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The significance of kelvin as a unit of absolute temperature
Is the Boltzmann constant really that important?Temperature below absolute zero?Thermodynamics, temperature below 0 KelvinAre there reasons for the discrepancies in absolute temp units - Kelvin vs. kelvins vs. degrees Kelvin?minimum possible absolute temperature in the universe?Physical significance of temperatureBase unit definition and realization under relativistic conditionsThermal expansion of solid body and calculation of work producedHumans have an average energy budget of $100$ Watts, but the power radiated from the body is $1000$ Watts?Is absolute zero still 0 Kelvin?Why is the length of the Kelvin unit of temperature equal to that of the Celsius unit?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
$$PV=nRT$$
Where $P$ is pressure in pascals ($textPa$)
$V$ is volume in cubic metres ($textm^3$)
$n$ is amount of substance in moles ($textmol$)
$R$ is the gas constant having units, joules per mole kelvin
($text J K^-1text mol^-1$)
$T$ is the temperature in kelvins ($text K$)
The melting and boiling points of pure water are $0 ^circtext C $ and $100 ^circtext C $ respectively. $1 ^circtext C$ is $dfrac 1 100^textth$ of the difference in temperature.
1 kelvin and 1 degree Celsius represent the same difference in temperature.
However, why is it that in any formula (such as the one above) involving temperature (not temperature difference), plugging in values will result in a nice and correct temperature?
For example, $P=100 mathrm$, $V=8.314 mathrm$, $n=1 mathrm$, $R=8.314 mathrm $
Plugging in the values will result in $T=100 $
$100$ kelvin.
But what is the significance of 100 kelvin? Why could it not have been any other unit in an absolute temperature scale? e.g. 100 degrees Rankine
In simple formulas such as
$$textDisplacement = Velocity times textTime$$
I can understand that metres would be obtained when metres per second is multiplied with seconds. It does not make sense to obtain inches from the multiplication between metres per second and seconds.
thermodynamics temperature units si-units
asked Jun 8 at 10:12
helpmehelpme
9710 bronze badges
$endgroup$
add a comment |
$begingroup$
$$PV=nRT$$
Where $P$ is pressure in pascals ($textPa$)
$V$ is volume in cubic metres ($textm^3$)
$n$ is amount of substance in moles ($textmol$)
$R$ is the gas constant having units, joules per mole kelvin
($text J K^-1text mol^-1$)
$T$ is the temperature in kelvins ($text K$)
The melting and boiling points of pure water are $0 ^circtext C $ and $100 ^circtext C $ respectively. $1 ^circtext C$ is $dfrac 1 100^textth$ of the difference in temperature.
1 kelvin and 1 degree Celsius represent the same difference in temperature.
However, why is it that in any formula (such as the one above) involving temperature (not temperature difference), plugging in values will result in a nice and correct temperature?
For example, $P=100 mathrm$, $V=8.314 mathrm$, $n=1 mathrm$, $R=8.314 mathrm $
Plugging in the values will result in $T=100 $
$100$ kelvin.
But what is the significance of 100 kelvin? Why could it not have been any other unit in an absolute temperature scale? e.g. 100 degrees Rankine
In simple formulas such as
$$textDisplacement = Velocity times textTime$$
I can understand that metres would be obtained when metres per second is multiplied with seconds. It does not make sense to obtain inches from the multiplication between metres per second and seconds.
thermodynamics temperature units si-units
asked Jun 8 at 10:12
helpmehelpme
9710 bronze badges
$endgroup$
5
$begingroup$
It may be useful to note that $R T$ has the units of energy, so the choice for units of T is really about what value you take $R$ to be.
$endgroup$
– jim
Jun 9 at 15:17
$begingroup$
I suggest you keep the units with the numbers. $R$ is not $8.314$. $R = 8.314 mathrmJcdotmathrmK^-1cdotmathrmmol^-1 = 1.104 mathrmcal_thcdotmathrm⁰R^-1cdotmathrmmol^-1$ (for example). That's true equals, because it is still the same constant with the same value, just expressed in different units.
$endgroup$
– Jan Hudec
Jun 10 at 13:43
1
$begingroup$
This post is relevant: Is the Boltzmann constant really that important?.
$endgroup$
– DanielSank
Jun 10 at 19:47
add a comment |
$begingroup$
$$PV=nRT$$
Where $P$ is pressure in pascals ($textPa$)
$V$ is volume in cubic metres ($textm^3$)
$n$ is amount of substance in moles ($textmol$)
$R$ is the gas constant having units, joules per mole kelvin
($text J K^-1text mol^-1$)
$T$ is the temperature in kelvins ($text K$)
The melting and boiling points of pure water are $0 ^circtext C $ and $100 ^circtext C $ respectively. $1 ^circtext C$ is $dfrac 1 100^textth$ of the difference in temperature.
1 kelvin and 1 degree Celsius represent the same difference in temperature.
However, why is it that in any formula (such as the one above) involving temperature (not temperature difference), plugging in values will result in a nice and correct temperature?
For example, $P=100 mathrm$, $V=8.314 mathrm$, $n=1 mathrm$, $R=8.314 mathrm $
Plugging in the values will result in $T=100 $
$100$ kelvin.
