Does the Coefficient of Thermal Expansion Apply to Both Expansion and Contraction?

[Jo MacDonald]

We did an activity today in class where we placed an empty 2L bottle in the Freezer, let it contract, then calculated the coeffcient of volumetric expansion using ΔV = V_oβΔT. We got pretty close to the accepted value. The question is...

Does the coefficient of thermal expansion only work for expansion? not contracting? Because if you let it contract, then the original volume is the room temp volume and you get a different coefficient than if you let it expand and use the cold volume as your original volume. For both cases, the ΔV and ΔT are the same. We got closer to the accepted value using the cold volume. Some of my students got it on the nose using the cold volume as their original volume.

In the past, I've used the coefficient of expansion for both warming and cooling, but I guess that doesn't work? What do you think?

Jo MacDonald

 

[berkeman]

Welcome to the PF.

Since you are talking about air, how do your equation and calculations compare to if you used the Ideal Gas Law? Or are they the same?

 

[haruspex]

That equation is only an approximation for small changes in temperature. A more general equation is Charles' Law: volume is proportional to absolute temperature, V₁/T₁=V₀/T₀. From this you can show
##\Delta V=V_0\frac{\Delta T}{T_0}##. So it is clear that β in your equation depends on initial temperature.

 

[Jo MacDonald]

We found the volumes both directly and with Charles law. We came up with similar initial (freezer) volumes both ways, so the difference is using the cold volume for your Vo out the warm volume.

It should contract and expand equally due to Charles law, but you get different coefficients depending on which volume you use as your starting volume. So does the coefficient of expansion work only in expanding?

 

[Jo MacDonald]

The temperature change was about 30 degrees Celsius. We use the expansion equation over much larger temperature ranges.

 

[haruspex]

you get different coefficients depending on which volume you use as your starting volume.

As I thought was clear from my previous post, the coefficient depends on the starting temperature. Specifically, β=1/T₀.

 

[russ_watters]

We did an activity today in class where we placed an empty 2L bottle in the Freezer, let it contract, then calculated the coeffcient of volumetric expansion using ΔV = V_oβΔT. We got pretty close to the accepted value. The question is...

Does the coefficient of thermal expansion only work for expansion? not contracting? Because if you let it contract, then the original volume is the room temp volume and you get a different coefficient than if you let it expand and use the cold volume as your original volume. For both cases, the ΔV and ΔT are the same. We got closer to the accepted value using the cold volume. Some of my students got it on the nose using the cold volume as their original volume.

In the past, I've used the coefficient of expansion for both warming and cooling, but I guess that doesn't work? What do you think?

Jo MacDonald

[edit: didn't read close enough]
So you noted a difference using different coefficients for different temperatures. This is normal/real. The way to reconcile it (at this level) is to use both such as by interpolating between them.

 

[Jo MacDonald]

OK, I just noticed in my textbook that the coefficients are given at 20 C. Since I've never seen any other numbers, I just assumed that they were constants. I had never heard that they were temperature dependent.

When haruspex said that the coefficient depended on the initial temperature, I thought he misunderstood what I was asking, because we had the same temperatures, just in opposite order (final, initial), but I get it now. Didn't realize the coefficients were not really CONSTANTS, that helped a lot. Now I can go back and explain to my students.

Interesting that we got better answers with the -9 C volume than the 20 C volume, but now I can tell them which volume to use and we will just get the percent error we get.

Thank you

 

[haruspex]

I gather your warm temperature was room temperature, say 20C, and your freezer temp 30C less, so about -10C.
On that basis, starting at room temperature you should have got the book answer, but in going from cold to warm you should have got 293/263 = 1.1 times as much.
If you got the book answer for the second and a lower number for the first then it may be to do with the physics of the container. That will partly resist shrinking, so the pressure in the bottle will fall. Charles' Law assumes constant pressure.
You might do better to use a rubber balloon, not inflated too much.

 

[Jo MacDonald]

So the coefficient should increase for lower temps, that makes sense to get you back to the warm Charles' Law volume.

 

[haruspex]

So the coefficient should increase for lower temps, that makes sense to get you back to the warm Charles' Law volume.

Yes.

 

[Jo MacDonald]

One last clarification. Since liquids and solids don't change as much as gases, would it be safe to assume that their expansion coefficients change less with changing temperatures? That is, it is safe to use the published one at 20 C for most all temperatures (within reason)?

 

[haruspex]

One last clarification. Since liquids and solids don't change as much as gases, would it be safe to assume that their expansion coefficients change less with changing temperatures? That is, it is safe to use the published one at 20 C for most all temperatures (within reason)?

They provably do still change with temperature. Suppose it is constant. Consider a sample length L warmed by ΔT. It now has length L(1+αΔT). Had we warmed it by 2ΔT the formula says it should now have length L(1+2αΔT), whereas if we take the previously warmed sample of length L(1+αΔT) and warm it by the second ΔT it says L(1+αΔT)², which is just that little bit more.

