Determining the mass of a free to move piston

[1729]

Homework Statement

A vertical cylinder contains 50.0 mg of hydrogen gas. A free to move piston with area 27.0 cm² is pressing down on the gas. The gas's temperature is equal to 25.0 degrees Celsius.
A 30 degrees Celsius increase in temperature occurs and the piston is elevated by 2.00 cm.
The molar mass of hydrogen gas is 2.02 g/mol.

The mass of the piston is equal to:
(a) 1.2 kg
(b) 1.8 kg
(c) 3.6 kg (correct answer)
(d) 4.0 kg

(source: Vlaamse Fysica Olympiade 2015, 1st Round
https://gyazo.com/f7c960fbef62943fb0432bcffd2468b8)

Homework Equations

pV=nRT (ideal gas law)
V=Ah (substitution)
p=F/A (substitution)

The Attempt at a Solution

Notice that the piston is free to move (isobaric process). The pressure exerted on the piston by the gas will remain constant during the experiment.
Suppose hydrogen gas is an ideal gas. We will devise a system of equation using the ideal gas law by considering the two situations.

Situation 1: $pV=nRT \Rightarrow Fh=\frac{mRT}{M}$
Situation 2: $pV=nRT \Rightarrow F(h+0.0200 \mathrm{\ m})=\frac{mR(T+30\mathrm{\ K})}{M}$
(note: T=298K, not 25 deg C)

By regarding the two situations as a system of equations, we obtain the solutions:
$$F=308.54 \mathrm{\ N} \land h=0.20 \mathrm{\ m}$$ There are two major forces which cause the piston to be at rest in both situations: gravity and the pressure exerted on the piston. By Newton's second law, the sum of the two forces is equal to zero Newton.
$$m=\frac{308.54 \mathrm{\ N}}{9.81 \mathrm{\frac{m}{s^2}}}=31.45 \mathrm{\ kg}$$ However, this is not a possible answer. What's wrong?

 

[JBA]

p = F/A. You have left the area A out of your equations.

 

[1729]

Hi, this is the result of two overlapping substitutions. Consider the following:
$$pV=\frac{F}{A}Ah=\frac{FAh}{A}=Fh$$

 

[gneill]

Hint: You're equations are dealing with absolute pressures inside the cylinder. But the situation is immersed in a background pressure...

What's the real pressure delivered downward by the piston?

 

[JBA]

Excellent point, V = h * A, sometimes I am right and then sometimes I simply miss the train.

However, in this case I have managed to catch it again: In you used h+.0200 m, now look at the next equation, there you used .20 m. if you correct that to .020 m then your answer is 3.145 kg. which is not the exact answer but gets you back within the same order of magnitude

 

[1729]

Hint: You're equations are dealing with absolute pressures inside the cylinder. But the situation is immersed in a background pressure...

What's the real pressure delivered downward by the piston?

Thanks for the tip. The atmospheric pressure (101,000 Pa) is delivered over an area of 0.0027 m².
How should I continue from this point? I've tried a similar solution before. The ideal gas law seems like a dead end for this to me and Charles's law is out of the question too since it would only give me the height.

Excellent point, V = h * A, sometimes I am right and then sometimes I simply miss the train.

However, in this case I have managed to catch it again: In you used h+.0200 m, now look at the next equation, there you used .20 m. if you correct that to .020 m then your answer is 3.145 kg. which is not the exact answer but gets you back within the same order of magnitude

That's correct. I've been trying to solve this problem for a few days now and I ended up with 3.145 kg last time I gave it a shot.

 

[gneill]

Thanks for the tip. The atmospheric pressure (101,000 Pa) is delivered over an area of 0.0027 m2.
How should I continue from this point? I've tried a similar solution before. The ideal gas law seems like a dead end for this to me and Charles's law is out of the question too since it would only give me the height.

You should be able to write an expression for the pressure involving the mass and the 1 atm. Then PV = nRT is the way to go. Try to devise a single equation that involves ΔV and ΔT. ΔT can be trivially found from the given values, and you should be able to do something with the change in height Δh for the ΔV term...

 

[Nathanael]

That's correct. I've been trying to solve this problem for a few days now and I ended up with 3.145 kg last time I gave it a shot.

The 0.2m was just a typo, you did the calculations correct in your OP.

How should I continue from this point?

Your work in the OP is good, and you are almost finished. (You've already done most of what Gneill is suggesting.) The last step is to consider that the force F you calculated is the weight of the piston plus the weight of the atmosphere above the piston.

 

[1729]

Your work in the OP is good, and you are almost finished. (You've already done most of what Gneill is suggesting.) The last step is to consider that the force F you calculated is the weight of the piston plus the weight of the atmosphere above the piston.

Thank you for your reply -- this is a very satisfying explanation! I hadn't thought of the atmospheric pressure exerting a force on the piston as well.
Just for the record: 308.54 N - 1.01*10^5 Pa * 0.0027 m² = mg --> m = 3.65 kg (I'm wondering because of the slight difference with the answer)

You should be able to write an expression for the pressure involving the mass and the 1 atm. Then PV = nRT is the way to go. Try to devise a single equation that involves ΔV and ΔT. ΔT can be trivially found from the given values, and you should be able to do something with the change in height Δh for the ΔV term...

Can you expand on this please? I don't think I fully understand what you're trying to tell me. I've thought about using $\frac{p_1V_1}{T_2}=\frac{p_2V_2}{T_2}$ which unfortunately reduces to Charles's law. ΔV is dependent of Δh and independent of ΔA, so I understand that part of your hint.

 

[gneill]

Can you expand on this please? I don't think I fully understand what you're trying to tell me. I've thought about using $\frac{p_1V_1}{T_2}=\frac{p_2V_2}{T_2}$ which unfortunately reduces to Charles's law. ΔV is dependent of Δh and independent of ΔA, so I understand that part of your hint.

You're looking to write things in terns of Δ quantities, so don't go the ratio route. Instead, write the gas law equation for the initial and final states, then subtract one from the other to evolve the Δ quantities for temperature and volume.

{You could also take advantage of the time saving shortcut afforded by the fact that for a linear equation like y = mx, you can also write Δy = mΔx directly. The ideal gas law equation with one variable held constant becomes such a linear equation }

 

[Nathanael]

Just for the record: 308.54 N - 1.01*10^5 Pa * 0.0027 m² = mg --> m = 3.65 kg (I'm wondering because of the slight difference with the answer)

If you use a slightly more accurate value for atmospheric pressure, 1.013*10^5 Pa, then the slight difference disappears (I get 3.57 kg).

 

[1729]

You're looking to write things in terns of Δ quantities, so don't go the ratio route. Instead, write the gas law equation for the initial and final states, then subtract one from the other to evolve the Δ quantities for temperature and volume.

{You could also take advantage of the time saving shortcut afforded by the fact that for a linear equation like y = mx, you can also write Δy = mΔx directly. The ideal gas law equation with one variable held constant becomes such a linear equation }

That's a very interesting way to think about it, it's new to me. It's a lot faster than what I did too! I have a few questions about this method that I'll ask tomorrow hopefully (it's 2.30am right now).
Just wanted to let you know I figured it out and it worked.

If you use a slightly more accurate value for atmospheric pressure, 1.013*10^5 Pa, then the slight difference disappears (I get 3.57 kg).

Thanks!