Is the Thermal/Mechanics Problem incoherent?

[jaumzaum]

Homework Statement

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Relevant Equations

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Hello guys. I am a Physics student and I think one of the questions I had to solve in an admission exam from a certain university is incoherent. Can you help me to understand if it is in fact incoherent or not, and if it needs annulment?

The question is:
Three moles of a certain ideal gas, whose molar heat at constant pressure is 5.00 cal / mol.K, is inside the cylinder in the figure below. The gas receives heat from a thermal source (not shown in the figure) in such a way that its temperature increases by 10.0 ° C. When absorbing heat, it is verified that the piston, adiabatic and of negligible mass, rises 2.00 meters. On the piston we have block 1 of mass m1 = 20.0 kg. Consider: g= 10 m/ s ^2 and 1.00 cal = 4.18 J.

a) Calculate the change in the internal energy (in joules) of the gas. (4 points)
b) At the end of the gas expansion, block 1 at rest on the horizontal surface AB, with negligible friction, is reached by block 2 with mass m2 = 10.0 kg and speed equal to 5.00 m / s. Calculate the recoil speed of block 2, knowing that the restitution coefficient is 0.800. (7 points)

I am concerned only with letter A.
Why I think it's wrong?
The question says the cylinder is adiabatic. I will also assume that the process ais reversible and that the transformation inside the cylinder is isobaric. If P is the pressure inside the cylinder:
$$W = P \Delta V = n R \Delta T = 3 \cdot 8,31 \cdot 10=249,3 J$$
$$Q = n C_P \Delta T = 3 \cdot 5 \cdot 4,18 \cdot 10 = 627 J$$
$$\Delta U = Q-W \approx 378 J$$
$$\gamma = Q/\Delta U \approx 5/3 $$
Typical of any monoatomic gas.

However, the work is also:
$$W=P \Delta V = (P_{atm } + mg/A) \Delta V = P_{atm} \Delta V + mgH = P_{atm} \Delta V + 400 = 249,3 J$$
$$P_{atm} \Delta V = -151,7 J$$

So this makes the question impossible, as we cannot have negative atmosferic pressure. The question didn't give the value of R either, nor the value of Pa. If we consider the process is not reversible, is it still possible that this would still be valid? I am not seeing any way this could be true.Can you guys help me? Thank you very much!

 

[Chestermiller]

The change in internal energy of an ideal gas, irrespective of process, is $$\Delta U=nC_v\Delta T$$. So n = 3, Cv=5.00-1.99=3.01, and $\Delta T=10 C$

 

[jaumzaum]

The change in internal energy of an ideal gas, irrespective of process, is $$\Delta U=nC_v\Delta T$$. So n = 3, Cv=5.00-1.99=3.01, and $\Delta T=10 C$

Thanks @Chestermiller. I know, and when I use that equation I also get 378 J. I just didn't use it because the exercise didn't give me Cv, or told that the gas was monoatomic. So I first calculated the work and then subtracted from Q. But both things are the same. The question is if that change of internal energy is compatible with the problem, given that the block with mass 20kg has moved 2m upwards and the cylinder is adiabatic. That would led to a negative pressure for me. And even though irreversible processes don't affect U, they can affect W and Q, so my question is if there can be some process in which the transformation could still be valid, or if the transformation is impossible considering all of the information the question gave above?

 

[Chestermiller]

Thanks @Chestermiller. I know, and when I use that equation I also get 378 J. I just didn't use it because the exercise didn't give me Cv, or told that the gas was monoatomic.

You didn't need either of these. All you needed to know was that $C_p-C_v=R$, and you knew both Cp and R

So I first calculated the work and then subtracted from Q. But both things are the same. The question is if that change of internal energy is compatible with the problem, given that the block with mass 20kg has moved 2m upwards and the cylinder is adiabatic.

The problem statement didn't say that the cylinder is adiabatic. It said that the piston is adiabatic.

That would led to a negative pressure for me. And even though irreversible processes don't affect U, they can affect W and Q, so my question is if there can be some process in which the transformation could still be valid, or if the transformation is impossible considering all of the information the question gave above?

Irrespective of whether the process is reversible or irreversible, the work in this problem is unique (assuming negligible piston friction), but they don't tell you the external pressure. Of course, the heat has to come out to $\Delta H$. You can use this in conjunction with the change in internal energy to get the work. Then you can then check to see whether there was any external pressure.

 

[Cutter Ketch]

The question says the cylinder is adiabatic.

No it doesn’t, unless it’s stated somewhere in the figure you didn’t show. In your problem statement all it says is that the piston is adiabatic meaning no heat is transferred to or from the piston. Obviously the system is not adiabatic. The first thing the problem mentions is that the gas is being heated.

 

[Chestermiller]

I see what you are saying. The problem statement is implying a negative (absolute) external pressure. That certainly is not physically realistic.

 

[jaumzaum]

@Chestermiller @Cutter Ketch Thanks very much!
Sorry for the "cylinder", I really meant "piston".

Now I get it, the variation of internal energy is 378 J no matter what. But W needs to be greater or equal to 400J, which leads to Q greater or equal to 778J. That said, the transformation needs to be irreversible. But it IS POSSIBLE.

Thank you very much!

 

[Chestermiller]

I agree with you; the problem statement is not self-consistent.

 

[Chestermiller]

Here is an irreversible scenario where everything comes out consistent:

1. The initial external pressure is zero (vacuum), rather than 1 atm
2. The initial height of gas in the cylinder is 38.66 m
3. The initial temperature is 300 K
4. Under these circumstances, the initial force exerted by the gas on the piston is 204.1 N (rather than 200 N for the mass), so that, initially, the piston is held down by a supplemental force of 4.1 N prior to being released.
5. The heating reservoir is at 310 K.

Try solving this problem to see how it plays out.