Mixing
Energy of room. Ideal gas law
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I have been following Blundel's "Concepts of thermal Physics" and I got to the derivation of the ideal gas law. And it all made sense, we made a couple of assumptions and approximations, but then I got to an exercise that doesn't make a lot of sense to me:
"If a room is initially at 18ºC and is then heated up to 25ºC what happens to the total energy of the air in the room?"
That seems a straight forward question and an attempt of solution is:
$$langle Erangle =uV=frac12Nmlangle v^2 rangle$$
as $langle v^2 rangle=3frack_B Tm$ we get:
$$langle Erangle =frac 32NK_BT$$
where $u$ is the energy density, $V$ is the volume of the room, and $langle Erangle$ is the mean kinetic energy of the gas in the room. After this we could just compare for both temperatures.
But doing this we are only considering kinetic energy. Throughout the derivation we ignored all kinds of motion a particle and considered only translational energy, but even doing this a particle still has 2 ways of having energy: kinetic and potential. We obviously didn't ignore the mass of a particle otherwise it wouldn't have kinetic energy either. so the particles still have mass, therefore it must have potential energy! I get that in the comparison of the energy at both temperatures, as the room stays still, the potential energy cancels out, but then saying that the mean energy of the room at a temperature $T$ is $langle Erangle =frac 32NK_BT$ is inclomplete!
How do I describe the total energy of a room at temperature $T$?
thermodynamics newtonian-gravity temperature potential-energy probability
edited 28 mins ago
Qmechanic♦
98.5k121741074
asked 3 hours ago
Duartejfs
467
 |Â
show 1 more comment
up vote
2
down vote
favorite
I have been following Blundel's "Concepts of thermal Physics" and I got to the derivation of the ideal gas law. And it all made sense, we made a couple of assumptions and approximations, but then I got to an exercise that doesn't make a lot of sense to me:
"If a room is initially at 18ºC and is then heated up to 25ºC what happens to the total energy of the air in the room?"
That seems a straight forward question and an attempt of solution is:
$$langle Erangle =uV=frac12Nmlangle v^2 rangle$$
as $langle v^2 rangle=3frack_B Tm$ we get:
$$langle Erangle =frac 32NK_BT$$
where $u$ is the energy density, $V$ is the volume of the room, and $langle Erangle$ is the mean kinetic energy of the gas in the room. After this we could just compare for both temperatures.
But doing this we are only considering kinetic energy. Throughout the derivation we ignored all kinds of motion a particle and considered only translational energy, but even doing this a particle still has 2 ways of having energy: kinetic and potential. We obviously didn't ignore the mass of a particle otherwise it wouldn't have kinetic energy either. so the particles still have mass, therefore it must have potential energy! I get that in the comparison of the energy at both temperatures, as the room stays still, the potential energy cancels out, but then saying that the mean energy of the room at a temperature $T$ is $langle Erangle =frac 32NK_BT$ is inclomplete!
How do I describe the total energy of a room at temperature $T$?
thermodynamics newtonian-gravity temperature potential-energy probability
edited 28 mins ago
Qmechanic♦
98.5k121741074
asked 3 hours ago
Duartejfs
467
And some air will leave the room...
– Pieter
3 hours ago
yes, but assume not. All the air stays in the same room always.
– Duartejfs
3 hours ago
You mean that air at a higher point in the room should have more energy than air at a lower point?
– Aaron Stevens
2 hours ago
1
Due to the room's size you are not considering the difference in potential energy between particles at differeht heights.
– Wolphram jonny
2 hours ago
@Wolphramjonny The height distribution does not change much with temperature. Potential energy does not matter. What changes is the rotational kinetic energy of air molecules.
– Pieter
24 mins ago
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have been following Blundel's "Concepts of thermal Physics" and I got to the derivation of the ideal gas law. And it all made sense, we made a couple of assumptions and approximations, but then I got to an exercise that doesn't make a lot of sense to me:
"If a room is initially at 18ºC and is then heated up to 25ºC what happens to the total energy of the air in the room?"
