Thermodynamics: Horizontal Tube w/ Sliding Piston - Q&A

[C.E]

A thermally insulating horizontal tube of cross-sectional area A is filled with a
volume 2V of gas at a pressure p. At the centre of the tube is a freely sliding piston
of mass M which seals the left and right hand sides of the tube from each other (i.e.
at equilibrium there is a volume V either side of the piston).

The piston is given a small displacement x to the left and then released. Assuming
that friction has a negligible effect over many cycles:

(i) Show that the force F on the piston is given by:

F=$$\frac{-2xpA^2}{\gamma V}$$

$$\gamma$$= C_p/ C_v

ii) and comment on the nature of the motion. {1}

(iii) Derive the expression for $$\omega$$ the frequency of oscillation.

3. I don't know how to start this, as the gas is not Ideal I have no idea what formulas to apply to it. I know that Q=0 and think the work done =0. Is this right? Can somebody please help me get started?

 

[Mapes]

If the piston is moved through a displacement against a resisting pressure, I wouldn't say that the work done is zero. But I agree that Q = 0 is a reasonable assumption. What would we call such a process, and what are the applicable equations?

 

[C.E]

Is the process is adiabatic? I have only been taught equations which apply to ideal gasses in adiabatic processes, are there others? Oh, I have learn't that change in internal energy= - work done though, I don't see how that is much help. Do I need to assume the gas is ideal?

 

[Mapes]

Yes, assume an adiabatic process on an ideal gas.

 

[C.E]

I have been shown the following.

$$pv^{\gamma}=constant$$ - 1

$$pT^{\gamma -1} =constant$$ - 2

Work=n$$C_v(T_1-T_2)$$ -3

and

Work= $$\frac{C_v}{R}(P_1V_1-P_2V_2)$$ - 4

I don't think 2, 3, 4 are relevant as we do not know anything about the temperature/ the temperature change, is this right? I have tried playing around with 1 substituting it into the following. Force = A(P1-p2) (where P1 is the pressure in the left side of the tube and P2 is the pressure in the right hand side) but I did not get anywhere, what do you suggest I try now?

 

[Mapes]

How would you approximate

$$(V-xA)^\gamma$$

when $x$ is small?

 

[C.E]

I would do the following, is this what you are thinking of?

$$(v-Ax)^{\gamma} = v^{\gamma}(1-\frac{Ax}{v})^{\gamma}$$

Therefore $$(v-Ax)^{\gamma}$$ can be approximated by the following:

$$v^{\gamma}(1 + \gamma \times A \times x/v)$$

=$$v^{\gamma}$$ + gamma $$v^{\gamma}$$Ax/v

 

[Mapes]

You need to get the $\gamma$ as a multiplicative factor rather than an exponent. Are you familiar with expanding a function as a Taylor series?

(Hint: A Taylor series would give $(v-Ax)^{\gamma}\approx v^\gamma-Ax\gamma$ above.)

 

[C.E]

I have seen Taylor series but not in much detail. Which point are you expanding the Taylor series about?

 

[Mapes]

V ("small displacement").

 

[C.E]

Is this the way to progress with the question?

$$\ (v-Ax)^{\gamma}\approx v^\gamma-Ax\gamma$$

and $$\ (v+Ax)^{\gamma}\approx v^\gamma+Ax\gamma$$

Therefore v2-v1 $$\approx$$ -2Ax$$\gamma$$

Is there a way to directly relate p1 to v1 and p2 to v2 given that I don't know the temperature or the number of moles of gas?

 

[Mapes]

Yes, your equation (1) above.

 

[C.E]

Using (1) I get that:

p1v1^(gamma)=p2v2^(gamma)

How can I use this?

 

[Mapes]

Plug in the volume after a small displacement. Use Taylor series approximation. What is the new pressure? How is this related to the force?

 

[C.E]

Am I plugging in the volume of the left or right hand side of the cylinder after the displacement?
force= pressure x area in this case A x pressure.

 

[Mapes]

Am I plugging in the volume of the left or right hand side of the cylinder after the displacement?
force= pressure x area in this case A x pressure.

Generally you would calculate the force related to the left side, calculate the force related to the right side, and add them together. However, you may notice that the system is symmetric for small displacements (i.e., the forces are equal from both sides and additive), which simplifies things.

 

[C.E]

for the left hand side I get:

p1= $$\ v^{\gamma}$$p/$$\ ( v^{\gamma}-Ax\gamma$$)


for the right hand side I get:

p2= $$\ v^{\gamma}$$p/$$\ (v^{\gamma}+Ax\gamma$$)


Is this right? Do I just add these and multiply by A?