 

[Chestermiller]

The correct definition of the coefficient of expansion is ##\beta=\frac{d(\ln{V})}{dT}=\frac{1}{V}\frac{dV}{dT}##. The equation ##\beta=\frac{1}{V_0}\frac{\Delta V}{\Delta T}## is only approximate, for beginners. With the correct definition, the value obtained for heating will be the same as the value obtained for cooling.

 

[haruspex]

The correct definition of the coefficient of expansion is ##\beta=\frac{d(\ln{V})}{dT}=\frac{1}{V}\frac{dV}{dT}##. The equation ##\beta=\frac{1}{V_0}\frac{\Delta V}{\Delta T}## is only approximate, for beginners. With the correct definition, the value obtained for heating will be the same as the value obtained for cooling.

That may be right for solids but it does not fit with Charles' Law for ideal gases. It would make the value of β depend on the two temperatures. If we suppose β is constant over some range T₀ to T₁ then we get ##\beta=\frac{\ln(T_1)-\ln(T_0)}{T_1-T_0}##.

 

[Chestermiller]

That may be right for solids but it does not fit with Charles' Law for ideal gases. It would make the value of β depend on the two temperatures. If we suppose β is constant over some range T₀ to T₁ then we get ##\beta=\frac{\ln(T_1)-\ln(T_0)}{T_1-T_0}##.

For an ideal gas, the coefficient of volume expansion is equal to 1/T. So it sure does fit with Charles' Law. Also, even for solids, the coefficient of volume expansion has some small temperature dependence.

 

[haruspex]

For an ideal gas, the coefficient of volume expansion is equal to 1/T.

OK, I didn't state it quite correctly: I meant that a constant β would not be consistent with Charles' Law.

In post #14 you wrote

With the correct definition, the value obtained

but how is that differential definition to be used in the experimental context of only knowing the volumes at two widely separated temperatures? To get a value for β we would need to solve the differential equation, and that requires knowing how β depends on temperature. If we take it as constant then, yes, you will get a result that is consistent for both heating and cooling, but only with those two specific temperatures as the endpoints.

The 1/T relationship is what I wrote in post #3.

 

[Chestermiller]

OK, I didn't state it quite correctly: I meant that a constant β would not be consistent with Charles' Law.

Who said anything about ##\beta## being constant? For real materials, ##\beta## is a function of temperature.

but how is that differential definition to be used in the experimental context of only knowing the volumes at two widely separated temperatures?

I said nothing about how it is measured experimentally or how it is applied in practice. But, to apply it in practice, you need to know ##\beta(T)## and write
$$\frac{V_2}{V_1}=e^{\int_{T_1}^{T_2}{\beta(T)dT}}$$For an ideal gas, this reduces to Charles' law $$\frac{V_2}{V_1}=\frac{T_2}{T_1}$$
So I stand by what I said. To be more precise, the thermodynamic definition of the coefficient of volume expansion is:
$$\beta(T)=\left(\frac{\partial \ln{V}}{\partial T}\right)_P$$
An example of how it is applied in practice is in calculating the effect of pressure on enthalpy of real gases:
$$\left(\frac{\partial H}{\partial P}\right)_T=V(1-\beta T)$$
In thermodynamics, the symbol ##\alpha## is often used for the coefficient of volume expansion, rather than ##\beta##. In the limit of an ideal gas (for which ##\beta=1/T##, the right hand side of this equation vanishes, such that the zero pressure dependence of enthalpy for an ideal gas is properly captured.

 

[haruspex]

I stand by what I said

All I am disputing is the implication in this statement:

With the correct definition, the value obtained

that the differential equations suffice to obtain a value for β in the context of the experiment conducted.

 

[Chestermiller]

That may be right for solids but it does not fit with Charles' Law for ideal gases. It would make the value of β depend on the two temperatures. If we suppose β is constant over some range T₀ to T₁ then we get ##\beta=\frac{\ln(T_1)-\ln(T_0)}{T_1-T_0}##.

but how is that differential definition to be used in the experimental context of only knowing the volumes at two widely separated temperatures? To get a value for β we would need to solve the differential equation, and that requires knowing how β depends on temperature.

For ideal gases, using the relationship I presented over a finite temperature range gives the exact value for ##\beta##, not at the arithmetic mean temperature, but at the well-known logarithmic mean absolute temperature $$T_{LM}=\frac{(T_2-T_1)}{\ln{(T_2/T_1)}}$$Maybe physicists don't use the logarithmic mean very much in their work, but we engineers, particularly in the area of heat transfer, use the logarithmic mean quite abundantly. Anyway, if the absolute temperatures T1 and T2 do not differ by more than a factor of 2, the logarithmic mean is very close to the arithmetic mean. Here is an example of the use of logarithmic means in heat transfer: https://physics.stackexchange.com/questions/379870/heat-transfer-area-across-a-surface
For experimental evaluations over finite temperature ranges, if we say that the measured value applies to the logarithmic mean temperature, we will be obtaining a very accurate value of ##\beta## even if it is function of temperature, and even for solids and liquids (since, in these cases, the absolute temperatures typically don't differ by a factor of 2 in the tests).