# Ideal gas law problem -- Pneumatic piston movement with air temperature changes

• JennyLee1989

#### JennyLee1989

Homework Statement
7) Figure 15.5 shows a 50 kg lead cylindrical piston which floats on 0.37 mol of compressed ideal air at 30°C. How far does the piston move if the temperature is increased to 300°C ?
Relevant Equations
pV=nRT
A=pi* r^2 I have come up with the change in height as 170 cm. My professor does not want to solve for the problem for a reason I do not understand. 170 cm is not part of the answer key. The answer according to the answer key is 65 cm.
My attempt is:

Initial temperature:
p=F/A; (50 *9.8) / (pi * 0.05^2); p = 62389 Pa;
pv1=nRT1; v1= (nRT1)/p; v1=(0.37*8.31*303)/62389; v1=0.01493m^3;
v1=h1*A; h1= 1.9m;

Final temperature:
pv2=nRT2; v2=(nRT2)/p;
h2*A=(nRT2)/p; h2 = (nRT2)/pA;
h2=(0.37*8.31*573)/(62389*0.00785); h2 = 3.6 m;

change in h = h2-h1; 3.6 - 1.9 = 1.7m= 170 cm.

Last edited by a moderator:
• Delta2

I would give zero credit for your answer. I give partial credit only until numbers go in. You've written a jungle of numbers with implausible precision. I would need a machete to hack through it.

Define your variables and use symbols in the calculation. Don't expect us to understand a lack of formatting: does T2 mean T2 or T2? (Or better still $T_2$ or $T^2$.) Put your numbers in only at the last step.

• Delta2
I checked the numbers and they look OK to me.

I tried again and found the homework statement implicitly assumes the system is surrounded by a standard atmosphere. With this assumption I found the height increases 65 cm

Last edited:
p=F/A; (50 *9.8) / (pi * 0.05^2); p = 62389 Pa;
You have used the wrong pressure.

In equilibrium:
(gas pressure) = (atmospheric pressure) + (pressure due to the lead weight).
The gas pressure is the same before and after heating because the piston can slide and always has atmospheric pressure acting on it top surface.

Your working/presentation is too hard to follow. I don't know if there are other mistakes. My strategy would be this:

1. Calculate (correctly) the intitial gas pressure as above. (Atmospheric pressure is 1.10x10⁵ Pa.)

2. Using PV = nRT, calculate the initial volume, V₁ and hence the initial height (h₁).

3. Since pressure and the amount (n moles) of gas are both constant, PV = nRT gives$$\frac{V_2}{V_1} = \frac{T_2}{T_1}$$ (work out this value, call it f (for 'factor'))

4. V is proportional to h, so h₂ = fh₁ so you can find h₂. Then you find h₂ – h₁.

• Delta2 and Lnewqban
I would actually recommend doing this in more or less the reverse order. It's not immediately obvious that you need to know h1 explicitly. But it will become so as one continues through the steps.

I would give zero credit for your answer. I give partial credit only until numbers go in. You've written a jungle of numbers with implausible precision. I would need a machete to hack through it.

Define your variables and use symbols in the calculation. Don't expect us to understand a lack of formatting: does T2 mean T2 or T2? (Or better still $T_2$ or $T^2$.) Put your numbers in only at the last step.
This was very discouraging. I'm just getting back into school. No need to insult my work please.

• Delta2
I would give zero credit for your answer. I give partial credit only until numbers go in. You've written a jungle of numbers with implausible precision. I would need a machete to hack through it.

Define your variables and use symbols in the calculation. Don't expect us to understand a lack of formatting: does T2 mean T2 or T2? (Or better still $T_2$ or $T^2$.) Put your numbers in only at the last step.
I think you are abit too harsh and strict on OP. Though indeed we should encourage students to learn to use ##\LaTeX## I don't think you encourage the right way.

The post isn't that bad even without##\LaTeX##, i think we all can understand from the context that when she writes T2 she means ##T_2## e.t.c

You have used the wrong pressure.

In equilibrium:
(gas pressure) = (atmospheric pressure) + (pressure due to the lead weight).
The gas pressure is the same before and after heating because the piston can slide and always has atmospheric pressure acting on it top surface.

Your working/presentation is too hard to follow. I don't know if there are other mistakes. My strategy would be this:

1. Calculate (correctly) the intitial gas pressure as above. (Atmospheric pressure is 1.10x10⁵ Pa.)

2. Using PV = nRT, calculate the initial volume, V₁ and hence the initial height (h₁).

3. Since pressure and the amount (n moles) of gas are both constant, PV = nRT gives$$\frac{V_2}{V_1} = \frac{T_2}{T_1}$$ (work out this value, call it f (for 'factor'))

4. V is proportional to h, so h₂ = fh₁ so you can find h₂. Then you find h₂ – h₁.

I should have been more clear with my work but what you've written has been helpful.
I came up with 61cm as my answer which I guess is close enough. I am wondering, where you wrote "pressure due to the lead weight," is that the pressure equation that I had above where I used Pressure = Force/Area?

I think you are abit too harsh and strict on OP. Though indeed we should encourage students to learn to use ##\LaTeX## I don't think you encourage the right way.

The post isn't that bad even without##\ LaTeX##, i think we all can understand from the context that when she writes T2 she means ##T_2## e.t.c
Thank you! Yes, I don't know what LaTeX is and I appreciate all the help I can get.

• Delta2
Thank you! Yes, I don't know what LaTeX is and I appreciate all the help I can get.
You are welcome!
Check the LaTeX Guide. At the bottom left corner of the text box where you write your message there should be the link to the LaTeX guide.

• JennyLee1989

However, it is important that you understand the importance of communication in homework, and clear thinking in problem solving. Working with numbers rather than symbols obscures rather than clarifies, and where a grader might understand that you dropped a sign and meant (2a - b) when you write (2a + b), there is no way she can be expected to figure out that 0.38765 really is supposed to be 0.91123.

This is especially important in trying to figure out a mistake. Clarity is even more important here.

• Delta2
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