Spring expansion on thermodynamic system

In summary, the problem involves a confined gas in an adiabatic cylinder with a frictionless piston activated by an elastic spring. The initial conditions are 300 K and 400 kPa, and the air is allowed to expand until the final volume is twice the initial volume. The final pressure and temperature are to be determined using only the first law of thermodynamics. Two different approaches are presented, one using an expression for external pressure and the other considering the spring as part of the system. It is unclear whether the spring remains compressed or not, but assuming it does, the process is equivalent to a free expansion with no work done.
  • #1
Hobold
83
1
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."

termo.png


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

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

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 [tex]V_1 = \frac{MRT_1}{p_1}[/tex]

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

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.
 
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  • #2
Hobold said:
...This problem is to be solved using only the first law of thermodynamics. Here is what I have done:

Considering [tex]\delta Q = dU + \delta W = mc_{vo}dT + pdV = 0 [/tex]
So:

[tex]dT = -pdV/mc_{v}[/tex]

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
 
  • #3
I'd say the expression for external pressure would be [tex]p = -\frac{F}{A} = -\frac{\kappa \Delta x}{A} = -\frac{\kappa \Delta V}{A^2} = -\frac{\kappa}{A^2}(V-V_1)[/tex], where kappa is a spring constant and A is piston's transversal area, but neither are given.
 
  • #4
Hobold said:
I'd say the expression for external pressure would be [tex]p = -\frac{F}{A} = -\frac{\kappa \Delta x}{A} = -\frac{\kappa \Delta V}{A^2} = -\frac{\kappa}{A^2}(V-V_1)[/tex], where kappa is a spring constant and A is piston's transversal area, but neither are given.
[itex]\Delta \vec{x} [/itex] has to be the displacement from the spring's equilibrium position. So the [itex]\Delta V[/itex] 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 ([itex]\Delta \vec{x} < 0[/itex]) 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
 
  • #5
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.
 
  • #6
Hobold said:
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
 
  • #7
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?
 
  • #8
Hobold said:
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
 

What is spring expansion on a thermodynamic system?

Spring expansion on a thermodynamic system refers to the increase in volume or size of a material when it is heated. This expansion is caused by the increase in kinetic energy of the particles within the material, which causes them to vibrate and take up more space.

How does spring expansion affect the behavior of a thermodynamic system?

Spring expansion can affect the behavior of a thermodynamic system in several ways. It can cause changes in pressure, density, and other thermodynamic properties. It can also lead to structural changes, such as in the case of metals, where thermal expansion can cause warping or bending.

What factors influence spring expansion on a thermodynamic system?

The extent of spring expansion on a thermodynamic system depends on several factors, including the type of material, the temperature change, and the original size and shape of the material. Some materials, such as metals, expand more than others, and larger temperature changes will result in greater expansion.

How is spring expansion measured in a thermodynamic system?

Spring expansion is typically measured using a device called a dilatometer, which can accurately measure changes in length or volume of a material as it is heated. This measurement is then used to determine the coefficient of thermal expansion, which describes the amount of expansion per unit change in temperature.

Why is spring expansion important in thermodynamics?

Spring expansion is an important concept in thermodynamics because it affects the behavior and properties of materials when they are exposed to changes in temperature. It is also a key factor in the design and functionality of many everyday objects, such as bridges and buildings, where thermal expansion must be taken into account to prevent structural damage.

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