Work done during adiabatic expansion on piston

[MysticDream]

TL;DR

Needing clarification on how much work is done in various cases

Does the work "extracted" from a gas (with the same initial properties) against a piston while expanding change based on the mass of the piston?
For example, I have a specific volume of compressed air inside a cylinder with a piston positioned horizontally with stops. The air temperature is 350 K and the pressure is 2000 Kpa. If I remove the stops and the gas expands pushing the piston until the pressure is lowered to 1500 Kpa, will the volume, temperature, and work done (up until that point) be the same regardless of the mass of the piston? What formula would be used to calculate this? In this case we could assume atmospheric pressure exists externally on the opposite side of the piston.

The way it's been explained to me is that regardless of the mass, the same amount of work would be done unless there was no mass (free expansion), in which case no work would be done, so it's all or nothing which doesn't seem right.

 

[Chestermiller]

TL;DR Summary: Needing clarification on how much work is done in various cases

Does the work "extracted" from a gas (with the same initial properties) against a piston while expanding change based on the mass of the piston?
For example, I have a specific volume of compressed air inside a cylinder with a piston positioned horizontally with stops. The air temperature is 350 K and the pressure is 2000 Kpa. If I remove the stops and the gas expands pushing the piston until the pressure is lowered to 1500 Kpa, will the volume, temperature, and work done (up until that point) be the same regardless of the mass of the piston?

Is the piston still moving at this point, or has it stopped moving? Let's see your Newton's 2nd law force balance equation on the piston

 

[MysticDream]

Is the piston still moving at this point, or has it stopped moving? Let's see your Newton's 2nd law force balance equation on the piston

Would it matter if I'm only interested in knowing how much work was done up until that point? We could say the piston stopped by hitting a stop which would transfer it's energy as heat and deformation on impact, or if it was released like a cannonball it would continue on until friction and/or gravity caused it to slow down and hit the ground. Either way, there is certain amount of work that the gas did on the piston which I'd like to be able to calculate (for any mass) and the resulting properties of the gas at the final pressure. As far as Newton's 2nd law, F=ma, but I'm not sure what the acceleration would be in the adiabatic case as the force is reducing over time. Any insight you could share, I'd appreciate.

 

[Chestermiller]

Would it matter if I'm only interested in knowing how much work was done up until that point? We could say the piston stopped by hitting a stop which would transfer it's energy as heat and deformation on impact, or if it was released like a cannonball it would continue on until friction and/or gravity caused it to slow down and hit the ground. Either way, there is certain amount of work that the gas did on the piston which I'd like to be able to calculate (for any mass) and the resulting properties of the gas at the final pressure. As far as Newton's 2nd law, F=ma, but I'm not sure what the acceleration would be in the adiabatic case as the force is reducing over time. Any insight you could share, I'd appreciate.

Well, let's just see. Let F(t) be the force that the gas exerts on the inside face of the piston at time t during the expansion, m be the mass of the piston, $P_{atm}$ be the pressure of the outside air on the outside face of the piston, and v(t) be the velocity of the piston at time t. Neglecting friction, Newton's 2nd law applied to the piston reads: $$F(t)-P_{atm}A=m\frac{dv}{dt}$$where A is the area of the piston. If we multiply this equation by $v(t)=\frac{dx}{dt}$ where x(t) is the location of the piston at time t, we obtain: $$F(t)\frac{dx}{dt}-P_{atm}\frac{dV}{dt}=mv\frac{dv}{dt}$$where V(t) is the volume of the gas at time t. If we integrate this equation between time t = 0 and arbitrary time t, we obtain: $$W_g(t)=\int_{0}^{x(t)}{Fdx}=P_{atm}(V(t)-V(0))+\frac{1}{2}mv^2(t)$$where W(t) is the amount of work done by the gas between time t=0 and tune t, If the piston has come to rest after it interacts with the stop (either by bouncing off and moving until its kinetic energy is dissipated by the gas, or by sticking to the stop, the ultimate amount of work done by the gas on the piston (no matter what its mass) is $$W=P_{atm}(V(\infty)-V(0))$$where $V(\infty)$ is the volume of gas corresponding to the stop.

 

[MysticDream]

Well, let's just see. Let F(t) be the force that the gas exerts on the inside face of the piston at time t during the expansion, m be the mass of the piston, $P_{atm}$ be the pressure of the outside air on the outside face of the piston, and v(t) be the velocity of the piston at time t. Neglecting friction, Newton's 2nd law applied to the piston reads: $$F(t)-P_{atm}A=m\frac{dv}{dt}$$where A is the area of the piston. If we multiply this equation by $v(t)=\frac{dx}{dt}$ where x(t) is the location of the piston at time t, we obtain: $$F(t)\frac{dx}{dt}-P_{atm}\frac{dV}{dt}=mv\frac{dv}{dt}$$where V(t) is the volume of the gas at time t. If we integrate this equation between time t = 0 and arbitrary time t, we obtain: $$W_g(t)=\int_{0}^{x(t)}{Fdx}=P_{atm}(V(t)-V(0))+\frac{1}{2}mv^2(t)$$where W(t) is the amount of work done by the gas between time t=0 and tune t, If the piston has come to rest after it interacts with the stop (either by bouncing off and moving until its kinetic energy is dissipated by the gas, or by sticking to the stop, the ultimate amount of work done by the gas on the piston (no matter what its mass) is $$W=P_{atm}(V(\infty)-V(0))$$where $V(\infty)$ is the volume of gas corresponding to the stop.

