Thermodynamics problem involving piston and spring

[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.

 

[Chestermiller]

OK. I'm going to try to do it using ehild's condition, rather than vacuum on the right hand side and see where that takes us. The force balance on the piston in this situation during the change is:$$P=P_1+\frac{k}{A}x$$where x is the displacement of the piston to the right (equal to the absolute change in length relative to the unextended state). In this situation, x is related to the volume change of the right hand chamber by:$$x=\frac{V-V_1}{A}$$ So if we combine these two equations, we get:$$P=P_1+\frac{k}{A^2}(V-V_1)$$ This equation applies at (V2, P2). So, here again, we get the exact same relationship between pressure and volume that we obtained with the assumption of vacuum on the right side of the piston:$$P=P_1+\frac{(P_2-P_1)}{(V_2-V_1)}(V-V_1)$$So the question is, why did ehild get different results from us for case A if we have the same P-V relation applying.

The we showed that the work done on the spring is: $$W=\frac{(P_2+P_1)}{2}(V_2-V_1)$$ This is also the energy stored in the spring. From the ideal gas law,$$P_2=\frac{V_1}{V_2}\frac{T_2}{T_1}P_1$$
In case A, $P_2=1.5P_1$, so $W=\frac{5}{4}P_1V_1$. This is not what ehild got. We need to resolve this.

See next post for resolution.

Chet

 

[Chestermiller]

OK. I see where I went wrong on ehild's version of the problem. The work done to compress the spring is the integral of (P-P1)dV, not the integral of PdV. So the work to compress the spring turns out to be $\frac{(P_2-P_1)}{2}(V_2-V_1)$. The answer then agree's with ehild's result for case A, and also agrees with the given answer. However, our answers to the other parts of the problem should be correct, because the focus there on the gas rather than the spring.

 

[Titan97]

The pressure on the right chamber is not mentioned. @Chestermiller solved this by assuming that there was vacuum in the right chamber. That's why we took $P_1A=kx_1$. With this, the answers is B,D

@ehild took the initial pressure on the right chamber to be $P_1$ and using this you can get option (A).

Am I correct?

 

[Chestermiller]

The pressure on the right chamber is not mentioned. @Chestermiller solved this by assuming that there was vacuum in the right chamber. That's why we took $P_1A=kx_1$. With this, the answers is B,D

@ehild took the initial pressure on the right chamber to be $P_1$ and using this you can get option (A).

Am I correct?

Yes. She noted that, if the pressure in the right chamber is always P1, then the spring can start out in its uncompressed configuration.

 

[ehild]

OK. I see where I went wrong on ehild's version of the problem. The work done to compress the spring is the integral of (P-P1)dV, not the integral of PdV. So the work to compress the spring turns out to be $\frac{(P_2-P_1)}{2}(V_2-V_1)$. The answer then agree's with ehild's result for case A, and also agrees with the given answer. However, our answers to the other parts of the problem should be correct, because the focus there on the gas rather than the spring.

Yes, the other parts was not influenced by the spring.
For a) see also my solution in #26.
This was a really weird problem!

 

[Chestermiller]

Yes, the other parts was not influenced by the spring.
For a) see also my solution in #26.
This was a really weird problem!

Thanks ehild.

Just to be clear, once Titan97 showed that, for any spring, the gas pressure varies linearly with the gas volume in this problem, it no longer became necessary to consider the specific spring constant for each case, since the slope and intercept of that linear relationship could be determined from the initial and final pressures and volumes. And, of course, since pressure is varying linearly with volume (a trapazoid on a P-V diagram), the work done by the gas would be equal to the average of the initial and final pressures times the volume change.

I agree that this was a really weird problem. But it was also very interesting and instructive. The P-V constraint in this problem is reminiscent of what one encounters in a polytropic process, where the volume and temperature are controlled in tandem such that, at any point along the process path, PVⁿ is constant. In our case, the presence of the spring constrains the P-V variation to a straight line.

Chet