Thermodynamics problem involving piston and spring

[Titan97]

Homework Statement

[/B]

Homework Equations

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

The Attempt at a Solution

. [/B]
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?

 

[Grinkle]

But since volume increases, the increases in pressure is nullified.

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.

 

[Chestermiller]

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

 

[Titan97]

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

 

[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?

 

[Chestermiller]

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

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.

 

[Chestermiller]

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.

 

[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

PA−P1A=k(x-x_1)

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

 

[Chestermiller]

@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?

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?

 

[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 $\Delta\text{KE}=0$.

 

[Chestermiller]

Yes, I do. Why are expressing P in terms of $x-x_1$ when you already have P in terms of $x$?

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

 

[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?

 

[Chestermiller]

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?

That relationship is correct. So the energy stored in the spring is equal to the work done by the gas.

Initial volume is $V_1=AL_1$ of L1 is the initial length of cylinder containing gas.
$V=AL_2$
$$V-V_1=A(L-L_1)=A\Delta L=A(x-x_1)$$

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?

 

[Titan97]

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

 

[Chestermiller]

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

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?

 

[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 $nC_vdT=\frac{nR}{\gamma-1}(3T_1-T_1)=3P_1V_1$

 

[Chestermiller]

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

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 $nC_vdT=\frac{nR}{\gamma-1}(3T_1-T_1)=3P_1V_1$

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

 

[Titan97]

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

 

[Chestermiller]

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

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

 

[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)

 

[Chestermiller]

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

$$Q=U+W=3P_1V_1+\frac{7}{3}P_1V_1=\frac{16}{3}P_1V_1$$

For this case, I get $\Delta U=\frac{9}{2}P_1V_1$. Please double check.
Otherwise, we're in total agreement.

Nice work.

Chet

 

[Titan97]

Oops. I used $V_2,T_2$ from second case. If $T_2=4T_1$ then $\Delta U=\frac{9}{2}P_1V_1$

 

[ehild]

Energy stored in spring is $\frac{5P_1V_1}{4}$. So (A) is wrong.

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

The spring constant is not the same in cases a,b and c,d. Using the appropriate spring constant, a) proves to be true.

 

[Chestermiller]

The spring constant is not the same in cases a,b and c,d. Using the appropriate spring constant, a) proves to be true.

Huh? Can you please elaborate.

 

[Titan97]

@ehild how can a spring "constant" change here?

 

[ehild]

Huh? Can you please elaborate.

It was said that the spring was relaxed initially. If P₁ V₁ and T₁ were the initial values of the state variables, the external pressure had to be P₁.
In case a, heating the gas slowly, the new volume is V₂=2V₁ and T₂ = 3T₁, and a gas is in equilibrium. The forces on the piston are balanced so P₁A + kx=P₂A. The change of volume is V₁=xA. From the ideal gas law P₂=1.5 P₁, So kx=0.5 AP₁, and k=0.5 A²P₁/V₁.
The elastic energy of the spring is 0.5kx² = 0.5 (0.5 A P₁)V₁/A= 0.25 P₁V₁.

In case C) the volume became V₃=3V₁ and the temperature T₃=4T₁. : P₃=4/3 P₁.
the force of the gas is balanced by the external pressure and the force of the spring. kx₃=1/3 P₁A and x₃=2V₂/A, so k=1/6 A² P₁/V₁.

 

[Chestermiller]

It was said that the spring was relaxed initially. If P₁ V₁ and T₁ were the initial values of the state variables, the external pressure had to be P₁.

HI ehild,

Actually, we were puzzled by the statement that the spring is initially at its unextended length initially (see post #6). We decided that this was an error in the problem statement, and the spring was actually compressed initially to balance the pressure in the left chamber. We also assumed that there was vacuum in the right chamber. That's what our analysis was based upon.

You have assumed that the right chamber was initially at P1, and was somehow subsequently held at that pressure for all time. I never considered this as a possibility. I thought that if there was a gas enclosed in the right chamber, its pressure would change, but we didn't know its volume. So we went with the assumption that the right chamber was under vacuum.

So, in short, we've solved two different problems. Maybe your assumption of constant pressure in the right chamber was what they intended.

Chet

 

[ehild]

HI ehild,

Actually, we were puzzled by the statement that the spring is initially at its unextended length initially (see post #6). We decided that this was an error in the problem statement, and the spring was actually compressed initially to balance the pressure in the left chamber. We also assumed that there was vacuum in the right chamber. That's what our analysis was based upon.

You have assumed that the right chamber was initially at P1, and was somehow subsequently held at that pressure for all time. I never considered this as a possibility. I thought that if there was a gas enclosed in the right chamber, its pressure would change, but we didn't know its volume. So we went with the assumption that the right chamber was under vacuum.

So, in short, we've solved two different problems. Maybe your assumption of constant pressure in the right chamber was what they intended.

Chet

The end of the cylinder may be open. I do not think it matters if initially the spring force balances the pressure of the gas or the external atmosphere.

 

[Chestermiller]

I do not think it matters if initially the force balancing the force of the pressure of the gas was the initial tension of the spring or force of the external atmosphere. The end of the cylinder may be open

Good point. I hadn't thought of that. Anyway, the OP got some good experience solving a problem (which may not have been the intended problem). Maybe you can take a look at what we did, based on our assumption, and see what you think.

Chet

 

[TSny]

Thanks, ehild. It did not occur to me that the region to the right of the piston could be interpreted as "open to the atmosphere", so to speak. That must have been the intended interpretation.

 

[ehild]

Thanks, ehild. It did not occur to me that the region to the right of the pendulum piston could be interpreted as "open to the atmosphere", so to speak. That must have been the intended interpretation.

Yes, the picture was misleading, but they emphasized that the spring was relaxed.

 

[ehild]

Good point. I hadn't thought of that. Anyway, the OP got some good experience solving a problem (which may not have been the intended problem). Maybe you can take a look at what we did, based on our assumption, and see what you think.

Chet

You answered b, c, d first. b included the temperature change only. c and d use the spring constant, and you answered a) with that spring constant, although the volume and temperature required an other k. It did not matter if the spring was relaxed or there was a constant ambient pressure.
The problem writer made and error with giving inconsistent data, although he did not state that it was the same spring in both experiments. Maybe, he just merged two problems together.

 

[TSny]

Yes, the picture was misleading, but they emphasized that the spring was relaxed.

Right. But I couldn't see how to make the problem consistent with a relaxed spring unless there was also a gas enclosed on the right side. But of course, then you would have to take account of the changing pressure of that gas. A mess.

 

[ehild]

Right. But I couldn't see how to make the problem consistent with a relaxed spring unless there was also a gas enclosed on the right side. But of course, then you would have to take account of the changing pressure of that gas. A mess.

Yes, it was a mess, but an open cylinder with a holder for the spring is a solution.

 

[TSny]

Yes, it was a mess, but an open cylinder with a holder for the spring is a solution.

I agree.