Position of piston related to ideal gas

[songoku]

Homework Statement

A free-moving piston of cross sectional $A$ and mass $m_0$ is attached to thermally insulated container as shown in the diagram. Monoatomic ideal gas of volume $V_0 (V_0 = A . L_0)$ is trapped inside the container with initial temperature $T_0$ and initial pressure $P_0$. Initially, piston is given speed $c$ where $c$ satisfies equation $\frac{m_0.c^2}{P_0.V_0}=\frac{29}{4}$ . Pressure and temperature of surrounding are always constant ($P_0$ and $T_0$ respectively).
a. Find the position when the piston is momentarily at rest in terms of $k L_0$
b. When oscillating, no energy lost to surroundings. Find the final position of piston in terms of $k L_0$

Relevant Equations

Ideal gas

a. The piston will be at rest when all its kinetic energy converted into work to push the gas, so:

$$\frac{1}{2}m_0 c^2=P_0. \Delta V$$
$$\frac{1}{2}m_0 \frac{29}{4} \frac{P_0.V_0}{m_0}=P_0.\Delta V$$
$$\frac{29}{8} V_0=\Delta V$$
$$\frac{29}{8} L_0 = L_0 - L$$
$$L=-\frac{21}{8} L_0$$

My answer does not make sense because it is negative. Where is my mistake?

Thanks

 

[haruspex]

Not my subject, but why isn't the work done $\int P.dV$?

 

[Chestermiller]

The trick to doing this problem is to treat the combination of the gas and piston as your "system." In the initial state, the system has kinetic energy. You have correctly approached the work done by the on the surroundings, but you have not included the change in internal energy of the gas from the initial state to the point where the kinetic energy of the piston first reaches zero. The first law of thermodynamics applied to this initial step in this process is as follows:
$$\Delta U+\Delta K=-P_0(V_1-V_0)$$where $V_1$ is the gas volume when the piston first stops and $$\Delta K=-m_0\frac{c^2}{2}$$

 

[songoku]

Not my subject, but why isn't the work done $\int P.dV$?

Not your subject but immediately spot my mistake

The trick to doing this problem is to treat the combination of the gas and piston as your "system." In the initial state, the system has kinetic energy. You have correctly approached the work done by the on the surroundings, but you have not included the change in internal energy of the gas from the initial state to the point where the kinetic energy of the piston first reaches zero. The first law of thermodynamics applied to this initial step in this process is as follows:
$$\Delta U+\Delta K=-P_0(V_1-V_0)$$where $V_1$ is the gas volume when the piston first stops and $$\Delta K=-m_0\frac{c^2}{2}$$

$$\Delta U+\Delta K=-P_0(V_1-V_0)$$
$$\frac{3}{2} (P_1 . V_1 - P_0.V_0) - \frac{1}{2} m_0 . \frac{29}{4} \frac{P_0.V_0}{m_0} = P_0.V_0 - P_1.V_1$$

I have 2 questions:
1. Do I use adiabatic equation to state $P_1$ in terms of $P_0, V_0 ~\text{and}~ V_1$?

2. Why the work done on the gas (the RHS part of the equation) is $-P_0.\Delta V$? I thought the initial pressure and final pressure of the gas will not be the same so maybe I need to use integration like suggested by @haruspex

Thanks

 

[Chestermiller]

Not your subject but immediately spot my mistake
ΔU+ΔK=−P0(V1−V0)
32(P1.V1−P0.V0)−12m0.294P0.V0m0=P0.V0−P1.V1

I have 2 questions:
1. Do I use adiabatic equation to state P1 in terms of P0,V0 and V1?

Yes, you do.

But you messed up on the algebra on the right hand sign. Both terms should have P0.

2. Why the work done on the gas (the RHS part of the equation) is −P0.ΔV? I thought the initial pressure and final pressure of the gas will not be the same so maybe I need to use integration like suggested by @haruspex

Thanks

If the system consists of the combination of the piston and the gas, the work on the right hand side is the work done by the outer face of the piston on the surroundings. The work done by the gas on the inside face of the piston is reflected in the change of internal energy; it is minus the change of internal energy. This is exactly the result of the integration suggested by @haruspex.

 

[Delta2]

Basically you have two works on the Piston:

1. The work from the gas which opposes the velocity and therefore should be taken as negative. This is the work of an adiabatic compression, or as Chet says from first law since $Q=0,W_{gas}=-\Delta U$
2. The work from the surroundings which have constant pressure $P_0$ and therefore this work is simply $W_{surroundings}=P_0(V_0-V_1)$ and should be taken as positive since it augments the velocity of the piston.