But what is the significance of 100 kelvin? Why could it not have been any other unit in an absolute temperature scale? e.g. 100 degrees Rankine
In simple formulas such as
$$textDisplacement = Velocity times textTime$$
I can understand that metres would be obtained when metres per second is multiplied with seconds. It does not make sense to obtain inches from the multiplication between metres per second and seconds.
thermodynamics temperature units si-units
asked Jun 8 at 10:12
helpmehelpme
9710 bronze badges
$endgroup$
$$PV=nRT$$
Where $P$ is pressure in pascals ($textPa$)
$V$ is volume in cubic metres ($textm^3$)
$n$ is amount of substance in moles ($textmol$)
$R$ is the gas constant having units, joules per mole kelvin
($text J K^-1text mol^-1$)
$T$ is the temperature in kelvins ($text K$)
The melting and boiling points of pure water are $0 ^circtext C $ and $100 ^circtext C $ respectively. $1 ^circtext C$ is $dfrac 1 100^textth$ of the difference in temperature.
1 kelvin and 1 degree Celsius represent the same difference in temperature.
However, why is it that in any formula (such as the one above) involving temperature (not temperature difference), plugging in values will result in a nice and correct temperature?
For example, $P=100 mathrm$, $V=8.314 mathrm$, $n=1 mathrm$, $R=8.314 mathrm $
Plugging in the values will result in $T=100 $
$100$ kelvin.
But what is the significance of 100 kelvin? Why could it not have been any other unit in an absolute temperature scale? e.g. 100 degrees Rankine
In simple formulas such as
$$textDisplacement = Velocity times textTime$$
I can understand that metres would be obtained when metres per second is multiplied with seconds. It does not make sense to obtain inches from the multiplication between metres per second and seconds.
thermodynamics temperature units si-units
thermodynamics temperature units si-units
asked Jun 8 at 10:12
helpmehelpme
9710 bronze badges
asked Jun 8 at 10:12
helpmehelpme
9710 bronze badges
edited Jun 9 at 16:06
helpme
asked Jun 8 at 10:12
helpmehelpme
9710 bronze badges
asked Jun 8 at 10:12
helpmehelpme
9710 bronze badges
asked Jun 8 at 10:12
helpmehelpme
9710 bronze badges
9710 bronze badges
5
$begingroup$
It may be useful to note that $R T$ has the units of energy, so the choice for units of T is really about what value you take $R$ to be.
$endgroup$
– jim
Jun 9 at 15:17
$begingroup$
I suggest you keep the units with the numbers. $R$ is not $8.314$. $R = 8.314 mathrmJcdotmathrmK^-1cdotmathrmmol^-1 = 1.104 mathrmcal_thcdotmathrm⁰R^-1cdotmathrmmol^-1$ (for example). That's true equals, because it is still the same constant with the same value, just expressed in different units.
$endgroup$
– Jan Hudec
Jun 10 at 13:43
1
$begingroup$
This post is relevant: Is the Boltzmann constant really that important?.
$endgroup$
– DanielSank
Jun 10 at 19:47
add a comment |
5
$begingroup$
It may be useful to note that $R T$ has the units of energy, so the choice for units of T is really about what value you take $R$ to be.
$endgroup$
– jim
Jun 9 at 15:17
$begingroup$
I suggest you keep the units with the numbers. $R$ is not $8.314$. $R = 8.314 mathrmJcdotmathrmK^-1cdotmathrmmol^-1 = 1.104 mathrmcal_thcdotmathrm⁰R^-1cdotmathrmmol^-1$ (for example). That's true equals, because it is still the same constant with the same value, just expressed in different units.
$endgroup$
– Jan Hudec
Jun 10 at 13:43
1
$begingroup$
This post is relevant: Is the Boltzmann constant really that important?.
$endgroup$
– DanielSank
Jun 10 at 19:47
5
5
$begingroup$
It may be useful to note that $R T$ has the units of energy, so the choice for units of T is really about what value you take $R$ to be.
$endgroup$
– jim
Jun 9 at 15:17
$begingroup$
It may be useful to note that $R T$ has the units of energy, so the choice for units of T is really about what value you take $R$ to be.
$endgroup$
– jim
Jun 9 at 15:17
$begingroup$
I suggest you keep the units with the numbers. $R$ is not $8.314$. $R = 8.314 mathrmJcdotmathrmK^-1cdotmathrmmol^-1 = 1.104 mathrmcal_thcdotmathrm⁰R^-1cdotmathrmmol^-1$ (for example). That's true equals, because it is still the same constant with the same value, just expressed in different units.
$endgroup$
– Jan Hudec
Jun 10 at 13:43
$begingroup$
I suggest you keep the units with the numbers. $R$ is not $8.314$. $R = 8.314 mathrmJcdotmathrmK^-1cdotmathrmmol^-1 = 1.104 mathrmcal_thcdotmathrm⁰R^-1cdotmathrmmol^-1$ (for example). That's true equals, because it is still the same constant with the same value, just expressed in different units.
$endgroup$
– Jan Hudec
Jun 10 at 13:43
1
1
$begingroup$
This post is relevant: Is the Boltzmann constant really that important?.
$endgroup$
– DanielSank
Jun 10 at 19:47
$begingroup$
This post is relevant: Is the Boltzmann constant really that important?.
