How Do You Calculate Ideal-Gas Temperature from Mercury Pressure Data?

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SUMMARY

The discussion focuses on calculating the ideal-gas temperature (Tg) of mercury using pressure data. The formula used is Tg(K) = 273.16 lim Pt ->0 (P/Ptp), where Ptp represents the reference pressures. Participants recommend evaluating the ratios of P/Ptp for each pressure pair and observing the trend as P approaches zero. The final calculated temperature is approximately 419.56 K.

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  • Understanding of ideal gas laws and temperature scales
  • Familiarity with pressure measurements and units
  • Basic graphing skills to analyze data trends
  • Knowledge of limits in calculus for evaluating P/Ptp
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  • Learn about the ideal gas law and its applications in thermodynamics
  • Study how to graph and interpret data trends in scientific experiments
  • Explore calculus concepts related to limits and their practical applications
  • Investigate the properties of mercury as a thermometric fluid
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Homework Statement



I have been given measurements of Ptp and corresponding values of P for mercury. My task is to calculate the idea-gas temperature Tg of the material to five significant figures.

Ptp| 1000.0|750.00|500.00|250.00|
P |1535.3 |1151.6| 767.82| 383.95|

Homework Equations



Tg(K) = 273.16 lim Pt ->0 (P/Ptp)

The Attempt at a Solution



Ok i have read up on the ideal gas temperature scale, and sort of grasp it. I have graphed the table and got and equations relating the line to be:

P = 1.5351x +0.21

I don't believe that this is the right way to go about this problem. I am just asking for some direction, as i have hit a wall.

Thanks Zac
 
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I'm assuming that the 'material' is mercury.

I recommend you evaluate the ratios P/Ptp for each of the four pairs. Take 6 sig figs, i.e. 5 decimal places. You should notice a simple quantitative trend. So what looks to be the limit as P approaches zero?

Say if you get stuck.
 
Thanks Philip, i managed to get it sorted, i found an approximation for Lim Ptp->0 (P/Ptp), when i graphed P/Ptp Vs Pt so then was able to approximate the limit to find out my temperature.
Thanks heaps
 
419.56 k ?
 
yep, exactly
 

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