Using the work-energy theorem on the piston $$0-\frac{1}{2}m_0c^2=W_{gas}+W_{surroundings}$$

 

[songoku]

Yes, you do.

But you messed up on the algebra on the right hand sign. Both terms should have P0.If the system consists of the combination of the piston and the gas, the work on the right hand side is the work done by the outer face of the piston on the surroundings. The work done by the gas on the inside face of the piston is reflected in the change of internal energy; it is minus the change of internal energy. This is exactly the result of the integration suggested by @haruspex.

Basically you have two works on the Piston:

1. The work from the gas which opposes the velocity and therefore should be taken as negative. This is the work of an adiabatic compression, or as Chet says from first law since $Q=0,W_{gas}=-\Delta U$
2. The work from the surroundings which have constant pressure $P_0$ and therefore this work is simply $W_{surroundings}=P_0(V_0-V_1)$ and should be taken as positive since it augments the velocity of the piston.

Using the work-energy theorem on the piston $$0-\frac{1}{2}m_0c^2=W_{gas}+W_{surroundings}$$

I understand and I think I can solve part (a). It is only algebra to state $V_1$ in terms of $V_0$

For part (b), my idea is to use energy. Let initial position of piston is $x$ (measured from equilibrium position of oscillation) and it moves as far as $\Delta L$ before coming to first stop.

Equation for energy:
$$KE_\text{oscillation} + PE_\text{oscillation} = \text{total energy}$$
$$\frac{1}{2}m_0c^2+\frac{1}{2}kx^2=\frac{1}{2}k(x+\Delta L)^2$$

I have problem in eliminating $k$ from that equation.

My questions:
1) Is my approach correct? If yes, how to eliminate $k$? I can use $k=\omega^2 .m_0$ but then there will be $\omega$ in the equation

2) What does $k$ in that equation mean? In case of oscillation of spring, $k$ is spring constant but I also encountered $k$ is being used in another oscillation problem, such as pendulum so I think $k$ should have some other meaning besides spring constant because it can be used for case not involving spring.

Thanks

 

[haruspex]

What does k in that equation mean?

The sudden introduction of k into the question without definition is very odd.
I suspect that the original question had k instead of that $\frac{29}4$ fraction. It's nothing to do with SHM.

 

[Delta2]

1) Is my approach correct? If yes, how to eliminate k? I can use k=ω2.m0 but then there will be ω in the equation

I don't think your approach is correct because I don't think the piston does harmonic oscillation. The Force F doesn't obey an equation of the form $F-F_0=-kx$ does it?

P.S I think we would need the gas to do isothermal process in order for the oscillation to be harmonic.

 

[songoku]

The sudden introduction of k into the question without definition is very odd.
I suspect that the original question had k instead of that $\frac{29}4$ fraction. It's nothing to do with SHM.

I have asked for clarification. $k$ in question (a) and (b) turns out to be a numerical constant (not related to any other $k$ such as Boltzmann constant or spring constant)

I don't think your approach is correct because I don't think the piston does harmonic oscillation. The Force F doesn't obey an equation of the form $F-F_0=-kx$ does it?

P.S I think we would need the gas to do isothermal process in order for the oscillation to be harmonic.

Oh I don't even check whether it satisfies SHM condition

So question (b) is wrong? Or it can be solved using certain method?

Thanks

 

[Delta2]

So question (b) is wrong? Or it can be solved using certain method?

I don't think the question is wrong. The method I had in mind was to find the function $g(x)$ such that $F-F_0=g(x)$ and then solve the ODE $$m_0\frac{d^2x}{dt^2}=F_0+g(x)$$. You might need to use the adiabatic equation to find g(x).

 

[songoku]

I don't think the question is wrong. The method I had in mind was to find the function $g(x)$ such that $F-F_0=g(x)$ and then solve the ODE $$m_0\frac{d^2x}{dt^2}=F_0+g(x)$$. You might need to use the adiabatic equation to find g(x).

What is $F$, $F_0$ and $g(x)$?

Thanks

 

[Delta2]

What is $F$, $F_0$ and $g(x)$?

Thanks

$F$ is the total force on the piston, $F_0=P_0A$ is the force from the surroundings and $g(x)$ is the force from the gas when piston is at position x.

 

[haruspex]

I went ahead with my assumption that k is supposed to represent $\frac{m_0c^2}{P_0V_0}$ and took $\gamma=5/3$.
Solving the integral, and writing $r=\frac L{L_0}$ I got a quintic relating r to k.
I would say this is why you are given a specific value for k, so that you can obtain an approximate solution. I get r about 0.8.

Part b makes no sense, though. If there are no losses, there is no final position. Maybe they meant to ask for the displacement at the other end of the oscillation.

