Spring expansion on thermodynamic system

[Hobold]

This is a very simple problem, though I have arrived at two different results and I would like some help to procceed (I'm not sure if this is the right section, though this is not a homework):

"2kg are confined in an adiabatic cilinder through a frictionless piston activated by an elastic spring, as in the picture. The piston is initially locked by an inner pin and keeps the air at 300 K and 400 kPa. The piston is then unlocked and moves freely, allowing the air to expand until the final volume is equal to double the initial volume. Determine both final pressure and temperature."

This problem is to be solved using only the first law of thermodynamics. Here is what I have done:

Considering $$\delta Q = dU + \delta W = mc_{vo}dT + pdV = 0$$ and air as a ideal gas, differentiating T = pV/(mR) you get $$dT = \frac{1}{mR}(pdV + Vdp)$$ and, with some algebraic manipulation and taking $$c_{p0} = R + c_{v0}$$, you get $$\frac{dp}{p} + \frac{dV}{V} \gamma = 0 \rightarrow pV^{\gamma} = \text{constant}, \ \gamma = c_{p0}/c_{v0}$$

We are allowed to use a table from which we can take c_p0 and c_v0.

The initial volume is easily found by doing $$V_1 = \frac{MRT_1}{p_1}$$

then, taking the expression found before, we get $$p_1 V_1^\gamma = p_2 V_2^\gamma \rightarrow p_2 = p_1(V_1/V_2)^\gamma$$

Finally, to find final temperature, we use Clapeyron's equation for ideal gas.

Though it is clear that T_2 < T_1. The solution my professor's PhD student proposed is much simpler: you consider the spring as a part of the system and this is nothing but a case of free expansion, then T_2 = T_1. I talked to him later and he told me that my solution is also right, depending on how you interpret the problem, so that's why I posted exactly the same question here. I want some clue whether my solution is right or not.

Thanks.

 

[Andrew Mason]

...This problem is to be solved using only the first law of thermodynamics. Here is what I have done:

Considering $$\delta Q = dU + \delta W = mc_{vo}dT + pdV = 0$$

So:

$$dT = -pdV/mc_{v}$$

P is the EXTERNAL pressure, not the internal pressure of the gas. What is the expression for external pressure? (hint: the spring has something to do with it).

AM

 

[Hobold]

I'd say the expression for external pressure would be $$p = -\frac{F}{A} = -\frac{\kappa \Delta x}{A} = -\frac{\kappa \Delta V}{A^2} = -\frac{\kappa}{A^2}(V-V_1)$$, where kappa is a spring constant and A is piston's transversal area, but neither are given.

 

[Andrew Mason]

I'd say the expression for external pressure would be $$p = -\frac{F}{A} = -\frac{\kappa \Delta x}{A} = -\frac{\kappa \Delta V}{A^2} = -\frac{\kappa}{A^2}(V-V_1)$$, where kappa is a spring constant and A is piston's transversal area, but neither are given.

$\Delta \vec{x}$ has to be the displacement from the spring's equilibrium position. So the $\Delta V$ is not the change in volume after the pin is lifted.

I am not sure why they give you a problem with a spring in it without giving you the spring constant and the initial displacement from equilibrium - unless the spring is initially compressed and remains compressed ($\Delta \vec{x} < 0$) at all times. If that was the case, then the air does not expand against an inward external force. This would be a free expansion so Tf = Ti. Does the problem say whether the spring remains compressed from its equilibrium position?

AM

 

[Hobold]

No, the problem doesn't say if the spring is compressed or not, but it does say that the piston is activated by the spring when the pin is removed.

 

[Andrew Mason]

No, the problem doesn't say if the spring is compressed or not, but it does say that the piston is activated by the spring when the pin is removed.

I think, then, that one is supposed to assume that the spring remains compressed for the entire expansion and that the piston is massless. The expanding gas does no work as there is no inward external pressure that the gas expands against. This makes the process thermodynamically equivalent to a free expansion of the gas: dQ = dU = dW = 0.

AM

 

[Hobold]

Thanks, got it.

One last question: so if the spring is locked in its equilibrium position, if the pin is removed, will the air work to stretch the spring?

 

[Andrew Mason]

Thanks, got it.

One last question: so if the spring is locked in its equilibrium position, if the pin is removed, will the air work to stretch the spring?

Only if the spring is at or stretched past its equilibrium position (so the air is doing work in further stretching the spring) OR if the piston has mass (so the air does work in accelerating the piston).

AM