# Thermodynamics problem involving piston and spring

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1. Feb 8, 2016

### Titan97

1. The problem statement, all variables and given/known data

2. Relevant equations

$$W=P\Delta V$$
$$\Delta U=nC_v\Delta T$$

3. The attempt at a solution.
The gas is slowly heated. The temperature increases and the pressure increases as well. But since volume increases, the increases in pressure is nullified. The process is isobaric.
Only option (B) is correct since $$\Delta U=n\frac{R}{\gamma-1}(3T_1-T_1)=3nRT_1=3P_1V_1$$

But the question has multiple answer. Those who answered (B) and those who answered (A),(B),(C) got marks. How is (A),(B),(C) the correct answer?

2. Feb 8, 2016

### Grinkle

I don't agree with the above reasoning. Write down the force the spring exerts on the surfaces it is attached to as a function of the deflection of the spring. Consider how the situation of the gas would be different if the lid were not attached to a spring at all and in addition if the lid were massless or we are saying there is no gravity involved.

3. Feb 8, 2016

### Staff: Mentor

Actually, this has nothing to do with the internal energy. It is strictly related to mechanics and the ideal gas law. If k is the spring constant, P is the pressure at any instant during the process, and A is the area of the piston, and the piston is essentially at equilibrium throughout the process, what is the relationship between k, x, P, and A?

Chet

4. Feb 8, 2016

### Titan97

@Chestermiller why can't I say that the process is isobaric? I got the answer by assuming the process is isobaric.

5. Feb 8, 2016

### Grinkle

If you write an equation that balances the forces on the lid as the deflection of the spring increases, you will derive the answer to whether the process is isobaric or not.

You are correct that thinking the process is isobaric is the crux of your confusion.

it may help to think if yourself in the can pushing up against the lid. As the spring deflects, is the force against your hands constant, or does it increase?

6. Feb 8, 2016

### Staff: Mentor

Fortuitous, I guess. To compress the spring, the pressure of the gas has to increase. Incidentally, this question is a little hokey because, the spring can't be in its relaxed state to begin with. Otherwise, there would be an unbalanced force on the piston initially. So, assume initially that the force of the spring initially just balances the force of the gas on the other side of the piston.

7. Feb 8, 2016

### Staff: Mentor

I can tell you what happened. They marked everybody correct, irrespective of their answers, because of the glitch I pointed out in the problem statement. So your answer wasn't really correct after all. Everyone was given credit even if they got the answer wrong. If this problem is done correctly, the pressure varies with the compression x of the spring from its unextended length as follows: $PA-P_1A=k(x-x_1)$, where x1 is the initial compression of the spring. Starting from this relation, the real correct answer to this problem is C.

8. Feb 8, 2016

### Titan97

@Chestermiller the question is a "more than one correct type". So you can have multiple correct answers. So the combinations that were given marks: B and A,B,C

Why $PA-P_1$? Are you considering pressure on both side of piston?

9. Feb 8, 2016

### Staff: Mentor

No. The assumption is that there is negligible pressure on the spring side of the piston.

Initially, before any heat is added, a force balance on the piston gives:
$$P_1A=kx_1$$
During the heating process, the pressure on the piston rises to P and the compression of the spring rises to x. So, at any time during the heating, a force balance on the piston gives:
$$PA=kx$$
If we subtract the first equation from the second, we obtain:
$$PA=P_1A+k(x-x_1)$$or, equivalently,
$$P=P_1+\frac{k}{A}(x-x_1)$$
Do you follow this so far?

10. Feb 9, 2016

### Titan97

Yes, I do. Why are expressing P in terms of $x-x_1$ when you already have P in terms of $x$?
Using work energy theorem, $$\int P(x)Adx-\int kxdx=0$$ assuming that $\Delta\text{KE}=0$.

11. Feb 9, 2016

### Staff: Mentor

Because all the choices for answers involve P's and V's, and not x's, k, and A. So we have to figure out a way of eliminating the x's, k, and A. Since in our problem, the piston moves to the right a distance $x-x_1$, the volume the piston sweeps out is related to the change in volume of the gas $V-V_1$. If A is the area of the piston, how is $V-V_1$ related to $x-x_1$? If we substitute this relationship into the equation $$P=P_1+\frac{k}{A}(x-x_1)$$what do we get?