That seems a straight forward question and an attempt of solution is:
$$langle Erangle =uV=frac12Nmlangle v^2 rangle$$
as $langle v^2 rangle=3frack_B Tm$ we get:
$$langle Erangle =frac 32NK_BT$$
where $u$ is the energy density, $V$ is the volume of the room, and $langle Erangle$ is the mean kinetic energy of the gas in the room. After this we could just compare for both temperatures.
But doing this we are only considering kinetic energy. Throughout the derivation we ignored all kinds of motion a particle and considered only translational energy, but even doing this a particle still has 2 ways of having energy: kinetic and potential. We obviously didn't ignore the mass of a particle otherwise it wouldn't have kinetic energy either. so the particles still have mass, therefore it must have potential energy! I get that in the comparison of the energy at both temperatures, as the room stays still, the potential energy cancels out, but then saying that the mean energy of the room at a temperature $T$ is $langle Erangle =frac 32NK_BT$ is inclomplete!
How do I describe the total energy of a room at temperature $T$?
thermodynamics newtonian-gravity temperature potential-energy probability
edited 28 mins ago
Qmechanic♦
98.5k121741074
asked 3 hours ago
Duartejfs
467
I have been following Blundel's "Concepts of thermal Physics" and I got to the derivation of the ideal gas law. And it all made sense, we made a couple of assumptions and approximations, but then I got to an exercise that doesn't make a lot of sense to me:
"If a room is initially at 18ºC and is then heated up to 25ºC what happens to the total energy of the air in the room?"
That seems a straight forward question and an attempt of solution is:
$$langle Erangle =uV=frac12Nmlangle v^2 rangle$$
as $langle v^2 rangle=3frack_B Tm$ we get:
$$langle Erangle =frac 32NK_BT$$
where $u$ is the energy density, $V$ is the volume of the room, and $langle Erangle$ is the mean kinetic energy of the gas in the room. After this we could just compare for both temperatures.
But doing this we are only considering kinetic energy. Throughout the derivation we ignored all kinds of motion a particle and considered only translational energy, but even doing this a particle still has 2 ways of having energy: kinetic and potential. We obviously didn't ignore the mass of a particle otherwise it wouldn't have kinetic energy either. so the particles still have mass, therefore it must have potential energy! I get that in the comparison of the energy at both temperatures, as the room stays still, the potential energy cancels out, but then saying that the mean energy of the room at a temperature $T$ is $langle Erangle =frac 32NK_BT$ is inclomplete!
How do I describe the total energy of a room at temperature $T$?
thermodynamics newtonian-gravity temperature potential-energy probability
thermodynamics newtonian-gravity temperature potential-energy probability
edited 28 mins ago
Qmechanic♦
98.5k121741074
asked 3 hours ago
Duartejfs
467
edited 28 mins ago
Qmechanic♦
98.5k121741074
asked 3 hours ago
Duartejfs
467
edited 28 mins ago
Qmechanic♦
98.5k121741074
edited 28 mins ago
Qmechanic♦
98.5k121741074
edited 28 mins ago
Qmechanic♦
98.5k121741074
98.5k121741074
asked 3 hours ago
Duartejfs
467
asked 3 hours ago
Duartejfs
467
asked 3 hours ago
Duartejfs
467
467
And some air will leave the room...
– Pieter
3 hours ago
yes, but assume not. All the air stays in the same room always.
– Duartejfs
3 hours ago
You mean that air at a higher point in the room should have more energy than air at a lower point?
– Aaron Stevens
2 hours ago
1
Due to the room's size you are not considering the difference in potential energy between particles at differeht heights.
– Wolphram jonny
2 hours ago
@Wolphramjonny The height distribution does not change much with temperature. Potential energy does not matter. What changes is the rotational kinetic energy of air molecules.
– Pieter
24 mins ago
 |Â
show 1 more comment
And some air will leave the room...
– Pieter
3 hours ago
yes, but assume not. All the air stays in the same room always.
– Duartejfs
3 hours ago
You mean that air at a higher point in the room should have more energy than air at a lower point?
