# Homework Help: Using the Ideal Gas temperature scale

1. Jul 7, 2013

### CAF123

1. The problem statement, all variables and given/known data
Below is some values of the observed values of pressure P of a gas in a constant volume gas thermometer at an unknown temperature and at the triple point of water as the mass of the gas used is reduced.

$P_{TP}$ (torr) | 100 | 200 | 300 | 400
$P$ (torr) | 127.9 | 256.5 | 385.8 | 516

Consider $\lim_{P_{TP} \rightarrow 0} (P/P_{TP})$ and then determine T.

2. Relevant equations

Ideal Gas scale

3. The attempt at a solution

Ideally, I would want to take the pressure at the triple point as close to possible to zero. However, with the given data, I closest I have is PTP = 100 torr, so I will have to extrapolate from the given ratios.

The 4 ratios calculated were 1.279, 1.2825, 1.286 and 1.29. The answer in the back of the book is precise so I am not really sure how I could precisely extrapolate this data to the origin. I could approximate the data as linear (i.e P proportional to PTP), but this appears to not be what is required.

Many thanks.

2. Jul 7, 2013

### Redbelly98

Staff Emeritus
The data do appear very nearly linear. If you go ahead and calculate a best-fit line to the data, how far off is your answer from the answer given in the book?

3. Jul 7, 2013

### CAF123

If I average the values I got, P ≈ 1.28PTP. Subbing into P/PTP, cancelling and then multiplying by 273.16 for the ideal scale, this then gives T = 349.64K and the correct answer is 348.35K. So it is close, but I feel given the precision of the answer, what I have done is not the required way.

Alternatively, using two points and computing the slope and then obtaining the linearisation gives $P = 1.286P_{TP} - 0.7$ which is a better approximation than the above. Then $$\frac{P}{P_{TP}} = \frac{1.286P_{TP} - 0.7}{P_{TP}} = 1.286 - \frac{0.7}{P_{TP}}$$ As I take increasingly smaller values of $P_{TP}$ (tending to zero), this tends to infinity. But this does not make sense.

Last edited: Jul 7, 2013
4. Jul 8, 2013

### CAF123

Can anyone help explain the above observation?

5. Jul 8, 2013

### voko

As suggested by Redbelly98, doing a linear fit via least squares is the way.

6. Jul 8, 2013

### CAF123

Thanks voko, the linear fit was $P = 1.2936P_{TP} - 1.85$, but I am still having the same problem as above. When I put this into $P/P_{TP}$, I obtain $$1.2936 - \frac{1.85}{P_{TP}}$$As $P_{TP} \rightarrow 0,\,\,P/P_{TP} \rightarrow \infty$, which means T tends to infinity, which doesn't make sense. Where did I go wrong?

7. Jul 8, 2013

### voko

Make your input variable x = 1, 2, 3, 4, and the measured variable y the corresponding ratios as per #1.

8. Jul 8, 2013

### CAF123

I don't think I understand what you mean. Do you mean set the expression $1.2936 - \frac{1.85}{P_{TP}}$ equal to the four ratios that I calculated?

9. Jul 8, 2013

### voko

Let your measured variable be $y = p/p_{TP}$. You have $y_i = \{1.279, \ 1.2825, \ 1.286, \ 1.29\}$. Let $x_i = \{1, \ 2, \ 3, \ 4\}$. Assume $y = ax + b$. Find $a$ and $b$ via least squares. You only need $b$ to solve this problem.

But then you need to be able to explain why that works :)

10. Jul 8, 2013

### CAF123

I can see that this works and gives the required answer, but I can't really see why you did it. Why was the method I did not work?
The equation is $$\frac{P}{P_{TP}} = 3.65 \times 10^{-3} x + 1.27525$$ Using the values of $x_i = \left\{1,2,3,4\right\}$ and $P_{TP_i} = \left\{100,200,300,400\right\},$ I obtain the values of pressure in the question. So I can see that it works but I am not sure why it works and the motivation behind it.

11. Jul 8, 2013

### voko

In your method, you were essentially building $p = a_0 + a_1 p_{PT}$, which you would then use in $\frac {a_0 + a_1 p_{PT}} {p_{PT}} |_{p_{PT} = 0}$. This will obviously never work no matter how you build $a_0 + a_1 p_{PT}$. Besides, we known that physically $p$ must go to zero, which it does not in this approximation. We could instead build $p = A_1 p_{PT} + A_2 p_{PT}^2$, then $\frac {A_1 p_{PT}+ A_2 p_{PT}^2} {p_{PT}} |_{p_{PT} = 0} = A_1$, which is OK. But then we could use just take $p/p_{PT} = A_1 + A_2 p_{PT} = ax + b$, where $p_{PT} = cx$.

