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At the end of the day what the thermometer is measuring as temperature is energy of the air molecules (which could come in the form of kinetic energy). Now, imagine the following scenario :

  1. Take a box with just one gas molecule (at speed x). It goes and hits the mercury of the thermometer. Thermometer will probably not record its proper temperature.
  2. Now fill the box with a million molecules (at the same speed x), and the thermometer records right some temperature.
  3. Now double the density (but keep the speed of individual air molecules the same). Fill the box with 2 million molecules, but keep the average speed of the molecules the same.

Will it record the same temperature?

My personal intuition is that the temperature recorded should increase, since more molecules are giving energy to that thermometer in the same amount of time.

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    $\begingroup$ I have posted a new answer. Hope it helps $\endgroup$
    – Bob D
    Commented Sep 9, 2020 at 20:12
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    $\begingroup$ A million (or two million) molecules isn’t really enough to give your box a measurable temperature. Unless it’s microscopically small, it will be a pretty good vacuum. Try using an Avogadro number of molecules (approx. 6E23). $\endgroup$
    – Mike Scott
    Commented Sep 10, 2020 at 7:32

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Assuming an ideal gas, if you are keeping the average speed of the molecules the same then you are holding the temperature of the gas constant. Assuming the container keeps the same volume, by the ideal gas law it must be that the pressure of the gas increases as you add more molecules.

Yes there are more molecules hitting the thermometer, but that also means there are more molecules hitting the thermometer. What I mean by this is that energy can be transferred to the thermometer at a higher rate due to collisions, but collisions at the same increased rate will then transfer that energy from the thermometer back to the gas (and the same vice versa). This is how thermal equilibrium works.

So no, just because you have more gas at the same temperature doesn't mean you will record a higher temperature. More gas will just mean fewer fluctuations about the same temperature.

In addressing points made in the comments, technically the temperature of the thermometer will be somewhere between the starting temperature of the thermometer and the starting temperature of the gas, and the final temperature of the thermometer will approach the starting temperature of the gas as more gas is let in. However, there will come a point where additional gas will make the final temperature indistinguishable from the initial gas temperature, and ideally the thermometer shouldn't influence the temperature of the gas.

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  • $\begingroup$ I've added a comment to the answer by Bob D. I think the same comment will work for your explanation. Pressure, I concede will increase. But the equilibrium kinetic energy of all the molecules will increase if I'm adding more molecules that are on the higher energy side to the system. That should warrant a temperature increase. $\endgroup$
    – Kshitij
    Commented Sep 9, 2020 at 16:01
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    $\begingroup$ @Kshitij You said the added molecules have the same average speed, i.e. the same temperature $\endgroup$
    – BioPhysicist
    Commented Sep 9, 2020 at 16:11
  • $\begingroup$ You are talking macroscopically. I'm referring to whats actually going on between the air molecules and the mercury. Initially the air molecules are at some temperature T (lets say), and mercury at some temperature T ' lets say (without losing generality, assume T > T ' ). Thermal equilibrium between the molecules and the mercury will decide a new temperature Y that will be between T and T ' which will be the recorded temperature. Addition of more molecules which are at temperature T, will lead to a new recorded temperature Y ' which is more towards T than Y? $\endgroup$
    – Kshitij
    Commented Sep 9, 2020 at 16:18
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    $\begingroup$ @BioPhysicist What you mean in 2nd paragraph is that some molecules with lower energy take some heat away. This means their energies increase after collision , won't this require a coefficient of restitution greater than 1. Can you tell how energy is transferred back to gas? $\endgroup$
    – Protein
    Commented Sep 9, 2020 at 16:48
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    $\begingroup$ @Protein Even in elastic collisions energy can be transferred between objects. $\endgroup$
    – BioPhysicist
    Commented Sep 9, 2020 at 17:42
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Now double the density (but keep the speed of individual air molecules the same). Fill the box with 2 million molecules, but keep the average speed of the molecules the same. Will it record the same temperature?

The actual temperature of the gas will be the same prior to making a reading, but the final temperature of the gas and thermometer after making the reading will depend on the heat capacity of the gas before and after doubling the number of molecules as compared to the heat capacity of the thermometer (fluid plus glass), which is fixed.

If both before and after doubling the number of molecules the heat capacity of the gas is much greater than the thermometer, then the actual temperature of the gas should not change due to the act of measurement, and the thermometer reading should not change.

Let's say the heat capacity of the gas is $C_H$ and the heat capacity of the thermometer, is $C_L$. Before making the measurement the actual temperature of the gas is $T_H$ (higher temperature) and the temperature of the thermometer before making the measurement is $T_L$, (lower temperature). Then the final equilibrium temperature $T$ of each will be

$$C_{H}(T_{H}-T)=C_{L}(T-T_L)$$

Now let the heat capacity of each before doubling the molecules be equal, or $C_{H}=C_{L}$, then

$$T=\frac{(T_{H}+T_L)}{2}$$

Now we double the number of gas molecules but the pre-measurement temperature is the same $T_H$. But the heat capacity heat of the gas is now double that of the thermometer, or $C_{H}=2C_L$. The final equilibrium is now

$$T=2/3T_{H}+1/3T_L$$

Which is higher than before doubling the number of molecules and is closer to the actual pre-reading temperature of the gas, $T_H$.