$endgroup$
– DanielSank
Jun 10 at 19:47
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You do not need to use the kelvins for the ideal gas formula. The important thing to know is that $PV propto nT$ where $T$ is the absolute temperature and can be expressed in any arbitrary linear scale (as long as the temperature is still absolute), including the kelvin and the rankine scale.
However, if you choose the constant of proportionality to be $R$ given in J/(K mol), then the temperature's units must be in kelvins for the equation to make sense. But you're not forced to use such a constant. You could pick up any other constant and you could adjust the temperature scale you're using. In particular you could have picked the constant that would make the temperature in rankine units.
Edit: Since the title of the question refers to "formulas", let me mention a case I've seen up to several papers. With regard to thermoelectric materials, there is a $ZT$ factor that's used to gauge the quality of the material to be used as a thermoelectric one, for example to make a thermoelectric generator. The important thing is that $T$ must be the absolute temperature. Unfortunately many papers mention that $T$ is given in kelvins, while it is not necessarily the case. Fortunately enough, I would say most (good) papers refer to $T$ as being the absolute temperature and do not mention the units.
answered Jun 8 at 13:19
thermomagnetic condensed bosonthermomagnetic condensed boson
2,9493 gold badges14 silver badges48 bronze badges
$endgroup$
3
$begingroup$
Follow-up question: Would the most natural absolute temperature unit be one where the constant of proportionality R=1?
$endgroup$
– helpme
Jun 9 at 4:20
2
$begingroup$
@helpme The answer is subjective and would only be "well suited" for the ideal gas relation. Not with, for example, the ZT relation I mention in my answer.
$endgroup$
– thermomagnetic condensed boson
Jun 9 at 7:44
6
$begingroup$
@helpme: "It's subjective" is probably not a satisfactory answer to your question. Thankfully there is a satisfactory answer and that answer is simply, no. :-) You'd instead want to use PV = NkT rather than PV = nRT, where k is the per-molecule normalized version of R, putting N in the natural units of "number of molecules". So the more natural choice would seem to be k = 1. And indeed this is what Planck units do... and if you continue this you get the Planck temperature as the natural unit.
$endgroup$
– Mehrdad
Jun 9 at 11:02
1
$begingroup$
The equation $PVpropto T$ (in the ideal gas limit) does not work for any scale. It only works for linear scales, that is scales where the reference points are linearly interpolated.
$endgroup$
– Diracology
Jun 10 at 17:59
$begingroup$
Good point @Diracology let me modify and improve the answer.
$endgroup$
– thermomagnetic condensed boson
Jun 10 at 19:32
add a comment |
$begingroup$
For example, 𝑃=100, 𝑉=8.314, 𝑛=1, 𝑅=8.314.
Plugging in the values will result in 𝑇=100
kelvin.
But what is the significance of 100 kelvin? Why could it not have been
any other unit in an absolute temperature scale? e.g. 100 degrees
Rankine
The answer is already contained in the first part of your question, when you write the units of the Gas Constant $R$. In order to get temperature in different units, one should use a different value and units for $R$. Similarly for a different thermometric scale.
Edit (improved the examples):
For example, in the case of Rankine degrees and using the same units for all other quantities, the perfect gas formula keeps the same form, but with a gas constant $R^prime=4.61915$. In the case of temperature, $theta$, in Celsius degrees, becomes:
$$
PV = ntilde Rtheta+ Q n,
$$
where $Q = 2.271 cdot 10^3 = 8.314*273.15$ J mol$^-1$, and $tilde R$ has the same numerical value of $R=8.314$, although now its units should be read as J mol$^-1$ $^circ$C$^-1$.
answered Jun 8 at 13:06
GiorgioPGiorgioP
6,2192 gold badges9 silver badges34 bronze badges
$endgroup$
$begingroup$
I don't think you should have $Rprime$ and $tilde R$ there. It is still the same constant, with the same value, even if expressed in different units (and then, you really should have units in the last equation, otherwise the dimensions don't match).
$endgroup$
– Jan Hudec
Jun 10 at 13:47
$begingroup$
@JanHudec I used different symbols for clarity. Even if the numerical values are the same, different units are used. I'll modify the way I am writing the last equation to avoid misunderstandings,
$endgroup$
– GiorgioP
Jun 10 at 17:46
add a comment |
$begingroup$
There is a point that the other answers haven't made, which I believe is important: Celsius and Farenheit both allow temperatures below (what they call) zero. If you plug those measurements into the formula, you get a value of (pressure times value) which is negative; this is not physically possible. Note also that negativeness cannot be compensated-for by a simple constant of proportionality (R). Kelvin at least has the virtue of never being negative-valued.
answered Jun 10 at 13:08
PMarPMar
111 bronze badge
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add a comment |
$begingroup$
Since a few weeks, the kelvin is a unit that is defined by setting Boltzmann's constant to exactly $1.380 649 cdot 10^-23$ J/K (see https://physics.nist.gov/cgi-bin/cuu/Value?k ).
For now, this does not change the practical temperature scale for calibrating thermometers around ambient temperatures: https://www.bipm.org/en/committees/cc/cct/guide-its90.html
answered Jun 10 at 19:47
PieterPieter
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You do not need to use the kelvins for the ideal gas formula. The important thing to know is that $PV propto nT$ where $T$ is the absolute temperature and can be expressed in any arbitrary linear scale (as long as the temperature is still absolute), including the kelvin and the rankine scale.
