# Thermodynamics problem - weight and piston connected by string

• Saitama
In summary: B%28105*8%29%2F%282%2F5%29*%281-%281%2B10%5E%28-2%29x%2F8%29%5E%28-2%2F5%29%29%3D10%5E5*10%5E%28-2%29*xI take ##P_{atm}\approx 10^5##The equation you have is correct. I ran into a similar problem when I tried to solve it using Wolfram Alpha. I think the issue is that you have
Saitama

## Homework Statement

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.)

## 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!

What simplifying assumptions could you make?

Pranav-Arora said:
confines a sample of air

Pranav-Arora said:
If I consider adiabatic, I need the adiabatic constant which is not mentioned in the problem statement.

Put these two quotes together.

1 person
DrClaude said:
Put these two quotes together.

Sorry, I can't really think of anything.

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

1 person
DrClaude said:
Can you figure out what the adiabatic constant for air could be?

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.

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

Pranav-Arora said:
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.
What is air mostly composed of?

1 person
DrClaude said:
What is air mostly composed of?

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##. :(

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.

1 person
DrClaude said:
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.

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?

Pranav-Arora said:
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?
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
Pranav-Arora said:
$$mgx+p_{atm}Ax=\frac{p_0V_0-P'(V_0+Ax)}{\gamma-1}$$
and
$$P'=\frac{p_0V_0^{\gamma}}{(V_0+Ax)^{\gamma}}$$

DrClaude said:
Could you clear up your thinking in writing

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.

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?

voko said:
Why is the work done by gravity negative?
Oops, very sorry.

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

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

voko said:
I do not see such a formula there.

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. :)

Clear indeed.

So what equation do you obtain in the end?

voko said:
Clear indeed.

So what equation do you obtain in the end?

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:
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?

voko said:
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?

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##.

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.

1 person
voko said:
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.

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!

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

1 person
Chestermiller said:
$$m\frac{d^2x}{dt^2}=mg-AP_0(1-\left(\frac{V_0}{V_0+Ax}\right)^γ)$$

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##. :(

Pranav-Arora said:
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.

It is not SHM per se. But for small ##x##, it is. Linearise the right-most term.

I am not sure why what you do for (c) is wrong. It looks correct but I will have another look later on. Perhaps some numeric errors again?

1 person
Pranav-Arora said:
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.

I didn't say it was SHM. I said it was analogous to SHM. As Voko implied, if you linearize about the equilibrium location x = 0.178 m, you get the differential equation for SHM. From this, you can actually determine the spring constant k for the cylinder/piston "spring." I recommend that you try it and see.

I confirmed your value for the equilibrium length, and, when I substituted this into your equation for the velocity, I got 1.27 m/s.

Incidentally, I assume there was a mistake with regard to the mass m in the problem statement. Apparently, m = 25 gm, not 25 kg.

Chet

Last edited:
Chestermiller said:
Oops. You're right. I made compensating errors with units to get the correct equilibrium location. I'll redo my calculations for the velocity.

Chet

I have used the same values given in the problem, what is wrong with my calculations?

Pranav-Arora said:
I have used the same values given in the problem, what is wrong with my calculations?
As I said in my previous edited posting, I redid my calculations, and obtained 0.178 m for the equilibrium position and 1.27 m/s for the velocity. This velocity agrees with your value.

Chet

1 person
Chestermiller said:
As I said in my previous edited posting, I redid my calculations, and obtained 0.178 m for the equilibrium position and 1.27 m/s for the velocity. This velocity agrees with your value.

Chet

Ah, then the given answer must be wrong.

Thanks for the help voko, Chestermiller and DrClaude!

## 1. How does the weight of the piston affect the pressure in the system?

The weight of the piston affects the pressure in the system by exerting a force on the gas inside the cylinder. As the weight increases, the force on the gas also increases, leading to an increase in pressure.

## 2. Can the weight of the piston be used to control the temperature of the system?

Yes, the weight of the piston can be used to control the temperature of the system. As the weight increases, the gas inside the cylinder is compressed, leading to an increase in temperature. Conversely, as the weight decreases, the gas expands and the temperature decreases.

## 3. What is the relationship between the weight of the piston and the volume of the gas?

The weight of the piston and the volume of the gas have an inverse relationship. As the weight increases, the gas is compressed and the volume decreases. As the weight decreases, the gas expands and the volume increases.

## 4. How does the length of the string connecting the weight and piston affect the system?

The length of the string has no direct effect on the system. However, it indirectly affects the system by determining the distance the weight can move and therefore the amount of work that can be done on the gas.

## 5. What is the purpose of using a weight and piston connected by a string in a thermodynamics problem?

The weight and piston connected by a string is a common setup used in thermodynamics problems to demonstrate the relationship between pressure, volume, and temperature in a gas. It allows for the manipulation of these variables through the use of the weight and string, making it a useful tool for understanding thermodynamic principles.

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