# Using the Ideal Gas temperature scale

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?
The gas used in the thermometer is real. It is not ideal and use of the ideal gas equatione: (T=273.16 P/Ptp) for small pressures gives approximately correct answers only. As P tends to zero the gas approaches ideal behaviour giving increasingly accurate values for the temperature. I think the graphs illustrate this trend nicely. Perhaps you could try extrapolating your graphs to P= zero when the gas behaves ideally. You can use the value of the intercept to find an accurate value for T.

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.
Note I said "a generic polynomial'. If you require that the fit pass through the origin, that will be a specialized polynomial and you can of course do that. In fact, you did just that.

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?
Not from a first degree polynomial, I am afraid. You would have to go for ##A_1 p_{TP} + A_2 p_{TP}^2##.

@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?
P/Ptp tends to the value you need to answer the question. As P tends to zero the gas tends to ideal behaviour where you can use the equation with greatest accuracy.

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Note I said "a generic polynomial'. If you require that the fit pass through the origin, that will be a specialized polynomial and you can of course do that. In fact, you did just that.

Not from a first degree polynomial, I am afraid. You would have to go for ##A_1 p_{TP} + A_2 p_{TP}^2##.
Using a polynomial of order 2 in Excel gives an equation y = 4E-5x2 + 1.2749x, which gives a more accurate number for T. (although looking at the data before the fit, it seems strange to approximate that quadratically) (see attached)

@Dadface: Yes, thanks, I extrapolated the data back and got the correct intercepts.

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Using a polynomial of order 2 in Excel gives an equation y = 4E-5x2 + 1.2749x, which gives a more accurate number for T. (although looking at the data before the fit, it seems strange to approximate that quadratically) (see attached)
The quadratic correction is indeed small, but not negligible given the accuracy of the data.