# Homework Help: Thermodynamics problem - weight and piston connected by string

1. Mar 6, 2014

### Saitama

1. The problem statement, all variables and given/known data
An easily moveable, thermally insulated piston of negligible mass confines a sample of air of volume $V_0=8\,\, \text{litre}$, at a temperature of $T_0=300 K$, into a thermally insulated cylinder of cross section $A=1\,\,dm^2$. The cylinder is fixed to a horizontal table. The pressure of the confined air is the same as the atmospheric pressure $p_0=105 \,\,Pa$. A piece of thread is attached to the centre of the piston, it lies in the line of the axis of the cylinder and leads through a pulley. A weight of mass $m=25\,\, kg$ is attached to the other end of the thread as shown in the figure. Holding the weight, any part of the thread is straight, but loose.
The weight is released without any push. To an accuracy of 1% determine the

a) temperature of the air in the cylinder, when the weight is at its lowermost position;

b) the acceleration of the piston at the lowermost position of the weight;

c) the greatest speed of the piston during the process. (The mass of the thread and the pulley and friction are negligible.)

2. Relevant equations

3. The attempt at a solution
I am thinking of using conservation of energy but I am not sure which expansion process should I consider. If I consider adiabatic, I need the adiabatic constant which is not mentioned in the problem statement. I can't think of any other process which takes place in a thermally insulated cylinder.

Any help is appreciated. Thanks!

2. Mar 6, 2014

### NoPoke

What simplifying assumptions could you make?

3. Mar 6, 2014

### Staff: Mentor

Put these two quotes together.

4. Mar 6, 2014

### Saitama

Sorry, I can't really think of anything.

5. Mar 6, 2014

### Staff: Mentor

Can you figure out what the adiabatic constant for air could be?

6. Mar 6, 2014

### Saitama

The question doesn't state the nature of gas (air) i.e is it monoatomic, diatomic or something else? I don't see how to find the adiabatic constant without this information.

7. Mar 6, 2014

### NoPoke

What gases will you find in air?
What is their composition (monatomic , diatomic, etc)?
If one type is predominant then what?

8. Mar 6, 2014

### Staff: Mentor

What is air mostly composed of?

9. Mar 6, 2014

### Saitama

Okay, I took it as diatomic so $\gamma=7/5$ (adiabatic constant). At the lowermost position of block, the kinetic energy of block is zero, so from conservation of energy:
$$mgx+p_{atm}Ax=\frac{p_0V_0-P'(V_0+Ax)}{\gamma-1}$$
and
$$P'=\frac{p_0V_0^{\gamma}}{(V_0+Ax)^{\gamma}}$$
where $p_{atm}$ is the pressure due to air outside and x is the displacement of block. Solving the above gives me a negative value for $x$. :(

10. Mar 6, 2014

### Staff: Mentor

I'm not sre where you are going with this. I would start by considering the final state, when the piston is at equilibrium again, in terms of the forces acting on it.

11. Mar 6, 2014

### Saitama

Why equilibrium? How does it justify that at equilibrium of piston, the block is at lowermost position? What's wrong with my equation for conservation of energy?

12. Mar 6, 2014

### Staff: Mentor

I'm sorry if I'm confusing you. I was neglecting the inertia of the block, which might not be a good approximation here.

Could you clear up your thinking in writing

13. Mar 6, 2014

### Saitama

Sure! :)

I set x=0 and t=0 when the string tightens. Let the displacement of block when it is at lowermost position be $x$. Since there is no change in kinetic energy, the sum of work done by gravity on block, work done in expansion by confined and the work done by the air outside on the piston is zero i.e
$$-mgx+\frac{p_0V_0-P'(V_0+Ax)}{\gamma-1}-p_{atm}x=0$$
which is the same equation I wrote above, please let me know if anything is still unclear.

14. Mar 6, 2014

### voko

Why is the work done by gravity negative?

And frankly, I am not convinced your middle term is correct. There are many ways to express work in adiabatic expansion, but I do not think this is one of them. Where does that come from?

15. Mar 6, 2014

### Saitama

Oops, very sorry.

But I still don't get the answer. :(

I used the following formula:
$$W=\frac{P_iV_i-P_fV_f}{\gamma-1}$$

16. Mar 6, 2014

### voko

17. Mar 6, 2014

### Saitama

Okay let me explain. The formula mentioned on that page is:
$$W=\frac{K(V_f^{1-\gamma}-V_i^{1-\gamma})}{1-\gamma}$$
Factoring out -1 from the denominator:
$$W=\frac{K(V_i^{1-\gamma}-V_f^{1-\gamma})}{\gamma-1}$$
Also, $K=PV^{\gamma}$, hence, $KV_f^{1-\gamma}=P_fV_f\cdot V_f^{1-\gamma}=P_fV_f$, similarly, $KV_i^{1-\gamma}=P_iV_i$. I hope everything is clear now. :)

18. Mar 6, 2014

### voko

Clear indeed.

So what equation do you obtain in the end?