Chet

12. Feb 9, 2016

### Titan97

Initial volume is $V_1=AL_1$ of L1 is the initial length of cylinder containing gas.
$V=AL_2$
$$V-V_1=A(L-L_1)=A\Delta L=A(x-x_1)$$
What about the work energy relation in post #10?

13. Feb 9, 2016

### Staff: Mentor

That relationship is correct. So the energy stored in the spring is equal to the work done by the gas.
If we substitute this equation into the equation $P=P_1+\frac{k}{A}(x-x_1)$, we get:
$$P=P_1+\frac{k}{A^2}(V-V_1)$$
According to this equation, because of the constraint imposed by the spring, the pressure of the gas during our process will be a linear function of the gas volume. Since this equation must also pass through the point (V2,P2), we can use this fact to determine k/A2 in terms of P1, V1, P2, and V2. What is this relationship?

14. Feb 9, 2016

### Titan97

$$P_2=P_1+\frac{k}{A^2}(V_2-V_1)$$
$$\frac{k}{A^2}=\frac{P_2-P_1}{V_2-V_1}$$

15. Feb 9, 2016

### Staff: Mentor

OK. If you take that and substitute it into the equation $P=P_1+\frac{k}{A^2}(V-V_1)$, what do you get?

16. Feb 9, 2016

### Titan97

$$P=P_1+\frac{P_2-P_1}{V_2-V_1}(V-V_1)$$

If $V_2=3V_1 , T_2=4T_1$,
$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$
$$\frac{P_1V_1}{T_1}=\frac{P_2\cdot 3V_1}{4T_1}$$
$$P_2=\frac{P_1V_1\cdot 4T_1}{3V_1T_1}$$
$$P_2=\frac{4}{3}P_1$$

By the way, (B) is anyway correct because the change in internal energy is always $nC_vdT=\frac{nR}{\gamma-1}(3T_1-T_1)=3P_1V_1$

17. Feb 9, 2016

### Staff: Mentor

You're right. I used nRΔT rather than nCvΔT for ΔU. I don't know what I was thinking. Senior Moment.

Getting back to our problem, if we start with the P-V equation you correctly obtained, the next step is to determine the work done by the gas in expanding from $V_1$ to $V_2$. Please don't substitute any of the cases into the equation yet. I would like to get the general result first. So what do you get when you integrate PdV between these two volumes?

Chet

Last edited: Feb 9, 2016
18. Feb 9, 2016

### Titan97

$$\int_{V_1}^{V_2}PdV=P_1(V_2-V_1)+\frac{P_2-P_1}{V_2-V_1}(\int_{V_1}^{V_2}VdV-V_1(V_2-V_1))$$
$$=P_1(V_2-V_1)+\frac{P_2-P_1}{V_2-V_1}\big(\frac{V_2^2-V_1^2-2V_1(V_2-V_1)}{2}\big)$$
$$=P_1(V_2-V_1)+(P_2-P_1)\frac{V_2+V_1-2V_1}{2}$$
$$=\frac{2P_1(V_2-V_1)+(P_2-P_1)(V_2-V_1)}{2}$$
$$=\frac{(P_1+P_2)(V_2-V_1)}{2}$$

19. Feb 9, 2016

### Staff: Mentor

Excellent. That's what I get.

Now, in terms of $P_1V_1$, determine the work W, the change in internal energy ΔU, and the heat added Q for each of the two changes they specify in the problem statement.

Then we'll be done, and we'll actually see which of the 4 choices (besides B) are correct.

Chet

20. Feb 9, 2016

### Titan97

If $V_2=3V_1$, and for post #16, $$W=\frac{7}{3}P_1V_1$$

$$Q=U+W=3P_1V_1+\frac{7}{3}P_1V_1=\frac{16}{3}P_1V_1$$
So (D) is wrong.

If $T_2=3T_1$ and $V_2=2V_1$, energy stored in spring = work done by gas
$$P_2=\frac{P_1V_1T_2}{V_2T_1}=\frac{3P_1}{2}$$
Energy stored in spring is $\frac{5P_1V_1}{4}$. So (A) is wrong.

Hence correct answer is : (B),(C)