However, if you choose the constant of proportionality to be $R$ given in J/(K mol), then the temperature's units must be in kelvins for the equation to make sense. But you're not forced to use such a constant. You could pick up any other constant and you could adjust the temperature scale you're using. In particular you could have picked the constant that would make the temperature in rankine units.
Edit: Since the title of the question refers to "formulas", let me mention a case I've seen up to several papers. With regard to thermoelectric materials, there is a $ZT$ factor that's used to gauge the quality of the material to be used as a thermoelectric one, for example to make a thermoelectric generator. The important thing is that $T$ must be the absolute temperature. Unfortunately many papers mention that $T$ is given in kelvins, while it is not necessarily the case. Fortunately enough, I would say most (good) papers refer to $T$ as being the absolute temperature and do not mention the units.
answered Jun 8 at 13:19
thermomagnetic condensed bosonthermomagnetic condensed boson
2,9493 gold badges14 silver badges48 bronze badges
$endgroup$
3
$begingroup$
Follow-up question: Would the most natural absolute temperature unit be one where the constant of proportionality R=1?
$endgroup$
– helpme
Jun 9 at 4:20
2
$begingroup$
@helpme The answer is subjective and would only be "well suited" for the ideal gas relation. Not with, for example, the ZT relation I mention in my answer.
$endgroup$
– thermomagnetic condensed boson
Jun 9 at 7:44
6
$begingroup$
@helpme: "It's subjective" is probably not a satisfactory answer to your question. Thankfully there is a satisfactory answer and that answer is simply, no. :-) You'd instead want to use PV = NkT rather than PV = nRT, where k is the per-molecule normalized version of R, putting N in the natural units of "number of molecules". So the more natural choice would seem to be k = 1. And indeed this is what Planck units do... and if you continue this you get the Planck temperature as the natural unit.
$endgroup$
– Mehrdad
Jun 9 at 11:02
1
$begingroup$
The equation $PVpropto T$ (in the ideal gas limit) does not work for any scale. It only works for linear scales, that is scales where the reference points are linearly interpolated.
$endgroup$
– Diracology
Jun 10 at 17:59
$begingroup$
Good point @Diracology let me modify and improve the answer.
$endgroup$
– thermomagnetic condensed boson
Jun 10 at 19:32
add a comment |
$begingroup$
You do not need to use the kelvins for the ideal gas formula. The important thing to know is that $PV propto nT$ where $T$ is the absolute temperature and can be expressed in any arbitrary linear scale (as long as the temperature is still absolute), including the kelvin and the rankine scale.
However, if you choose the constant of proportionality to be $R$ given in J/(K mol), then the temperature's units must be in kelvins for the equation to make sense. But you're not forced to use such a constant. You could pick up any other constant and you could adjust the temperature scale you're using. In particular you could have picked the constant that would make the temperature in rankine units.
Edit: Since the title of the question refers to "formulas", let me mention a case I've seen up to several papers. With regard to thermoelectric materials, there is a $ZT$ factor that's used to gauge the quality of the material to be used as a thermoelectric one, for example to make a thermoelectric generator. The important thing is that $T$ must be the absolute temperature. Unfortunately many papers mention that $T$ is given in kelvins, while it is not necessarily the case. Fortunately enough, I would say most (good) papers refer to $T$ as being the absolute temperature and do not mention the units.
answered Jun 8 at 13:19
thermomagnetic condensed bosonthermomagnetic condensed boson
2,9493 gold badges14 silver badges48 bronze badges
$endgroup$
3
$begingroup$
Follow-up question: Would the most natural absolute temperature unit be one where the constant of proportionality R=1?
$endgroup$
– helpme
Jun 9 at 4:20
2
$begingroup$
@helpme The answer is subjective and would only be "well suited" for the ideal gas relation. Not with, for example, the ZT relation I mention in my answer.
$endgroup$
– thermomagnetic condensed boson
Jun 9 at 7:44
6
$begingroup$
@helpme: "It's subjective" is probably not a satisfactory answer to your question. Thankfully there is a satisfactory answer and that answer is simply, no. :-) You'd instead want to use PV = NkT rather than PV = nRT, where k is the per-molecule normalized version of R, putting N in the natural units of "number of molecules". So the more natural choice would seem to be k = 1. And indeed this is what Planck units do... and if you continue this you get the Planck temperature as the natural unit.
$endgroup$
– Mehrdad
Jun 9 at 11:02
1
$begingroup$
The equation $PVpropto T$ (in the ideal gas limit) does not work for any scale. It only works for linear scales, that is scales where the reference points are linearly interpolated.
$endgroup$
– Diracology
Jun 10 at 17:59
$begingroup$
Good point @Diracology let me modify and improve the answer.
$endgroup$
– thermomagnetic condensed boson
Jun 10 at 19:32
add a comment |
$begingroup$
You do not need to use the kelvins for the ideal gas formula. The important thing to know is that $PV propto nT$ where $T$ is the absolute temperature and can be expressed in any arbitrary linear scale (as long as the temperature is still absolute), including the kelvin and the rankine scale.
However, if you choose the constant of proportionality to be $R$ given in J/(K mol), then the temperature's units must be in kelvins for the equation to make sense. But you're not forced to use such a constant. You could pick up any other constant and you could adjust the temperature scale you're using. In particular you could have picked the constant that would make the temperature in rankine units.
