Why must we use absolute temperature for the Ideal Gas Law?

[member 731016]

Homework Statement

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Relevant Equations

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For this problem,

The solution is,

However, why must we use absolute temperature for the ideal gas law (i.e why can we not use Celsius for T)

Many thanks!

 

[Grelbr42]

The origin of the Celsius scale was chosen by https://en.wikipedia.org/wiki/Anders_Celsius

The origin of the Fahrenheit scale was chosen by https://en.wikipedia.org/wiki/Daniel_Gabriel_Fahrenheit

They chose the origin of the temperature scales based on convenience and application to problems they were working on. Things like constructing a thermometer that was useful for everyday experience.

The ideal gas law needs to consider ratios of temperatures under different conditions. If you use either C. or F. then you have this choice of origin. If you changed the origin you could get different results for the same physical situation. For example, suppose you arbitrarily invented the ChiralSuperfields temperature scale, which was just the Celsius number plus 10. If you used the ChiralSuperfields temperature scale in the calculation, you would get different numbers when you took ratios of temperatures. The same exact situation would give you different results. That is clearly wrong.

Only choosing a temperature scale with its origin at absolute zero will let you avoid that. That is because absolute zero is a physical thing. One aspect of how it is physical is, it's the temperature that works for the origin of the temperature in the ideal gas law.

The size of the degree in these two systems is different. That does not matter. It only changes the units of the results you report. The R in PV = nRT also has units, so you only need to get the value of R in the other units. Or, if you are doing ratios of temperatures the units will divide out.

 

[kuruman]

However, why must we use absolute temperature for the ideal gas law (i.e why can we not use Celsius for T)

What would happen if you use the Celsius scale and the temperature drops to zero Celsius in the room you are in? If the ideal gas law went by the Celsius scale, the pressure in the room will be reduced to zero (at constant volume), you will find yourself in a perfect vacuum and most likely explode.

 

[TeethWhitener]

What would happen if you use the Celsius scale and the temperature drops to zero Celsius in the room you are in? If the ideal gas law went by the Celsius scale, the pressure in the room will be reduced to zero (at constant volume), you will find yourself in a perfect vacuum and most likely explode.

And if you open the freezer, you get sucked into a wormhole!

 

[Lnewqban]

There are no negative Kelvin degrees.

Please, see:
https://en.wikipedia.org/wiki/Kelvin#Lord_Kelvin

 

[DrClaude]

There are no negative Kelvin degrees.

Please, see:
https://en.wikipedia.org/wiki/Kelvin#Lord_Kelvin

There are no Kelvin "degrees". The kelvin is a unit, like the newton or the joule, and there is no reason to affix it the term degree.

Second, temperatures in kelvin can be negative, and then correspond to a system that is "hotter" than any system with a positive temperature, see https://en.wikipedia.org/wiki/Negative_temperature

 

[nasu]

While the explanations about the meaning and usefulness of the Kelvin scale are probably useful for the OP, they seem to imply that the Kelvin scale is required for the gas law. The answer to the OP should be that the law as is written assumes temperature in Kelvin scale and this is why you need to use the temperature in Kelvin scale. If it were written for other system of units, then you have to use these units. So we need to use these units simply because the formula was devised for these units. But the law can be formulated for any unit system. Actually, one of the first formulations (if not the first) by Clapeyron in 1834 does not mention the absolute scale. The formula he gives is $pv=R (267+t)$.

And same is true for any other formula. If you want to get the force in newtons from newton's second law written for SI units, you need to use mass in kg and acceleration in m/s². But there is no deep meaning of using these units. If you want to use other units you just have to use the formula devised for that system of units.

 

[jbriggs444]

Suppose that you use a meter stick that is ruled with a label "10 cm" at the left end and a label "110 cm" at the right end.

You line up the left end of your meter stick with the left edge of a rectangular piece of plywood. You see that the right edge is at 40 cm.

You line up the left end of your meter stick with the bottom edge of the plywood. You see that the top edge is at 70 cm.

You compute the area of the plywood as 2800 cm².

Your friend computes the area as 1800 cm².

Which one of you is correct? Why?

This is what happens when you use numeric results from a scale that does not have a suitably defined zero point.

 

[TonyStewart]

The celsius scale defines zero as the transition between ice and pure H2O water.

The Kelvin scale defines zero as the lower limit of temperature without any heat. There is a lot of volume in outer space without much heat. You can call that -273.15 'C or 0 'K but in terms of ratios, you must use a scale that starts from "absolute zero".

An ideal gas molecule is just a point charge without heat and consumes no volume, also doesn't move, collide, attract or repulse other molecules.

 

[nasu]

Suppose that you use a meter stick that is ruled with a label "10 cm" at the left end and a label "110 cm" at the right end.

You line up the left end of your meter stick with the left edge of a rectangular piece of plywood. You see that the right edge is at 40 cm.

You line up the left end of your meter stick with the bottom edge of the plywood. You see that the top edge is at 70 cm.

You compute the area of the plywood as 2800 cm².

Your friend computes the area as 1800 cm².

Which one of you is correct? Why?

This is what happens when you use numeric results from a scale that does not have a suitably defined zero point.

This would happen if the formula used is not appropriate for the system used for measurements. For the person with the "unusual" meter stick, the area should be calculated as $A=(h-10)(w-10)$ where h and w are the height and width of the rectangle as shown by his meter stick. If he uses the right formula he gets the same area of 1800 cm as the other person. He can get the right formula by experiments.
I did not say that is not easier to use a scale with a suitable zero point. Just that it is not a necessary condition.

And the same applies to the gas law written for temperatures in degrees Celsius. Clapeyron used experimental results to find a relationship between the quantities. He put togeteher several numerical terms into just one which he called "R". These people who found the gas laws for the first time did not use absolute scale to start with. But they got the results right.

 

[Tom.G]

I'm not dis-agreeing with it, but this logic seems dangerously close to a perfect circle!

Only choosing a temperature scale with its origin at absolute zero will let you avoid that. That is because absolute zero is a physical thing. One aspect of how it is physical is, it's the temperature that works for the origin of the temperature in the ideal gas law.