– Aaron Stevens
2 hours ago
1
Due to the room's size you are not considering the difference in potential energy between particles at differeht heights.
– Wolphram jonny
2 hours ago
@Wolphramjonny The height distribution does not change much with temperature. Potential energy does not matter. What changes is the rotational kinetic energy of air molecules.
– Pieter
24 mins ago
And some air will leave the room...
– Pieter
3 hours ago
And some air will leave the room...
– Pieter
3 hours ago
yes, but assume not. All the air stays in the same room always.
– Duartejfs
3 hours ago
yes, but assume not. All the air stays in the same room always.
– Duartejfs
3 hours ago
You mean that air at a higher point in the room should have more energy than air at a lower point?
– Aaron Stevens
2 hours ago
You mean that air at a higher point in the room should have more energy than air at a lower point?
– Aaron Stevens
2 hours ago
1
1
Due to the room's size you are not considering the difference in potential energy between particles at differeht heights.
– Wolphram jonny
2 hours ago
Due to the room's size you are not considering the difference in potential energy between particles at differeht heights.
– Wolphram jonny
2 hours ago
@Wolphramjonny The height distribution does not change much with temperature. Potential energy does not matter. What changes is the rotational kinetic energy of air molecules.
– Pieter
24 mins ago
@Wolphramjonny The height distribution does not change much with temperature. Potential energy does not matter. What changes is the rotational kinetic energy of air molecules.
– Pieter
24 mins ago
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Let's look at the ratios between the average kinetic energy (for one particle) and the change in potential energy between the bottom and top of our container.
$$fracKDelta U=fracfrac 32 k_BTmgh$$
Air is made of many types of particles, but nitrogen is the most common, so let's work with that. The mass of an $rm N_2$ molecule is $4.65times10^-26 rmkg$, and let's say the height of the room is about $3 rm m$. Then for $T=295 rm K$ (around the middle of your temperature range) we have
$$fracfrac 32 k_BTmghapprox 15$$
So the kinetic energy is much larger than the potential difference between the top and bottom of the room, which is why we tend to ignore it. You do have do consider potential energy when looking at larger heights, like when analyzing the atmosphere for example.
answered 2 hours ago
Aaron Stevens
5,2232828
add a comment |Â
up vote
2
down vote
This is where we get into some of the complexities of how we actually define the 'total' energy. As far as gravity is concerned, physics only depends on changes in the potential energy, not its absolute value; we can add a constant to the potential energy and get the same predictions! So the 'total potential energy' is not an entirely well-defined quantity.
As far as ideal gas theory is concerned, we treat gas particles as rigid balls and ignore any intermolecular interactions. So in this approximation, we ignore any electromagnetic forces between particles. (If we wanted to add them, we would reach something like the van der Waals model which I'm sure will come up later in the book you're working through, or one of many others!)
But what about gravitational potential energy? This is defined up to a constant, but there might still be differences in gravitational potential energy between the two situations, because the distribution of air density changes.
Just how much?
Since we're assuming uniform temperature, and assuming a uniform gravitational field, we can calculate the distribution of density in hydrostatic equilibrium. In hydrostatic equilibrium, the pressure and density are functions of height, but we still take them to be related by the ideal gas law $mP(z)=rho(z) k_BT$. We have $$fracdPdz=-rho(z)g=-fracmgk_BTp(z)$$ with the solution $$rho(z)=rho_0expleft(-fracmgzk_BTright)$$We set the density of air at ground level, $rho_0$ by requiring that the total mass of particles in the gas is constant: $$int_textroom rho(z) dV = M$$ If we assume the room has floor area $A$ and height $h$, this becomes $$M=Aint_0^h rho(z) dz = fracArho_0 k_B Tmgleft(1-expleft(-fracmghk_BTright)right)$$We can rearrange this to find $rho_0$ as a function of temperature: $$rho_0=fracMmgAk_B T left(1-expleft(-fracmghk_BTright)right)$$
Having found the density as a function of height, we can calculate the potential energy of the gas.