Edit: Since the title of the question refers to "formulas", let me mention a case I've seen up to several papers. With regard to thermoelectric materials, there is a $ZT$ factor that's used to gauge the quality of the material to be used as a thermoelectric one, for example to make a thermoelectric generator. The important thing is that $T$ must be the absolute temperature. Unfortunately many papers mention that $T$ is given in kelvins, while it is not necessarily the case. Fortunately enough, I would say most (good) papers refer to $T$ as being the absolute temperature and do not mention the units.
answered Jun 8 at 13:19
thermomagnetic condensed bosonthermomagnetic condensed boson
2,9493 gold badges14 silver badges48 bronze badges
$endgroup$
You do not need to use the kelvins for the ideal gas formula. The important thing to know is that $PV propto nT$ where $T$ is the absolute temperature and can be expressed in any arbitrary linear scale (as long as the temperature is still absolute), including the kelvin and the rankine scale.
However, if you choose the constant of proportionality to be $R$ given in J/(K mol), then the temperature's units must be in kelvins for the equation to make sense. But you're not forced to use such a constant. You could pick up any other constant and you could adjust the temperature scale you're using. In particular you could have picked the constant that would make the temperature in rankine units.
Edit: Since the title of the question refers to "formulas", let me mention a case I've seen up to several papers. With regard to thermoelectric materials, there is a $ZT$ factor that's used to gauge the quality of the material to be used as a thermoelectric one, for example to make a thermoelectric generator. The important thing is that $T$ must be the absolute temperature. Unfortunately many papers mention that $T$ is given in kelvins, while it is not necessarily the case. Fortunately enough, I would say most (good) papers refer to $T$ as being the absolute temperature and do not mention the units.
answered Jun 8 at 13:19
thermomagnetic condensed bosonthermomagnetic condensed boson
2,9493 gold badges14 silver badges48 bronze badges
edited Jun 10 at 19:33
answered Jun 8 at 13:19
thermomagnetic condensed bosonthermomagnetic condensed boson
2,9493 gold badges14 silver badges48 bronze badges
answered Jun 8 at 13:19
thermomagnetic condensed bosonthermomagnetic condensed boson
2,9493 gold badges14 silver badges48 bronze badges
answered Jun 8 at 13:19
thermomagnetic condensed bosonthermomagnetic condensed boson
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2,9493 gold badges14 silver badges48 bronze badges
3
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Follow-up question: Would the most natural absolute temperature unit be one where the constant of proportionality R=1?
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– helpme
Jun 9 at 4:20
2
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@helpme The answer is subjective and would only be "well suited" for the ideal gas relation. Not with, for example, the ZT relation I mention in my answer.
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– thermomagnetic condensed boson
Jun 9 at 7:44
6
$begingroup$
@helpme: "It's subjective" is probably not a satisfactory answer to your question. Thankfully there is a satisfactory answer and that answer is simply, no. :-) You'd instead want to use PV = NkT rather than PV = nRT, where k is the per-molecule normalized version of R, putting N in the natural units of "number of molecules". So the more natural choice would seem to be k = 1. And indeed this is what Planck units do... and if you continue this you get the Planck temperature as the natural unit.
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– Mehrdad
Jun 9 at 11:02
1
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The equation $PVpropto T$ (in the ideal gas limit) does not work for any scale. It only works for linear scales, that is scales where the reference points are linearly interpolated.
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– Diracology
Jun 10 at 17:59
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Good point @Diracology let me modify and improve the answer.
$endgroup$
– thermomagnetic condensed boson
Jun 10 at 19:32
add a comment |
3
$begingroup$
Follow-up question: Would the most natural absolute temperature unit be one where the constant of proportionality R=1?
$endgroup$
– helpme
Jun 9 at 4:20
2
$begingroup$
@helpme The answer is subjective and would only be "well suited" for the ideal gas relation. Not with, for example, the ZT relation I mention in my answer.
$endgroup$
– thermomagnetic condensed boson
Jun 9 at 7:44
6
$begingroup$
@helpme: "It's subjective" is probably not a satisfactory answer to your question. Thankfully there is a satisfactory answer and that answer is simply, no. :-) You'd instead want to use PV = NkT rather than PV = nRT, where k is the per-molecule normalized version of R, putting N in the natural units of "number of molecules". So the more natural choice would seem to be k = 1. And indeed this is what Planck units do... and if you continue this you get the Planck temperature as the natural unit.
$endgroup$
– Mehrdad
Jun 9 at 11:02
1
$begingroup$
The equation $PVpropto T$ (in the ideal gas limit) does not work for any scale. It only works for linear scales, that is scales where the reference points are linearly interpolated.
$endgroup$
– Diracology
Jun 10 at 17:59
$begingroup$
Good point @Diracology let me modify and improve the answer.
$endgroup$
– thermomagnetic condensed boson
Jun 10 at 19:32
3
3
$begingroup$
Follow-up question: Would the most natural absolute temperature unit be one where the constant of proportionality R=1?
$endgroup$
– helpme
Jun 9 at 4:20
$begingroup$
Follow-up question: Would the most natural absolute temperature unit be one where the constant of proportionality R=1?
