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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.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?