The gravitational potential energy density of a piece of gas at height $z$ above the zero-potential-energy reference level is just $u(z)=rho(z) g z$. So the total potential energy of the whole room is $$U=int_textroom u(z),dV=int_textroom rho(z) g zthinspace dV=int_textroomgzrho_0expleft(-fracmgzk_BTright)dV$$ and with our area $A$ and height $h$, this becomes $$U=Agrho_0int_0^h zexpleft(-fracmgzk_BTright),dz$$
To calculate this, we can use a standard integral result: $$int_0^h z exp left(-fraczCright)=C^2left(1-left(1+frachCright)expleft(-frachCright)right)$$which we can use with $C=frack_B Tmg$to find (deep breath)... $$U=fracA rho_0 k_B Tmcdotfrack_B Tmgleft(1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)right)$$
We can combine this with our earlier result for $rho_0$: $$U=fracMk_B Tmcdotfrac1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)1-expleft(-fracmghk_BTright)$$
Rearranging this slightly to clean it up, we get $$beginalign*U&=fracMk_B Tmcdotfracexpleft(fracmghk_B Tright)-left(1+fracmghk_B Tright)expleft(fracmghk_BTright)-1\&=fracMk_B Tmleft(1-fracmghk_BTfrac1expleft(fracmghk_BTright)-1right)\&=fracMk_B Tm-fracMghexpleft(fracmghk_BTright)-1\&=Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1endalign*$$ where in the last line we've used that the total mass is $M=Nm$.
So in answer to your question, the total energy of the ideal gas in a uniform temperature room of area (A) and height (z) in hydrostatic equilibrium would be (assuming I haven't made a mistake in the derivation above)...
$$E=frac52Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1$$
plus a constant.
answered 35 mins ago
canonicalmomentum
213
New contributor
canonicalmomentum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let's look at the ratios between the average kinetic energy (for one particle) and the change in potential energy between the bottom and top of our container.
$$fracKDelta U=fracfrac 32 k_BTmgh$$
Air is made of many types of particles, but nitrogen is the most common, so let's work with that. The mass of an $rm N_2$ molecule is $4.65times10^-26 rmkg$, and let's say the height of the room is about $3 rm m$. Then for $T=295 rm K$ (around the middle of your temperature range) we have
$$fracfrac 32 k_BTmghapprox 15$$
So the kinetic energy is much larger than the potential difference between the top and bottom of the room, which is why we tend to ignore it. You do have do consider potential energy when looking at larger heights, like when analyzing the atmosphere for example.
answered 2 hours ago
Aaron Stevens
5,2232828
add a comment |Â
up vote
2
down vote
accepted
Let's look at the ratios between the average kinetic energy (for one particle) and the change in potential energy between the bottom and top of our container.
$$fracKDelta U=fracfrac 32 k_BTmgh$$
Air is made of many types of particles, but nitrogen is the most common, so let's work with that. The mass of an $rm N_2$ molecule is $4.65times10^-26 rmkg$, and let's say the height of the room is about $3 rm m$. Then for $T=295 rm K$ (around the middle of your temperature range) we have
$$fracfrac 32 k_BTmghapprox 15$$
So the kinetic energy is much larger than the potential difference between the top and bottom of the room, which is why we tend to ignore it. You do have do consider potential energy when looking at larger heights, like when analyzing the atmosphere for example.
answered 2 hours ago
Aaron Stevens
5,2232828
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let's look at the ratios between the average kinetic energy (for one particle) and the change in potential energy between the bottom and top of our container.
$$fracKDelta U=fracfrac 32 k_BTmgh$$
Air is made of many types of particles, but nitrogen is the most common, so let's work with that. The mass of an $rm N_2$ molecule is $4.65times10^-26 rmkg$, and let's say the height of the room is about $3 rm m$. Then for $T=295 rm K$ (around the middle of your temperature range) we have
$$fracfrac 32 k_BTmghapprox 15$$
So the kinetic energy is much larger than the potential difference between the top and bottom of the room, which is why we tend to ignore it. You do have do consider potential energy when looking at larger heights, like when analyzing the atmosphere for example.
