Why do we use external pressure to calculate work in gas cylinders?

AI Thread Summary
In a gas-filled cylinder with a piston, the calculation of work done is based on external pressure rather than internal gas pressure. This is because internal pressures fluctuate during processes, except in isobaric conditions, while external pressure remains constant, making it easier to understand the work done by the surroundings on the system. The internal forces within the gas cancel each other out, allowing the piston to move due to the net effect of external pressure. When considering work done by the gas during expansion, the relationship is expressed as dW = pressure x area x distance, leading to the integral work equation Work = ∫ V1 to V2 pdV. For isothermal expansion, this can be further simplified using the ideal gas law, resulting in Work = nRT ln(V2/V1).
sodium.dioxid
Messages
50
Reaction score
0
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?
 
Chemistry news on Phys.org
I think it would be because all the internal forces cancel each other out.
 
Woopydalan said:
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?
 
sodium.dioxid said:
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.
 
Infinitum said:
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
 
Last edited:
I want to test a humidity sensor with one or more saturated salt solutions. The table salt that I have on hand contains one of two anticaking agents, calcium silicate or sodium aluminosilicate. Will the presence of either of these additives (or iodine for that matter) significantly affect the equilibrium humidity? I searched and all the how-to-do-it guides did not address this question. One research paper I found reported that at 1.5% w/w calcium silicate increased the deliquescent point by...
Back
Top