The more molecules you add the closer the final equilibrium temperature of the thermometer is to $T_H$, until the heat capacity of the gas is so much higher than the thermometer that the addition of more gas at the same pre-measurement temperature has no effect on the final temperature of either one. If there are 1000 times more molecules than the original then $$T=\frac{1000T_H}{1001}+\frac{T_L}{1000}≈T_H$$

The above said, keep in mind that in order for a thermometer to measure the actual temperature of something (i.e., the temperature that something has before applying the device), the heat capacity of what is being measured must be much greater than the heat capacity of the liquid (and glass) of the thermometer so that the temperature of what is being measured does not change.

It is kind of a cardinal rule in making any measurement that the act of making the measurement should have a minimum effect on what is being measured. If doubling the amount of gas having the same pre-reading temperature gives you a different temperature reading, then perhaps you are using the wrong temperature measuring device.

Hope this helps.

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Temperature of a gas is a measure of average speed of molecules. Since average speed of gas molecules is constant so is the temperature.

The increased density would increase thermal conductivity and not temperature (as avg speed of molecules is kept constant).The thermometer will attain equilibrium temperature faster and thus record temperature of gas in lesser time but the temperature recorded will be the same.

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Your question seems to be about what the equilibrium temperature recorded by the thermometer will be after being placed with the gas.

Your intuition that 'the temperature recorded should increase, since more molecules are giving energy to that thermometer in the same amount of time' entirely depends on the thermometer starting at a colder temperature than the gas. In this case the thermometer will cool the gas with fewer molecules by a larger amount before reaching equilibrium.

If the thermometer starts and the same temperature of the gas there would be no net change. The thermometer will read the same in both instances.

If the thermometer started at a higher temperature than the gas then the eventual reading of the thermometer would be lower if there were more gas molecules, not higher. The thermometer would bring the gas with fewer molecules to a higher temperature before reaching equilibrium.

Doing some napkin math, your example is almost meaningless because there are so few gas molecules. The temperature changes from thermal radiation would outweigh any change due to the gas. If you want to talk about how the temperature reading would be affected, you need to specify what thermometer you are using. 1 million or 2 million gas molecules would have little impact on the thermometer. At this point the container obviously becomes important, as does the method of inserting the thermometer, the method of adding gas atoms to make sure they are all moving the same average speed, thermal radiation, etc.

Just the 0.61 g of mercury in a normal thermometer has about 1.83E21 molecules. With a specific heat of 0.14 J/gC it would take 0.0854 J to raise the temperature by 1 degree celsius. If your gas was a mixture of air, 1 million molecules would be about 3E-17 grams. At a specific heat of 1.02 J/gC that means for just the mercury in the thermometer to go up by 1 degree it would have to absorb 0.0854 J, which would decrease the temperature of the gas by 2.8E15 (2,800,000,000,000,000) degrees celsius.

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  • $\begingroup$ I understand your concern, but here we can arbitrarily decrease the size of the mercury thermometer to suit the experiment. Feel free to take 5 Avogadro number of gas molecules and then make them 10 Avogadro number of gas molecules likewise for the two cases. Also, citing your comment "your example is almost meaningless because there are so few gas molecules" implicitly agrees with the fact that temperature measurements change with the number of molecules (or likewise, regular experimental thermometers like mercury ones only work within a defined range of air density or pressure.) $\endgroup$
    – Kshitij
    Commented Sep 11, 2020 at 9:08
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The denser gas should bring the thermometer to the same temperature, but faster.

When there is only one molecule in the box, it doesn't have enough total energy to affect the thermometer. This is a thermometer failing. Our intention is to bring the thermometer to the equilibrium temperature without affecting the gas we're measuring. When there are a whole lot of gas molecules, that's true enough that the error doesn't matter. When there are only a few, the error gets big.

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You seem to be talking about a traditional thermometer, such as a mercury thermometer (there are other devices that could be considered to be thermometers, such as infrared sensors). The idea behind a traditional thermometer is that it is a device whose temperature is easily seen. When such a device is allowed to reach thermal equilibrium with another substance, the thermometer's temperature is the same as the substance, and so we can determine the temperature of the substance by examining the thermometer.

Thermal equilibrium means that when molecules from the thermometer and the substance collide, there is on average no net transfer of energy. Sometimes energy will be transferred to the thermometer, and sometimes energy will be transferred from the thermometer, and overall they cancel out. Having more molecules means more collisions, but since each collision is on average not transferring energy, the increase in collisions will not affect the temperature.

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