# Costant volume gas thermometer [closed]

I have two constant volume gas thermometers, one with an ideal gas $pv=RT$ and the other following the Van der Waals equation $(p+\frac{a}{v^2})(v-b)=RT$. They are calibrated using the freezing point and the boiling point of water. Will they show the same temperatures?

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## closed as off-topic by Michael Brown, Nathaniel, Dilaton, user1504, akhmeteliJul 10 '13 at 3:16

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There will be a difference. However I am not sure how big this will be and whether it would be bigger than the uncertainty of the measurements. It will also matter what kind of gas the thermometers use, since helium can be approximated by the ideal gas law much better than bigger molecules like carbon-dioxide or methane. – fibonatic Jul 8 '13 at 13:48

Yes they will show the same temperature.

For the ideal gas:

The temperature as a function of pressure is given by $$T=\frac VR P$$

As you can see this is a linear graph($V/R$ is constant). If you calibrate it using water's boiling and freezing point, and make hundred divisions in the range, you will get accurate values of temperature for a linear change in pressure.

For the gas following Van der Waal's equation:

The temperature as a function of pressure is given by $$T=\frac{V-b}{R} P + \frac{(V-b)a}{RV^2}$$. As you can see this is also a linear function(of the form $T=mP+c$). Calibrating this with water's boiling point and freezing point will give the same temperature readings for linear change in pressure, just as before.

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