# Thermal Expansion Problem

1. Aug 16, 2005

### maverick280857

Hello everyone

My query is regarding the following problem:

A barometer reads 75.0 cm on a steel scale. The room temperature is 30 degrees C. The scale is correctly graduated for 0 degrees C. The coefficient of linear expansion of steel is $\alpha = 1.2 * 10^{-5}$ per degree C and the coefficient of volume expansion of mercury is $\gamma = 1.8 * 10^{-4}$ per degree C. Find the correct atmospheric pressure.

My solution does not match the book's solution, which goes as follows:

the length of the 75 cm steel wire at 0 degrees C will become $l_{\theta}$ where,

$$l_{\theta} = (75cm)(1 + \alpha(30))$$

The length of the mercury column at 30 degrees C is $l_{\theta}$. Suppose the length of the mercury column if it were at 0 degrees C is $l_{0}$. Then,

$$l_{\theta} = l_{0}(1 +\frac{\gamma}{3}(30))$$

So

$$l_{0}(1 + \frac{\gamma}{3}(30)) = (75 cm)(1 + \alpha(30))$$

or $$l_{0} = 75cm \frac{1 + \alpha(30)}{1 + \frac{\gamma}{3}(30)}$$ = 74.89 cm (after the binomial approximation)

My solution

Suppose the volume of mercury at the higher temperature is $V_{Hg}$, the cross-sectional area of the mercury column (equal to the cross-sectional area of the steel tube) is $A_{v}$ and the height of the mercury column is $h_{Hg}$.

$$V_{Hg} = V_{Hg,0}(1 + \gamma_{Hg}\Delta T)$$

$$A_{v} = A_{v,0}(1 + \beta_{v}\Delta T)$$

where $\gamma_{Hg}$ and $\beta_{v}$ are the coefficients of volume expansion of mercury and of area expansion of the vessel (steel tube).

Also
$$V_{Hg} = A_{v}h_{Hg}$$
and
$$V_{Hg,0} = A_{v,0}h_{Hg,0}$$

$$A_{v}h_{Hg} = A_{v,0}h_{Hg,0}(1 + \gamma_{Hg}\Delta T)$$
so
$$A_{v,0}h_{Hg}(1 + \beta_{v}\Delta T) = A_{v,0}h_{Hg,0}(1 + \gamma_{Hg}\Delta T)$$
so
$$h_{Hg} = h_{Hg,0}\frac{(1 + \gamma_{Hg}\Delta T)}{(1 + \beta_{v}\Delta T)}$$

which after the binomial approximation yields

$$h_{Hg} = h_{Hg,0}(1 + \gamma_{Hg}\Delta T)(1 - \beta_{v}\Delta T)$$

If we neglect the term second order in $(\Delta T)^2$, then

$$h_{Hg} = h_{Hg,0}(1 + (\gamma_{Hg}-\beta_{v})\Delta T)$$

But $$\beta_{v} = \frac{\alpha_{v}}{2}$$. Instead of the factor 1/3 I get 1/2. Where I am I going wrong?

Thanks and cheers
Vivek

Last edited: Aug 16, 2005
2. Aug 16, 2005

### mukundpa

1.)
The atm. pressure is given by hDg (where D is the density of Hg.) and hence is independent of the area of the tube.

2.)
The correction should be in the length of the scale and in the density of Hg.

so the correct height is h = h0(1 + alfa x 30) as in the first solution and for the density m/V = m/V0(1 + gamma x 30)
gives D =D0(1 - gamma x 30) [after binomial appro.]

hence h'D0g = hDg = h0(1 + alfa x 30)D0(1 - gamma x 30)g
or h' = 75(1 + alfa x 30)(1 - gamma x 30)

3. Aug 16, 2005

### Staff: Mentor

I read that as the mercury level indicates 75 cm at 30°C. So if one corrects for length at 0°C, the steel would contract and the indication of height would be greater. However, one must also correct for the height of the mercury which also decreases with temperature.

$$l_{\theta} = 75cm = l_o(1 + \alpha(30))$$, where $$\theta$$ = 30°C

The V = Vo * (1 + $${\gamma} \Delta T$$) =

Vo (1 + $${\alpha} \Delta T$$)3

expand the cubic term.

Last edited: Aug 16, 2005
4. Aug 17, 2005

### mukundpa

Most humbly, I think that by correcting the length of the scale we are measuring the actual height of the mercury column at 30 degree C and the density of the mercury is taken at 30 degree C, will give the atm. pressure at that time.

regards

5. Aug 17, 2005

### maverick280857

mukundpa, this is solved problem 17 of Dr. HC Verma's book volume II (the chapter on thermal expansion).

6. Aug 17, 2005

### mukundpa

no matter where it is solved and who has solved. I just want to learn. If there is any misconception please point it out.
Thanks in anticipation.

7. Aug 18, 2005

### maverick280857

Hey that was just to let you know where the problem is from. Since you're from India you might have heard of this book. It was to let you know that you can refer to the book's solution to make sure what the author wants to convey. As otherwise, it is an exercise in futility.

8. Aug 19, 2005

### Staff: Mentor

This part of the book's solution is correct.

This is not correct:
(1) It treats the mercury column as a solid bar of mercury of fixed mass, not as a fluid (with a reservoir) that assumes the shape of its container. It ignores the expansion of the container housing the mercury, presumably a glass tube.
(2) It is irrelevant: As mukundpa points out, what must be accounted for is the change in the mercury's density.

For the reasons above, this is incorrect.

It seems like you are also trying to account for the changing height of the mercury column. This is incorrect, but at least you recognize that the height would depend on the expansion of the container, not just the mercury. But you are assuming that the mass of the column of mercury is fixed. Not so; mercury can fill the tube as needed to balance atmospheric pressure. (And why do you assume that the tube is a steel tube? I would have assumed glass. It's the ruler that's steel.) I suggest you review my comments above and rework this problem, forgetting the book's answer.

I agree with mukundpa. The pressure of the column of mercury equals $\rho g h$. Find the correct height of the column (by adjusting for the expansion of the steel ruler) and adjust for the reduced density of the mercury and you will find the atmospheric pressure in standard units.

9. Aug 20, 2005

### maverick280857

Thank you Doc!! I get it now. But hold on a minute:

You say that the first part of the book's solution is correct. But the second equation is wrong right? I think the second equation refers to the reading on the scale and so they treat the expansion as being linear, with a linear coefficient given by $\frac{\gamma}{3}$.

The way I first thought was that the area of cross-section of the tube is the same as the area of cross-section of the meniscus of mercury, so the tube "dictates" the area.

Cheers
Vivek

Last edited: Aug 20, 2005
10. Aug 20, 2005

### Staff: Mentor

When I say that the second equation is "wrong", I mean that it's wrong to apply it to this problem. There is nothing wrong with estimating the linear coefficient of expansion by dividing the volume coefficient by three. But trying to use the linear expansion of the mercury column to determine its height is incorrect. (If the mercury column were a solid bar of fixed mass, then the height would change per the linear expansion coefficient. But that's a different problem.)

Your thinking here is correct. The area of the tube does dictate the cross-sectional area of the mercury, since mercury is a liquid. But it doesn't dictate the height: that depends on the density of mercury and on atmospheric pressure only.

11. Aug 24, 2005

### maverick280857

Thanks.

Where can I find a mathematical treatment of thermal stress, strain and situations like a sphere inside a spherical cavity carved in a cubical container, stress in two rods in series/parallel when heated? In particular I am looking for a complete description of the constraints and how the stress components work. If you are aware of any book/website, please do let me know.

Cheers
vivek