answered 2 hours ago
Aaron Stevens
5,2232828
Let's look at the ratios between the average kinetic energy (for one particle) and the change in potential energy between the bottom and top of our container.
$$fracKDelta U=fracfrac 32 k_BTmgh$$
Air is made of many types of particles, but nitrogen is the most common, so let's work with that. The mass of an $rm N_2$ molecule is $4.65times10^-26 rmkg$, and let's say the height of the room is about $3 rm m$. Then for $T=295 rm K$ (around the middle of your temperature range) we have
$$fracfrac 32 k_BTmghapprox 15$$
So the kinetic energy is much larger than the potential difference between the top and bottom of the room, which is why we tend to ignore it. You do have do consider potential energy when looking at larger heights, like when analyzing the atmosphere for example.
answered 2 hours ago
Aaron Stevens
5,2232828
answered 2 hours ago
Aaron Stevens
5,2232828
answered 2 hours ago
Aaron Stevens
5,2232828
answered 2 hours ago
Aaron Stevens
5,2232828
5,2232828
add a comment |Â
add a comment |Â
up vote
2
down vote
This is where we get into some of the complexities of how we actually define the 'total' energy. As far as gravity is concerned, physics only depends on changes in the potential energy, not its absolute value; we can add a constant to the potential energy and get the same predictions! So the 'total potential energy' is not an entirely well-defined quantity.
As far as ideal gas theory is concerned, we treat gas particles as rigid balls and ignore any intermolecular interactions. So in this approximation, we ignore any electromagnetic forces between particles. (If we wanted to add them, we would reach something like the van der Waals model which I'm sure will come up later in the book you're working through, or one of many others!)
But what about gravitational potential energy? This is defined up to a constant, but there might still be differences in gravitational potential energy between the two situations, because the distribution of air density changes.
Just how much?
Since we're assuming uniform temperature, and assuming a uniform gravitational field, we can calculate the distribution of density in hydrostatic equilibrium. In hydrostatic equilibrium, the pressure and density are functions of height, but we still take them to be related by the ideal gas law $mP(z)=rho(z) k_BT$. We have $$fracdPdz=-rho(z)g=-fracmgk_BTp(z)$$ with the solution $$rho(z)=rho_0expleft(-fracmgzk_BTright)$$We set the density of air at ground level, $rho_0$ by requiring that the total mass of particles in the gas is constant: $$int_textroom rho(z) dV = M$$ If we assume the room has floor area $A$ and height $h$, this becomes $$M=Aint_0^h rho(z) dz = fracArho_0 k_B Tmgleft(1-expleft(-fracmghk_BTright)right)$$We can rearrange this to find $rho_0$ as a function of temperature: $$rho_0=fracMmgAk_B T left(1-expleft(-fracmghk_BTright)right)$$
Having found the density as a function of height, we can calculate the potential energy of the gas.
The gravitational potential energy density of a piece of gas at height $z$ above the zero-potential-energy reference level is just $u(z)=rho(z) g z$. So the total potential energy of the whole room is $$U=int_textroom u(z),dV=int_textroom rho(z) g zthinspace dV=int_textroomgzrho_0expleft(-fracmgzk_BTright)dV$$ and with our area $A$ and height $h$, this becomes $$U=Agrho_0int_0^h zexpleft(-fracmgzk_BTright),dz$$
To calculate this, we can use a standard integral result: $$int_0^h z exp left(-fraczCright)=C^2left(1-left(1+frachCright)expleft(-frachCright)right)$$which we can use with $C=frack_B Tmg$to find (deep breath)... $$U=fracA rho_0 k_B Tmcdotfrack_B Tmgleft(1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)right)$$
We can combine this with our earlier result for $rho_0$: $$U=fracMk_B Tmcdotfrac1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)1-expleft(-fracmghk_BTright)$$
Rearranging this slightly to clean it up, we get $$beginalign*U&=fracMk_B Tmcdotfracexpleft(fracmghk_B Tright)-left(1+fracmghk_B Tright)expleft(fracmghk_BTright)-1\&=fracMk_B Tmleft(1-fracmghk_BTfrac1expleft(fracmghk_BTright)-1right)\&=fracMk_B Tm-fracMghexpleft(fracmghk_BTright)-1\&=Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1endalign*$$ where in the last line we've used that the total mass is $M=Nm$.