Perhaps more intuitively, and it is hinted in the problem, $p/p_{PT}$ is well behaved as the mass approaches zero, plus it is what we really need, so it makes sense to approximate that directly.

12. Jul 8, 2013

### CAF123

Okay, so my method does not work because the end result does not conform with what we expect physically.

So $x = p_{TP}/c$, where $a = cA_2$ and you arbitrarily chose c such that when you divide each of the given triple points by c, you get 1,2,3 and 4?

Really nice method

Forgetting what the answer is supposed to be, in my first attempt taking p ≈ 1.28pTP, in the end when I evaluate p/pTP, I obviously do get some constant $A_1$ (namely 1.28). This gave quite a close result. Is there anything fundamentally incorrect here or does it suffice as a naive approximation?

Thanks.

13. Jul 9, 2013

### voko

In fact, there could be any sort of linear dependency between $x$ and $p_{TP}$. This reflects the fact that units are ultimately arbitrary.

I would not talk about anything fundamentally incorrect, but it does ignore the obvious trend in the data.

14. Jul 9, 2013

### CAF123

Yes, I see that the scaling of x can be arbritary.

For example, by taking y = P/PTP and $x_i = \left\{100,200,300,400\right\}$ (i.e c = 1), I obtain the same result (for b in y = ax + b).

Is there an underlying reason why finding p = ... and then subbing into p/pTP gives non-physical results?

15. Jul 9, 2013

### voko

The question assumes that finding $p$ as a function of $p_{TP}$ would always produce something incorrect. That is not so. As shown in an earlier post, you can find such a dependency, but it must have $p = 0$ when $p_{TP} = 0$.

16. Jul 9, 2013

### CAF123

I see. Why is it that we must have $p = 0$ at $p_{TP} = 0$? In my book it shows that as $P_{TP} \rightarrow 0$, all gases converge to a limiting value (373.15K) when concerning the steam point of water, but the external pressure is 1 atm ≠ 0?

17. Jul 9, 2013

### voko

Recall that the ideal gas scale works by assuming the ideal gas law becomes an increasingly better approximation of a real gas as pressure approaches zero.

Perhaps I misunderstand you, but what does the "external pressure" have to do with the ideal has scale?

18. Jul 9, 2013

### CAF123

Okay, this is expressed mathematically in my book as $$T = 273.16 \times \lim_{P_{TP} \rightarrow 0} \frac{P}{P_{TP}} K$$ Why is it they have defined T in terms of $P_{TP}$ going to zero and not $P$ going to zero (other than the fact that as the triple point goes to zero, all gases agree on a particular temperature)? (i.e you mention above about the pressure going to zero rather than the pressure at the triple point going to zero)

Actually, it is external pressure so indeed has nothing to do with the ideal gas scale.

19. Jul 9, 2013

Hello CAF. Try plotting a graph of P/Ptp against P and extrapolate to P=0. Using the same axes plot P/Ptp against Ptp and and extrapolate to Ptp=0. By doing this you might get a clearer picture of the procedure.

20. Jul 9, 2013

### voko

For that to be a finite limit, $P$ must necessarily go to zero.

Either one would be fine. The procedure is that the pressure is measured at two temperatures, then some gas is removed, and the process repeated. That makes pressure decrease no matter what temperature.

21. Jul 9, 2013

### CAF123

Thank you voko, it makes sense to me. If I wanted to obtain the condition $p \rightarrow 0$ as $p_{TP} \rightarrow 0$ with a linear model and the correct answer, then I would need that $P = 1.27525P_{TP}$. Is there anyway to extract this from my previous linear fit $P = 1.2936P_{TP} - 1.85$?

@Dadface I drew the graphs via Excel and I have attached it. I am not entirely sure what you wanted me to realize - is it just that $P/P_{TP}$ tends to a (very small, but necessarily postive) finite value as both p and p_{TP} tend to zero?

#### Attached Files:

• ###### Pressure Graph.png
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3.2 KB
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22. Jul 9, 2013

### voko

Try adding another datum, with both pressures zero.

23. Jul 9, 2013

### CAF123

So instead of having four data points I have five data points with (0,0) as well. If I try a linear fit, I will end up with a non zero P intercept, so did you have something else in mind for me to try?

24. Jul 9, 2013

### voko

On a second thought, I think the real issue here is that one cannot expect that a generic polynomial fit for experimental data will produce an intercept that is zero exactly. Even if we add additional data.

To make it zero exactly, we just have to let it be zero exactly, i.e., look for $p = (...) p_{TP}$.

25. Jul 9, 2013

### CAF123

Why is this the case? When I type in 'linear fit through origin' into google, there appears (although I cannot be sure since I have no prior experience) to be a way using MATLAB.

Okay, but it is not possible to obtain the value 1.27525 as the gradient? The best I could do would be to average like I did in #1?