Thanks, I see you have a good understanding of the subject. That gives me the amount of work done, but what about in the case of varying amounts of mass, and how would I calculate the final volume and temperature of the gas? I recall reading one of your comments on a thread here where you said that when less work was done, the temperature of the gas would be higher. This is what I'm confused about. Initially my understanding was that given a volume of compressed gas at a specific temperature, after expanding to a specific volume and doing work, the temperature will always be the same no matter what the mass of the piston is. I now suspect this is wrong as it doesn't seem as though a tiny mass would change the gas properties as much as a very large mass. I'd like to know given this initial volume of gas at a specific temperature, how much energy can be extracted from it if I know what I want the final pressure and temp to be? Are there cases where the mass isn't large enough and so not enough work can be done after expanding to a certain volume and/or pressure resulting in a higher gas temperature?

 

[Chestermiller]

Thanks, I see you have a good understanding of the subject. That gives me the amount of work done, but what about in the case of varying amounts of mass, and how would I calculate the final volume and temperature of the gas?

In my previous post, we say that the amount of work done by the gas is independent of the mass of the piston, provided the piston is at rest in the final thermodynamic equilibrium state.

To calculate the final volume and temperature of the gas, we use the first law of thermodynamics:
$$\Delta U=-W=-P_{ext }(V_2-V_1)$$

I recall reading one of your comments on a thread here where you said that when less work was done, the temperature of the gas would be higher.

You can see from the above equation that if the gas does less work, the internal energy U decreases less, and the temperature decrease will be less, so the final temperature will be higher.

This is what I'm confused about. Initially my understanding was that given a volume of compressed gas at a specific temperature, after expanding to a specific volume and doing work, the temperature will always be the same no matter what the mass of the piston is.

Yes. This is correct. What's the problem. We already showed that, provided the final state is a thermodynamic equilibrium state, such that the piston is no longer moving, the work depends only on the external pressure and the volume increase.

I now suspect this is wrong as it doesn't seem as though a tiny mass would change the gas properties as much as a very large mass.

No.

I'd like to know given this initial volume of gas at a specific temperature, how much energy can be extracted from it if I know what I want the final pressure and temp to be?

If you were specifying the final volume., assuming ideal gas behavior, we have $$\Delta U=\frac{P_1V_1}{RT_1}C_v(T_2-T_1)=-P_{ext}(V_2-V_1)$$so $$\frac{T_2}{T_1}=1-\frac{P_{ext}(V_2-V_1)R}{C_vP_1V_1}$$

Are there cases where the mass isn't large enough and so not enough work can be done after expanding to a certain volume and/or pressure resulting in a higher gas temperature?

The piston mass is irrelevant. The temperature will always drop.

 

[MysticDream]

In my previous post, we say that the amount of work done by the gas is independent of the mass of the piston, provided the piston is at rest in the final thermodynamic equilibrium state.

To calculate the final volume and temperature of the gas, we use the first law of thermodynamics:
$$\Delta U=-W=-P_{ext }(V_2-V_1)$$

You can see from the above equation that if the gas does less work, the internal energy U decreases less, and the temperature decrease will be less, so the final temperature will be higher.

Yes. This is correct. What's the problem. We already showed that, provided the final state is a thermodynamic equilibrium state, such that the piston is no longer moving, the work depends only on the external pressure and the volume increase.

No.

If you were specifying the final volume., assuming ideal gas behavior, we have $$\Delta U=\frac{P_1V_1}{RT_1}C_v(T_2-T_1)=-P_{ext}(V_2-V_1)$$so $$\frac{T_2}{T_1}=1-\frac{P_{ext}(V_2-V_1)R}{C_vP_1V_1}$$

The piston mass is irrelevant. The temperature will always drop.

This helps, thanks.

Just to clarify then, if $$P_{ext}$$ was double, then for the same volume change, double the work would be done, lowering the pressure and temperature for the same end volume?

I had been using this formula to determine how much work a specific compressed volume of gas would do:

$$\frac{(V_1P_1-V_2P_2)}{(\gamma-1)}$$

 

[Chestermiller]

This helps, thanks.

Just to clarify then, if $$P_{ext}$$ was double, then for the same volume change, double the work would be done, lowering the pressure and temperature for the same end volume?

Yes, as long as $P_{ext}$ is still less than the initial gas pressure P1.

I had been using this formula to determine how much work a specific compressed volume of gas would do:

$$\frac{(V_1P_1-V_2P_2)}{(\gamma-1)}$$

OK, but understand that, if V2 is specified, that does not mean that the final gas pressure P2 is equal to $P_{ext}$. So we have $$W=\frac{(P_1V_1-P_2V_2)}{(\gamma-1)}=P_{ext}(V_2-V_1)$$From this, you can solve for P2.

 

[MysticDream]

Yes, as long as $P_{ext}$ is still less than the initial gas pressure P1.

OK, but understand that, if V2 is specified, that does not mean that the final gas pressure P2 is equal to $P_{ext}$. So we have $$W=\frac{(P_1V_1-P_2V_2)}{(\gamma-1)}=P_{ext}(V_2-V_1)$$From this, you can solve for P2.

Thanks a lot!