So in answer to your question, the total energy of the ideal gas in a uniform temperature room of area (A) and height (z) in hydrostatic equilibrium would be (assuming I haven't made a mistake in the derivation above)...
$$E=frac52Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1$$
plus a constant.
answered 35 mins ago
canonicalmomentum
213
New contributor
canonicalmomentum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
This is where we get into some of the complexities of how we actually define the 'total' energy. As far as gravity is concerned, physics only depends on changes in the potential energy, not its absolute value; we can add a constant to the potential energy and get the same predictions! So the 'total potential energy' is not an entirely well-defined quantity.
As far as ideal gas theory is concerned, we treat gas particles as rigid balls and ignore any intermolecular interactions. So in this approximation, we ignore any electromagnetic forces between particles. (If we wanted to add them, we would reach something like the van der Waals model which I'm sure will come up later in the book you're working through, or one of many others!)
But what about gravitational potential energy? This is defined up to a constant, but there might still be differences in gravitational potential energy between the two situations, because the distribution of air density changes.
Just how much?
Since we're assuming uniform temperature, and assuming a uniform gravitational field, we can calculate the distribution of density in hydrostatic equilibrium. In hydrostatic equilibrium, the pressure and density are functions of height, but we still take them to be related by the ideal gas law $mP(z)=rho(z) k_BT$. We have $$fracdPdz=-rho(z)g=-fracmgk_BTp(z)$$ with the solution $$rho(z)=rho_0expleft(-fracmgzk_BTright)$$We set the density of air at ground level, $rho_0$ by requiring that the total mass of particles in the gas is constant: $$int_textroom rho(z) dV = M$$ If we assume the room has floor area $A$ and height $h$, this becomes $$M=Aint_0^h rho(z) dz = fracArho_0 k_B Tmgleft(1-expleft(-fracmghk_BTright)right)$$We can rearrange this to find $rho_0$ as a function of temperature: $$rho_0=fracMmgAk_B T left(1-expleft(-fracmghk_BTright)right)$$
Having found the density as a function of height, we can calculate the potential energy of the gas.
The gravitational potential energy density of a piece of gas at height $z$ above the zero-potential-energy reference level is just $u(z)=rho(z) g z$. So the total potential energy of the whole room is $$U=int_textroom u(z),dV=int_textroom rho(z) g zthinspace dV=int_textroomgzrho_0expleft(-fracmgzk_BTright)dV$$ and with our area $A$ and height $h$, this becomes $$U=Agrho_0int_0^h zexpleft(-fracmgzk_BTright),dz$$
To calculate this, we can use a standard integral result: $$int_0^h z exp left(-fraczCright)=C^2left(1-left(1+frachCright)expleft(-frachCright)right)$$which we can use with $C=frack_B Tmg$to find (deep breath)... $$U=fracA rho_0 k_B Tmcdotfrack_B Tmgleft(1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)right)$$
We can combine this with our earlier result for $rho_0$: $$U=fracMk_B Tmcdotfrac1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)1-expleft(-fracmghk_BTright)$$
Rearranging this slightly to clean it up, we get $$beginalign*U&=fracMk_B Tmcdotfracexpleft(fracmghk_B Tright)-left(1+fracmghk_B Tright)expleft(fracmghk_BTright)-1\&=fracMk_B Tmleft(1-fracmghk_BTfrac1expleft(fracmghk_BTright)-1right)\&=fracMk_B Tm-fracMghexpleft(fracmghk_BTright)-1\&=Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1endalign*$$ where in the last line we've used that the total mass is $M=Nm$.