19. Mar 6, 2014

### Saitama

This is what I have:
$$mgx+\frac{p_0V_0-P'(V_0+Ax)}{\gamma-1}-p_{atm}x=0$$
where
$$P'=\frac{p_0V_0^{\gamma}}{(V_0+Ax)^{\gamma}}$$
I simplify the middle term in the energy equation:
\begin{align*} \frac{p_0V_0-P'(V_0+Ax)}{\gamma-1} & = \frac{1}{\gamma-1}\left(p_0V_0-\frac{p_0V_0^{\gamma}}{(V_0+Ax)^{\gamma}}(V_0+Ax)\right)\\ & = \frac{p_0V_0}{\gamma-1}\left(1-\left(1+\frac{Ax}{V_0}\right)^{1-\gamma}\right)\\ \end{align*}
So the energy equation becomes:
$$mgx+\frac{p_0V_0}{\gamma-1}\left(1-\left(1+\frac{Ax}{V_0}\right)^{1-\gamma}\right)=p_{atm}Ax$$
Plugging this in Wolfram Alpha doesn't give me the answer. :(

http://www.wolframalpha.com/input/?i=25*9.8*x+(105*8)/(2/5)*(1-(1+10^(-2)x/8)^(-2/5))=10^5*10^(-2)*x

I take $P_{atm}\approx 10^5 \,\,Pa$.

Last edited: Mar 6, 2014
20. Mar 6, 2014

### voko

There are a couple of errors. In one case, you use 105 for pressure. More seriously, you use 8 for the volume - is it really 8 cubic metres?

21. Mar 6, 2014

### Saitama

Ah, I am very sorry, I made the correction but I still don't get the right answer.

I replaced 8 litres with 0.008 $m^3$. http://www.wolframalpha.com/input/?...*(1-(1+10^(-2)x/0.008)^(-2/5))=10^5*10^(-2)*x

Is there something wrong with the pressure? It is given in the problem statement that $p_0$ is $105\,\,Pa$.

22. Mar 6, 2014

### voko

The problem states that the initial pressure inside the cylinder is equal to atmospheric. So you have to use the same number for both at the very least. And 105 Pa is probably an error, where 10^5 Pa was meant.

23. Mar 6, 2014

### Saitama

Sorry for doing so many mistakes, and the typo in the problem statement is due to me, I should have checked again.

I will be attempting the other two parts tomorrow, I need to leave now. Thanks for all the help voko!

24. Mar 6, 2014

### Staff: Mentor

This is basically going to be analogous to a SHM problem, except that the spring (in this case, the gas in the cylinder) is nonlinear. If T is the tension in the string, then a force balance on the piston gives:

T = (P0-P)A, where P is the pressure in the cylinder, and P0=Patm=105 Pa.

A force balance on the mass gives:

$$mg-T=m\frac{d^2x}{dt^2}$$

The gas is expanded and compressed adiabatically, so

PVγ=P0V0γ

or, equivalently,
$$P=P_0\left(\frac{V_0}{V_0+Ax}\right)^γ$$
Combining the previous equations gives:
$$m\frac{d^2x}{dt^2}=mg-AP_0(1-\left(\frac{V_0}{V_0+Ax}\right)^γ)$$
If we multiply this equation by v = dx/dt and integrate from 0 to x, we obtain:
$$m\frac{v^2}{2}=mgx-AP_0x+\frac{P_0V_0}{γ-1}\left(1-\left(\frac{V_0}{V_0+Ax}\right)^{γ-1}\right)$$

This confirms Panov-Arora's result for the case in which v = 0. This corresponds to the lowest point reached by the mass. The point of maximum velocity is where the acceleration= 0 (this is also the equilibrium point). This is where:
$$AP_0(1-\left(\frac{V_0}{V_0+Ax}\right)^γ)=mg$$

Chet

25. Mar 6, 2014

### Saitama

I got the exact same equation when I tried the problem with Newton's second law but I fail to see how this is SHM. Please help.

For the b) part, I solved the problem using Newton laws. And the equation I used is similar to the Chestermiller.

For the c) part, I used energy conservation. But first, I found $x$ from the Newton's second law. The velocity is maximum at the equilibrium point. From newton's second law:
$$m\frac{d^2x}{dt^2}=mg-Ap_0\left(1-\left(1+\frac{Ax}{V_0}\right)^{-\gamma}\right)=0$$
Solving for $x$ gives $x=0.177847\,\,m$.
From energy conservation,
$$mgx+\frac{p_0V_0}{\gamma-1}\left(1-\left(1+\frac{Ax}{V_0}\right)^{1-\gamma}\right)-p_{atm}Ax=\frac{1}{2}mv^2$$
But solving for $v$ doesn't give me the right answer.

http://www.wolframalpha.com/input/?...2/5))-10^5*10^(-2)*x=1/2+*+25*y^2,+x=0.177847

The given answer is $0.4\,\,m/s$ but I get $1.2\,\, m/s$. :(