 PM   #1
 

Calculating work done by gas?


In a cylinder (with a piston) containing gas, why do we use the external pressure, instead of the pressure of the gas, to calculate work?
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Jun13-12, 08:57 PM   #2
 
I think it would be because all the internal forces cancel each other out.
Jun13-12, 09:31 PM   #3
 
Quote by Woopydalan View Post
I think it would be because all the internal forces cancel each other out.
Then why does the piston move up due to internal pressure?
Jun14-12, 04:59 AM   #4
 

Calculating work done by gas?


Quote by sodium.dioxid View Post
In a cylinder (with a piston) containing gas, why do we use the external pressure, instead of the pressure of the gas, to calculate work?
Internal pressure changes in all processes except isobaric. External pressure does not. Also, the work done on the system by the surrounding is easier to understand in terms of external pressure.
Jun15-12, 03:51 AM   #5
 
Quote by Infinitum View Post
Internal pressure changes in all processes except isobaric. External pressure does not. Also, the work done on the system by the surrounding is easier to understand in terms of external pressure.
And if the original question is on the work done by the system.
That is, the pressure of the gas in the cylinder expanding against the piston.
Then dW = Force x distance = pressure x Area x distance = p dV
Work = ∫ V1 to V2 pdV
And with PV = n RT
Work = nRT ln V2/V1
For isothermal expansion
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