So in answer to your question, the total energy of the ideal gas in a uniform temperature room of area (A) and height (z) in hydrostatic equilibrium would be (assuming I haven't made a mistake in the derivation above)...
$$E=frac52Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1$$
plus a constant.
answered 35 mins ago
canonicalmomentum
213
New contributor
canonicalmomentum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is where we get into some of the complexities of how we actually define the 'total' energy. As far as gravity is concerned, physics only depends on changes in the potential energy, not its absolute value; we can add a constant to the potential energy and get the same predictions! So the 'total potential energy' is not an entirely well-defined quantity.
As far as ideal gas theory is concerned, we treat gas particles as rigid balls and ignore any intermolecular interactions. So in this approximation, we ignore any electromagnetic forces between particles. (If we wanted to add them, we would reach something like the van der Waals model which I'm sure will come up later in the book you're working through, or one of many others!)
But what about gravitational potential energy? This is defined up to a constant, but there might still be differences in gravitational potential energy between the two situations, because the distribution of air density changes.
Just how much?
Since we're assuming uniform temperature, and assuming a uniform gravitational field, we can calculate the distribution of density in hydrostatic equilibrium. In hydrostatic equilibrium, the pressure and density are functions of height, but we still take them to be related by the ideal gas law $mP(z)=rho(z) k_BT$. We have $$fracdPdz=-rho(z)g=-fracmgk_BTp(z)$$ with the solution $$rho(z)=rho_0expleft(-fracmgzk_BTright)$$We set the density of air at ground level, $rho_0$ by requiring that the total mass of particles in the gas is constant: $$int_textroom rho(z) dV = M$$ If we assume the room has floor area $A$ and height $h$, this becomes $$M=Aint_0^h rho(z) dz = fracArho_0 k_B Tmgleft(1-expleft(-fracmghk_BTright)right)$$We can rearrange this to find $rho_0$ as a function of temperature: $$rho_0=fracMmgAk_B T left(1-expleft(-fracmghk_BTright)right)$$
Having found the density as a function of height, we can calculate the potential energy of the gas.
The gravitational potential energy density of a piece of gas at height $z$ above the zero-potential-energy reference level is just $u(z)=rho(z) g z$. So the total potential energy of the whole room is $$U=int_textroom u(z),dV=int_textroom rho(z) g zthinspace dV=int_textroomgzrho_0expleft(-fracmgzk_BTright)dV$$ and with our area $A$ and height $h$, this becomes $$U=Agrho_0int_0^h zexpleft(-fracmgzk_BTright),dz$$
To calculate this, we can use a standard integral result: $$int_0^h z exp left(-fraczCright)=C^2left(1-left(1+frachCright)expleft(-frachCright)right)$$which we can use with $C=frack_B Tmg$to find (deep breath)... $$U=fracA rho_0 k_B Tmcdotfrack_B Tmgleft(1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)right)$$
We can combine this with our earlier result for $rho_0$: $$U=fracMk_B Tmcdotfrac1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)1-expleft(-fracmghk_BTright)$$
Rearranging this slightly to clean it up, we get $$beginalign*U&=fracMk_B Tmcdotfracexpleft(fracmghk_B Tright)-left(1+fracmghk_B Tright)expleft(fracmghk_BTright)-1\&=fracMk_B Tmleft(1-fracmghk_BTfrac1expleft(fracmghk_BTright)-1right)\&=fracMk_B Tm-fracMghexpleft(fracmghk_BTright)-1\&=Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1endalign*$$ where in the last line we've used that the total mass is $M=Nm$.
So in answer to your question, the total energy of the ideal gas in a uniform temperature room of area (A) and height (z) in hydrostatic equilibrium would be (assuming I haven't made a mistake in the derivation above)...
$$E=frac52Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1$$
plus a constant.
answered 35 mins ago
canonicalmomentum
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This is where we get into some of the complexities of how we actually define the 'total' energy. As far as gravity is concerned, physics only depends on changes in the potential energy, not its absolute value; we can add a constant to the potential energy and get the same predictions! So the 'total potential energy' is not an entirely well-defined quantity.