 

[Chestermiller]

I went ahead with my assumption that k is supposed to represent $\frac{m_0c^2}{P_0V_0}$ and took $\gamma=5/3$.
Solving the integral, and writing $r=\frac L{L_0}$ I got a quintic relating r to k.
I would say this is why you are given a specific value for k, so that you can obtain an approximate solution. I get r about 0.8.

Part b makes no sense, though. If there are no losses, there is no final position. Maybe they meant to ask for the displacement at the other end of the oscillation.

Maybe I made an error in my math, but, for part a, I got a value for r in the range 0.1 to 0.2.

Part b makes perfect sense to me. However, none of the suggestions offered so far is correct.

In part a, we were forced to make the approximation that, during this initial compression, the gas undergoes a reversible path. This assumes that viscous dissipation during this initial part of the oscillation is negligible. However, due to viscous dissipation, over the much longer time scale, the oscillation of the piston will be damped out and it will eventually come to rest. The final state of the gas will be one of thermodynamic equilibrium. And part b can be solved precisely, without invoking the approximation made in part a. In this final state, the pressure of the gas will again match the pressure of the surroundings.

I think I gave enough hints for now. OP, see if you can set up the first law equation that covers the overall transition between state 0 and state 2 (at the end of state b); whatever happened in reaching state a will not be significant in this part of the problem.

 

[haruspex]

due to viscous dissipation

Silly me. Thanks.

 

[haruspex]

Maybe I made an error in my math, but, for part a, I got a value for r in the range 0.1 to 0.2.

To be clear, I am using r for the remaining volume as a fraction of the initial volume. It is immediately clear this must exceed 1/k.
I did find a sign error in my working, though. I still get a value between 0.8 and 0.9, but now it is a rational root of the quintic(!). This is very persuasive, since it shows the 29/4 was carefully chosen.

 

[Chestermiller]

To be clear, I am using r for the remaining volume as a fraction of the initial volume. It is immediately clear this must exceed 1/k.
I did find a sign error in my working, though. I still get a value between 0.8 and 0.9, but now it is a rational root of the quintic(!). This is very persuasive, since it shows the 29/4 was carefully chosen.

I'll go back and check my arithmetic.

 

[Chestermiller]

I get $r=\frac{L}{L_0}=\frac{1}{8}$

 

[haruspex]

I get $r=\frac{L}{L_0}=\frac{1}{8}$

Ok, I found it. I had inadvertently switched which end I was measuring from.
So now we agree on the answer.
Also, where I wrote that r must exceed 1/k I should have written it must exceed $\frac 1{k+1}=\frac 4{33}$, which 1/8 does, just.

Not sure how the solver is expected to spot the rational root.

 

[songoku]

I think I gave enough hints for now. OP, see if you can set up the first law equation that covers the overall transition between state 0 and state 2 (at the end of state b); whatever happened in reaching state a will not be significant in this part of the problem.

State 0 is state of piston in question (a)?

If by first law equation you mean 1st law of thermodynamics, the formula is $\Delta U=Q+W$ where $W$ is the work done on the gas by the surroundings. Because this is adiabatic process, $Q = 0$ and $W$ will be the work done by the piston + work done by atmospheric pressure.

$$\Delta U=W_\text{piston}+W_\text{atmospheric}$$
$$\frac{3}{2}P_0 (V-V_0)=\frac{1}{2}m_0 c^2-P_0(V-V_0)$$

where $V$ is final volume of ideal gas when the piston stops moving.

Is this correct approach for question (b)?

Thanks

 

[Chestermiller]

State 0 is state of piston in question (a)?

If by first law equation you mean 1st law of thermodynamics, the formula is $\Delta U=Q+W$ where $W$ is the work done on the gas by the surroundings. Because this is adiabatic process, $Q = 0$ and $W$ will be the work done by the piston + work done by atmospheric pressure.

$$\Delta U=W_\text{piston}+W_\text{atmospheric}$$
$$\frac{3}{2}P_0 (V-V_0)=\frac{1}{2}m_0 c^2-P_0(V-V_0)$$

where $V$ is final volume of ideal gas when the piston stops moving.

Is this correct approach for question (b)?

Thanks

I'm not sure about your first equation, which, for the system consisting of gas plus piston, should read $\Delta U+\Delta K=-W$, but the rest is correct.

 

[songoku]

I'm not sure about your first equation, which, for the system consisting of gas plus piston, should read $\Delta U+\Delta K=-W$, but the rest is correct.

Ah sorry, the system I take is only the ideal gas and the surroundings are the piston and atmosphere.

I will try to solve the algebra, I will ask again if I find difficulties. For now, I will end this thread.

Thank you very much for all the help and explanation haruspex, Delta2, Chestermiller