$endgroup$
– helpme
Jun 9 at 4:20
2
2
$begingroup$
@helpme The answer is subjective and would only be "well suited" for the ideal gas relation. Not with, for example, the ZT relation I mention in my answer.
$endgroup$
– thermomagnetic condensed boson
Jun 9 at 7:44
$begingroup$
@helpme The answer is subjective and would only be "well suited" for the ideal gas relation. Not with, for example, the ZT relation I mention in my answer.
$endgroup$
– thermomagnetic condensed boson
Jun 9 at 7:44
6
6
$begingroup$
@helpme: "It's subjective" is probably not a satisfactory answer to your question. Thankfully there is a satisfactory answer and that answer is simply, no. :-) You'd instead want to use PV = NkT rather than PV = nRT, where k is the per-molecule normalized version of R, putting N in the natural units of "number of molecules". So the more natural choice would seem to be k = 1. And indeed this is what Planck units do... and if you continue this you get the Planck temperature as the natural unit.
$endgroup$
– Mehrdad
Jun 9 at 11:02
$begingroup$
@helpme: "It's subjective" is probably not a satisfactory answer to your question. Thankfully there is a satisfactory answer and that answer is simply, no. :-) You'd instead want to use PV = NkT rather than PV = nRT, where k is the per-molecule normalized version of R, putting N in the natural units of "number of molecules". So the more natural choice would seem to be k = 1. And indeed this is what Planck units do... and if you continue this you get the Planck temperature as the natural unit.
$endgroup$
– Mehrdad
Jun 9 at 11:02
1
1
$begingroup$
The equation $PVpropto T$ (in the ideal gas limit) does not work for any scale. It only works for linear scales, that is scales where the reference points are linearly interpolated.
$endgroup$
– Diracology
Jun 10 at 17:59
$begingroup$
The equation $PVpropto T$ (in the ideal gas limit) does not work for any scale. It only works for linear scales, that is scales where the reference points are linearly interpolated.
$endgroup$
– Diracology
Jun 10 at 17:59
$begingroup$
Good point @Diracology let me modify and improve the answer.
$endgroup$
– thermomagnetic condensed boson
Jun 10 at 19:32
$begingroup$
Good point @Diracology let me modify and improve the answer.
$endgroup$
– thermomagnetic condensed boson
Jun 10 at 19:32
add a comment |
$begingroup$
For example, 𝑃=100, 𝑉=8.314, 𝑛=1, 𝑅=8.314.
Plugging in the values will result in 𝑇=100
kelvin.
But what is the significance of 100 kelvin? Why could it not have been
any other unit in an absolute temperature scale? e.g. 100 degrees
Rankine
The answer is already contained in the first part of your question, when you write the units of the Gas Constant $R$. In order to get temperature in different units, one should use a different value and units for $R$. Similarly for a different thermometric scale.
Edit (improved the examples):
For example, in the case of Rankine degrees and using the same units for all other quantities, the perfect gas formula keeps the same form, but with a gas constant $R^prime=4.61915$. In the case of temperature, $theta$, in Celsius degrees, becomes:
$$
PV = ntilde Rtheta+ Q n,
$$
where $Q = 2.271 cdot 10^3 = 8.314*273.15$ J mol$^-1$, and $tilde R$ has the same numerical value of $R=8.314$, although now its units should be read as J mol$^-1$ $^circ$C$^-1$.
answered Jun 8 at 13:06
GiorgioPGiorgioP
6,2192 gold badges9 silver badges34 bronze badges
$endgroup$
$begingroup$
I don't think you should have $Rprime$ and $tilde R$ there. It is still the same constant, with the same value, even if expressed in different units (and then, you really should have units in the last equation, otherwise the dimensions don't match).
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– Jan Hudec
Jun 10 at 13:47
$begingroup$
@JanHudec I used different symbols for clarity. Even if the numerical values are the same, different units are used. I'll modify the way I am writing the last equation to avoid misunderstandings,
$endgroup$
– GiorgioP
Jun 10 at 17:46
add a comment |
$begingroup$
For example, 𝑃=100, 𝑉=8.314, 𝑛=1, 𝑅=8.314.
Plugging in the values will result in 𝑇=100
kelvin.
But what is the significance of 100 kelvin? Why could it not have been
any other unit in an absolute temperature scale? e.g. 100 degrees
Rankine
The answer is already contained in the first part of your question, when you write the units of the Gas Constant $R$. In order to get temperature in different units, one should use a different value and units for $R$. Similarly for a different thermometric scale.
Edit (improved the examples):
For example, in the case of Rankine degrees and using the same units for all other quantities, the perfect gas formula keeps the same form, but with a gas constant $R^prime=4.61915$. In the case of temperature, $theta$, in Celsius degrees, becomes:
$$
PV = ntilde Rtheta+ Q n,
$$
where $Q = 2.271 cdot 10^3 = 8.314*273.15$ J mol$^-1$, and $tilde R$ has the same numerical value of $R=8.314$, although now its units should be read as J mol$^-1$ $^circ$C$^-1$.
answered Jun 8 at 13:06
GiorgioPGiorgioP
6,2192 gold badges9 silver badges34 bronze badges
$endgroup$
$begingroup$
I don't think you should have $Rprime$ and $tilde R$ there. It is still the same constant, with the same value, even if expressed in different units (and then, you really should have units in the last equation, otherwise the dimensions don't match).
