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?

# [Physics] Costant volume gas thermometer

homework-and-exercisesideal-gasthermodynamics

## Best Answer

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.