Work done by a gas in a piston

[goggles31]

Let's assume that weights are placed on a massless piston and below the piston is a gas. If we remove all the weights at once, the work done by the gas should be the difference between the forces exerted by the gas and the atmosphere multiplied by the change in volume. However, we know that the pressure exerted by a gas is inversely proportional to its volume and hence the work done is given by the area under a curve. Is it okay to assume that the pressure of the gas instantaneously changes to atmospheric pressure when the weights are removed, as in Figure 4.10 of the link given?

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html

 

[BiGyElLoWhAt]

I don't think that seems very reasonable. Try solving $PV^{\gamma}=nRT$

 

[Chestermiller]

Let's assume that weights are placed on a massless piston and below the piston is a gas. If we remove all the weights at once, the work done by the gas should be the difference between the forces exerted by the gas and the atmosphere multiplied by the change in volume. However, we know that the pressure exerted by a gas is inversely proportional to its volume and hence the work done is given by the area under a curve. Is it okay to assume that the pressure of the gas instantaneously changes to atmospheric pressure when the weights are removed, as in Figure 4.10 of the link given?

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html

In an irreversible expansion, the pressure exerted by a gas on the piston is not inversely proportional to its volume. The gas pressure is not even uniform spatially within the cylinder during an irreversible expansion. Plus, the force is affected by viscous (dissipative) stresses in the gas such that the rate of change of volume also affects the force. However, the work that the gas does on the piston can be determined if we know the external force applied to the gas at the piston face (because the gas pressure matches the "external pressure" at the piston face).

For more details on this, see my two Physics Forums Insights articles: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/ and https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

 

[Chestermiller]

I don't think that seems very reasonable. Try solving $PV^{\gamma}=nRT$

This equation is not valid for an irreversible expansion, or any other expansion for that matter. If the expansion is reversible, then the ideal gas law applies (exactly as it is always written).

 

[BiGyElLoWhAt]

https://en.wikipedia.org/wiki/Adiab...-V_relation_for_adiabatic_heating_and_cooling
It seems the RHS was wrong in that equation. It should be equal to a constant.

 

[BiGyElLoWhAt]

Also, in the link, the processes are explicitly reversible.

 

[Chestermiller]

Also, in the link, the processes are explicitly reversible.

Not for the cases where finite weights are suddenly removed from the piston.

 

[Chestermiller]

https://en.wikipedia.org/wiki/Adiab...-V_relation_for_adiabatic_heating_and_cooling
It seems the RHS was wrong in that equation. It should be equal to a constant.

That is only valid for a reversible quasistatic process in which the pressure and the temperature are varied in tandem in such a way that the exponent is a constant. It can not apply to an irreversible process.

 

[BiGyElLoWhAt]

Hmm... I see. Removing each small weight is what they're referring to as reversible. Sorry about that.

 

[Chestermiller]

Hmm... I see. Removing each small weight is what they're referring to as reversible. Sorry about that.

No problem. These things are always very confusing.