$endgroup$
– Jan Hudec
Jun 10 at 13:47
$begingroup$
@JanHudec I used different symbols for clarity. Even if the numerical values are the same, different units are used. I'll modify the way I am writing the last equation to avoid misunderstandings,
$endgroup$
– GiorgioP
Jun 10 at 17:46
add a comment |
$begingroup$
For example, 𝑃=100, 𝑉=8.314, 𝑛=1, 𝑅=8.314.
Plugging in the values will result in 𝑇=100
kelvin.
But what is the significance of 100 kelvin? Why could it not have been
any other unit in an absolute temperature scale? e.g. 100 degrees
Rankine
The answer is already contained in the first part of your question, when you write the units of the Gas Constant $R$. In order to get temperature in different units, one should use a different value and units for $R$. Similarly for a different thermometric scale.
Edit (improved the examples):
For example, in the case of Rankine degrees and using the same units for all other quantities, the perfect gas formula keeps the same form, but with a gas constant $R^prime=4.61915$. In the case of temperature, $theta$, in Celsius degrees, becomes:
$$
PV = ntilde Rtheta+ Q n,
$$
where $Q = 2.271 cdot 10^3 = 8.314*273.15$ J mol$^-1$, and $tilde R$ has the same numerical value of $R=8.314$, although now its units should be read as J mol$^-1$ $^circ$C$^-1$.
answered Jun 8 at 13:06
GiorgioPGiorgioP
6,2192 gold badges9 silver badges34 bronze badges
$endgroup$
For example, 𝑃=100, 𝑉=8.314, 𝑛=1, 𝑅=8.314.
Plugging in the values will result in 𝑇=100
kelvin.
But what is the significance of 100 kelvin? Why could it not have been
any other unit in an absolute temperature scale? e.g. 100 degrees
Rankine
The answer is already contained in the first part of your question, when you write the units of the Gas Constant $R$. In order to get temperature in different units, one should use a different value and units for $R$. Similarly for a different thermometric scale.
Edit (improved the examples):
For example, in the case of Rankine degrees and using the same units for all other quantities, the perfect gas formula keeps the same form, but with a gas constant $R^prime=4.61915$. In the case of temperature, $theta$, in Celsius degrees, becomes:
$$
PV = ntilde Rtheta+ Q n,
$$
where $Q = 2.271 cdot 10^3 = 8.314*273.15$ J mol$^-1$, and $tilde R$ has the same numerical value of $R=8.314$, although now its units should be read as J mol$^-1$ $^circ$C$^-1$.
answered Jun 8 at 13:06
GiorgioPGiorgioP
6,2192 gold badges9 silver badges34 bronze badges
edited Jun 10 at 17:49
answered Jun 8 at 13:06
GiorgioPGiorgioP
6,2192 gold badges9 silver badges34 bronze badges
answered Jun 8 at 13:06
GiorgioPGiorgioP
6,2192 gold badges9 silver badges34 bronze badges
answered Jun 8 at 13:06
GiorgioPGiorgioP
6,2192 gold badges9 silver badges34 bronze badges
6,2192 gold badges9 silver badges34 bronze badges
$begingroup$
I don't think you should have $Rprime$ and $tilde R$ there. It is still the same constant, with the same value, even if expressed in different units (and then, you really should have units in the last equation, otherwise the dimensions don't match).
$endgroup$
– Jan Hudec
Jun 10 at 13:47
$begingroup$
@JanHudec I used different symbols for clarity. Even if the numerical values are the same, different units are used. I'll modify the way I am writing the last equation to avoid misunderstandings,
$endgroup$
– GiorgioP
Jun 10 at 17:46
add a comment |
$begingroup$
I don't think you should have $Rprime$ and $tilde R$ there. It is still the same constant, with the same value, even if expressed in different units (and then, you really should have units in the last equation, otherwise the dimensions don't match).
$endgroup$
– Jan Hudec
Jun 10 at 13:47
$begingroup$
@JanHudec I used different symbols for clarity. Even if the numerical values are the same, different units are used. I'll modify the way I am writing the last equation to avoid misunderstandings,
$endgroup$
– GiorgioP
Jun 10 at 17:46
$begingroup$
I don't think you should have $Rprime$ and $tilde R$ there. It is still the same constant, with the same value, even if expressed in different units (and then, you really should have units in the last equation, otherwise the dimensions don't match).
$endgroup$
– Jan Hudec
Jun 10 at 13:47
$begingroup$
I don't think you should have $Rprime$ and $tilde R$ there. It is still the same constant, with the same value, even if expressed in different units (and then, you really should have units in the last equation, otherwise the dimensions don't match).