 

[Mapes]

It would be a little easier if the equations were arranged so that the two-term parts are in the numerator rather than the denominator. Adding them as they are doesn't really get you anywhere, I don't think.

 

[C.E]

Adding the two equations I get:
p1($$\ v^{\gamma}-Ax \gamma$$) + p2($$\ v^{\gamma}+Ax \gamma$$)=2p$$v^{\gamma}$$
Is this right? How do I show this is equivalent to the expression given in the question?

 

[Mapes]

I recommend solving for p1 and p2 in a way that has the two-term parts in the numerators, rather than the denominators. This will make it easier to find their difference.

 

[C.E]

Sorry, I don't know what you mean by "the two-term parts in the numerators, rather than the denominators". What do you mean? How do I go about solving in this way?

 

[Mapes]

Instead of expressing v1 as v-Ax, how about expressing v as v1+Ax? I think this will be easier. (I realize that this is different from my earlier comment, but I hadn't worked the entire problem yet. I don't have the answer in front of me either!)

 

[C.E]

I still can't get it to work, any more hints?

 

[Mapes]

Use $p_1v_1^\gamma=pv^\gamma$, expressing $v$ in terms of $v_1$. Perform a Taylor series expansion to remove $\gamma$ as an exponent. Do the same thing for $p_2$. Determine the force from the difference between the two pressures.

 

[C.E]

This is as far as I keep getting. Is this right? What do I do now?

p1$$\ v1^{\gamma$$ = p$$\ v^{\gamma$$

By the Taylor approximation,

p1$$\ v1^{\gamma$$ + p1Ax$$\gamma$$=p$$\ v1^{\gamma}$$

so p1=p+pAx$$\frac{gamma}{v^{\gamma}}$$

Similarly

p$$\ v^{\gamma$$=p2$$\ v2^{\gamma$$=p$$\ (v2-Ax)^{\gamma$$

By the Taylor approximation.

p$$\ v2^{\gamma$$ - pAx$$\gamma$$=p2$$\ v2^{\gamma}}$$

so p2= p-pAx$$\frac{\gamma}{v^{\gamma}}$$

Hence P2-p1= $$\frac{-\gamma2pAx}{v^{\gamma}}$$

 

[Mapes]

Looks good. Is there a typo in the answer given in your first post? I believe $\gamma$ should be in the numerator, as you have it here.

 

[C.E]

No, the question given in my first post is exactly as it appeared on the sheet. Do you think the question is wrong? (My denominator is also different to theirs as I have v^gamma and they just have v x gamma).

For part b is the motion simple harmonic and for c is $$\omega$$=sqrt(the coefficient of x found in part a)?

 

[Mapes]

No, the question given in my first post is exactly as it appeared on the sheet. Do you think the question is wrong? (My denominator is also different to theirs as I have v^gamma and they just have v x gamma).

I think the answer should have a $\gamma$ in the numerator rather than the denominator. But your $v^\gamma$ should just be $v$; check your Taylor series calculations again.

 

[C.E]

Are the following Taylor approximations correct?

p$$\ v2^{\gamma$$ - pAx$$\gamma$$=p2$$\ v2^{\gamma}}$$

and p1$$\ v1^{\gamma$$ + p1Ax$$\gamma$$=p$$\ v1^{\gamma}$$

 

[Mapes]

I don't see how you're getting that; take a look at the Wikipedia article on Taylor series expansions, perhaps. The goal is to make $\gamma$ a multiplicative rather than an exponential factor (as we discussed in our posts #8-11).

 

[C.E]

I got if from when you said "$(v-Ax)^{\gamma}\approx v^\gamma-Ax\gamma$" in post 8.

 

[Mapes]

Ah, that was my own typo. Sorry about that. So you should now have

$$F=-kx=-\frac{2\gamma p A^2}{V}x$$

which doesn't quite match the given answer, but I suspect we're right and the given answer wrong. Please update if you find another solution.

 

[C.E]

I have read about Taylor expansions on wikipedia and am still a little confused. What are the Taylor expansions I should be using for (v2-Ax)^gamma and (v1+Ax)^gamma? How did you find them?

 

[Mapes]

The idea here is that

$$f(a+b)\approx f(a)+b\,f^\prime(a)$$

for small b. In this case

$$f(a)=a^\gamma$$

So

$$f(a+b)\approx a^\gamma+\gamma b a^{\gamma-1}=a^\gamma\left(1+\frac{\gamma b}{a}\right)$$

as you figured out from your knowledge that [itex](1+b)^\gamma\approx 1+\gamma b[/tex] (and I messed up).