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The following statement explains the main source of error in a constant volume gas thermometer:

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Figure 1 : Constant volume gas thermometer

In constant volume gas thermometer, the gas in the capillary tube [$BC$] connecting the bulb [$A$] and the manometer [$CD$] remains outside the heat bath. The temperature of this part is different from the main bulk of the gas and this introduces some error.

Later, the author says that "Callendar's compensated constant pressure thermometer" avoids this problem by a special design as given in the diagram below:

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Figure 2 : Callendar's compensated constant pressure thermometer

Here, bulbs $A$ and $C$, and connecting tubes $cd$ and $ef$ are of same volumes. $M$ is the manometer whose arms are connected to $cd$ and $ef$. Bulb $B$ is filled with mercury. And mercury can be taken out through the valve $S$. Both bulbs $A$ and $C$ are filled with the same ideal gas of equal amounts. When all the parts of the thermometer are at the same temperature the level of mercury in the arms on the manometer $M$ remains equal.

The following statement explains the procedure to determine temperature using this type of thermometer:

To measure the temperature of a heat bath (at a temperature larger than the ice point), the bulb $A$ is placed in the heat bath and the bulbs $B$ and $C$ are placed in melting-ice baths.

I also understood the mathematical derivation for the following formula:

$$T=\frac{V}{V-v'}T_0$$

where, $T$ is the temperature of the heat bath, $V$ is the volume of bulbs $A$ and $C$, $v'$ is the volume of mercury taken out of the bulb $B$ and $T_0$ is the temperature of the ice bath.

I don't understand why we must keep the bulb $B$ in the ice bath along with bulb $C$? While doing so, it seems we're again inducing errors due to non constant temperature of the gas sample as it happened in the constant volume gas thermometer. Or in other words, I feel the temperature of the gas in bulb $A$, tube $cd$ and bulb $B$ to vary drastically which will again lead to errors.

So, what is the reason behind keeping the mercury bulb $B$ in ice bath? Why not keep it in some other temperature, say, either room temperature, or the heat bath in which bulb $A$ is placed?


All quotes and images are from the book "Concepts of Physics" by Dr. H.C.Verma, chapter 23, "Heat and Temperature".

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  • $\begingroup$ I wonder if it is an error in Dr Verma's book. It would make more sense for A & B to be kept in the melting ice bath while C is kept in the thermal bath. But the description at toppr also has A in thermal bath and B,C compensated. Temperature of capillaries cd and ef must be same but can be different from that of A, B, C. $\endgroup$ Commented Mar 31, 2020 at 20:57
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    $\begingroup$ @sammygerbil: Thanks for the comment! "It would make more sense for A & B to be kept in the melting ice bath while C is kept in the thermal bath." - Doing so, the pressure in bulb $C$ will be greater than the pressure in bulb $A$, and hence the level of mercury in the left arm of the manometer will be higher than that of the right arm. Even after removing mercury from bulb $B$, we cannot attain the equilibrium point (when the levels of mercury is same in both arms of the manometer) as this would further decrease the pressure in bulb $A$ (or bulb $A+$ tube $cd+$bulb $B$ to be more precise). $\endgroup$
    – Vishnu
    Commented Apr 1, 2020 at 3:22
  • $\begingroup$ Yes, you are right. That would not work. $\endgroup$ Commented Apr 1, 2020 at 13:03

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Even I too didn't understand how calendar's thermometer overcomes error. But if we keep bulb $B$ at room temperature, the liquid mercury filled in bulb $B$ will evaporate and mercury vapors are formed. This would alter the pressure in thermometric part of thermometer as there are two different kinds of gases in $A$, cd and $B$ (one the gas filled in capillary tube cd and bulb $A$ and another mercury vapor). Because of this we surely cannot keep bulb $B$ in room temperature. But I still don't understand how it reduces error.

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