$endgroup$
– Jan Hudec
Jun 10 at 13:47
$begingroup$
@JanHudec I used different symbols for clarity. Even if the numerical values are the same, different units are used. I'll modify the way I am writing the last equation to avoid misunderstandings,
$endgroup$
– GiorgioP
Jun 10 at 17:46
$begingroup$
@JanHudec I used different symbols for clarity. Even if the numerical values are the same, different units are used. I'll modify the way I am writing the last equation to avoid misunderstandings,
$endgroup$
– GiorgioP
Jun 10 at 17:46
add a comment |
$begingroup$
There is a point that the other answers haven't made, which I believe is important: Celsius and Farenheit both allow temperatures below (what they call) zero. If you plug those measurements into the formula, you get a value of (pressure times value) which is negative; this is not physically possible. Note also that negativeness cannot be compensated-for by a simple constant of proportionality (R). Kelvin at least has the virtue of never being negative-valued.
answered Jun 10 at 13:08
PMarPMar
111 bronze badge
$endgroup$
add a comment |
$begingroup$
There is a point that the other answers haven't made, which I believe is important: Celsius and Farenheit both allow temperatures below (what they call) zero. If you plug those measurements into the formula, you get a value of (pressure times value) which is negative; this is not physically possible. Note also that negativeness cannot be compensated-for by a simple constant of proportionality (R). Kelvin at least has the virtue of never being negative-valued.
answered Jun 10 at 13:08
PMarPMar
111 bronze badge
$endgroup$
add a comment |
$begingroup$
There is a point that the other answers haven't made, which I believe is important: Celsius and Farenheit both allow temperatures below (what they call) zero. If you plug those measurements into the formula, you get a value of (pressure times value) which is negative; this is not physically possible. Note also that negativeness cannot be compensated-for by a simple constant of proportionality (R). Kelvin at least has the virtue of never being negative-valued.
answered Jun 10 at 13:08
PMarPMar
111 bronze badge
$endgroup$
There is a point that the other answers haven't made, which I believe is important: Celsius and Farenheit both allow temperatures below (what they call) zero. If you plug those measurements into the formula, you get a value of (pressure times value) which is negative; this is not physically possible. Note also that negativeness cannot be compensated-for by a simple constant of proportionality (R). Kelvin at least has the virtue of never being negative-valued.
answered Jun 10 at 13:08
PMarPMar
111 bronze badge
answered Jun 10 at 13:08
PMarPMar
111 bronze badge
answered Jun 10 at 13:08
PMarPMar
111 bronze badge
answered Jun 10 at 13:08
PMarPMar
111 bronze badge
111 bronze badge
add a comment |
add a comment |
$begingroup$
Since a few weeks, the kelvin is a unit that is defined by setting Boltzmann's constant to exactly $1.380 649 cdot 10^-23$ J/K (see https://physics.nist.gov/cgi-bin/cuu/Value?k ).
For now, this does not change the practical temperature scale for calibrating thermometers around ambient temperatures: https://www.bipm.org/en/committees/cc/cct/guide-its90.html
answered Jun 10 at 19:47
PieterPieter
9,6893 gold badges17 silver badges39 bronze badges
$endgroup$
add a comment |
$begingroup$
Since a few weeks, the kelvin is a unit that is defined by setting Boltzmann's constant to exactly $1.380 649 cdot 10^-23$ J/K (see https://physics.nist.gov/cgi-bin/cuu/Value?k ).
For now, this does not change the practical temperature scale for calibrating thermometers around ambient temperatures: https://www.bipm.org/en/committees/cc/cct/guide-its90.html
answered Jun 10 at 19:47
PieterPieter
9,6893 gold badges17 silver badges39 bronze badges
$endgroup$
add a comment |
$begingroup$
Since a few weeks, the kelvin is a unit that is defined by setting Boltzmann's constant to exactly $1.380 649 cdot 10^-23$ J/K (see https://physics.nist.gov/cgi-bin/cuu/Value?k ).
For now, this does not change the practical temperature scale for calibrating thermometers around ambient temperatures: https://www.bipm.org/en/committees/cc/cct/guide-its90.html
answered Jun 10 at 19:47
PieterPieter
9,6893 gold badges17 silver badges39 bronze badges
$endgroup$
Since a few weeks, the kelvin is a unit that is defined by setting Boltzmann's constant to exactly $1.380 649 cdot 10^-23$ J/K (see https://physics.nist.gov/cgi-bin/cuu/Value?k ).
For now, this does not change the practical temperature scale for calibrating thermometers around ambient temperatures: https://www.bipm.org/en/committees/cc/cct/guide-its90.html
answered Jun 10 at 19:47
PieterPieter
9,6893 gold badges17 silver badges39 bronze badges
edited Jun 10 at 19:58
answered Jun 10 at 19:47
PieterPieter
9,6893 gold badges17 silver badges39 bronze badges
answered Jun 10 at 19:47
PieterPieter
9,6893 gold badges17 silver badges39 bronze badges
answered Jun 10 at 19:47
PieterPieter
9,6893 gold badges17 silver badges39 bronze badges
9,6893 gold badges17 silver badges39 bronze badges
add a comment |
add a comment |
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5
$begingroup$
It may be useful to note that $R T$ has the units of energy, so the choice for units of T is really about what value you take $R$ to be.
$endgroup$
– jim
Jun 9 at 15:17
$begingroup$
I suggest you keep the units with the numbers. $R$ is not $8.314$. $R = 8.314 mathrmJcdotmathrmK^-1cdotmathrmmol^-1 = 1.104 mathrmcal_thcdotmathrm⁰R^-1cdotmathrmmol^-1$ (for example). That's true equals, because it is still the same constant with the same value, just expressed in different units.
$endgroup$
– Jan Hudec
Jun 10 at 13:43
1
$begingroup$
This post is relevant: Is the Boltzmann constant really that important?.
$endgroup$
– DanielSank
Jun 10 at 19:47