As far as ideal gas theory is concerned, we treat gas particles as rigid balls and ignore any intermolecular interactions. So in this approximation, we ignore any electromagnetic forces between particles. (If we wanted to add them, we would reach something like the van der Waals model which I'm sure will come up later in the book you're working through, or one of many others!)
But what about gravitational potential energy? This is defined up to a constant, but there might still be differences in gravitational potential energy between the two situations, because the distribution of air density changes.
Just how much?
Since we're assuming uniform temperature, and assuming a uniform gravitational field, we can calculate the distribution of density in hydrostatic equilibrium. In hydrostatic equilibrium, the pressure and density are functions of height, but we still take them to be related by the ideal gas law $mP(z)=rho(z) k_BT$. We have $$fracdPdz=-rho(z)g=-fracmgk_BTp(z)$$ with the solution $$rho(z)=rho_0expleft(-fracmgzk_BTright)$$We set the density of air at ground level, $rho_0$ by requiring that the total mass of particles in the gas is constant: $$int_textroom rho(z) dV = M$$ If we assume the room has floor area $A$ and height $h$, this becomes $$M=Aint_0^h rho(z) dz = fracArho_0 k_B Tmgleft(1-expleft(-fracmghk_BTright)right)$$We can rearrange this to find $rho_0$ as a function of temperature: $$rho_0=fracMmgAk_B T left(1-expleft(-fracmghk_BTright)right)$$
Having found the density as a function of height, we can calculate the potential energy of the gas.
The gravitational potential energy density of a piece of gas at height $z$ above the zero-potential-energy reference level is just $u(z)=rho(z) g z$. So the total potential energy of the whole room is $$U=int_textroom u(z),dV=int_textroom rho(z) g zthinspace dV=int_textroomgzrho_0expleft(-fracmgzk_BTright)dV$$ and with our area $A$ and height $h$, this becomes $$U=Agrho_0int_0^h zexpleft(-fracmgzk_BTright),dz$$
To calculate this, we can use a standard integral result: $$int_0^h z exp left(-fraczCright)=C^2left(1-left(1+frachCright)expleft(-frachCright)right)$$which we can use with $C=frack_B Tmg$to find (deep breath)... $$U=fracA rho_0 k_B Tmcdotfrack_B Tmgleft(1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)right)$$
We can combine this with our earlier result for $rho_0$: $$U=fracMk_B Tmcdotfrac1-left(1+fracmghk_B Tright)expleft(-fracmghk_B Tright)1-expleft(-fracmghk_BTright)$$
Rearranging this slightly to clean it up, we get $$beginalign*U&=fracMk_B Tmcdotfracexpleft(fracmghk_B Tright)-left(1+fracmghk_B Tright)expleft(fracmghk_BTright)-1\&=fracMk_B Tmleft(1-fracmghk_BTfrac1expleft(fracmghk_BTright)-1right)\&=fracMk_B Tm-fracMghexpleft(fracmghk_BTright)-1\&=Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1endalign*$$ where in the last line we've used that the total mass is $M=Nm$.
So in answer to your question, the total energy of the ideal gas in a uniform temperature room of area (A) and height (z) in hydrostatic equilibrium would be (assuming I haven't made a mistake in the derivation above)...
$$E=frac52Nk_B T - fracNmghexpleft(fracmghk_B Tright)-1$$
plus a constant.
answered 35 mins ago
canonicalmomentum
213
New contributor
canonicalmomentum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 35 mins ago
canonicalmomentum
213
New contributor
canonicalmomentum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 35 mins ago
canonicalmomentum
213
answered 35 mins ago
canonicalmomentum
213
213
New contributor
canonicalmomentum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
canonicalmomentum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
canonicalmomentum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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And some air will leave the room...
– Pieter
3 hours ago
yes, but assume not. All the air stays in the same room always.
– Duartejfs
3 hours ago
You mean that air at a higher point in the room should have more energy than air at a lower point?
– Aaron Stevens
2 hours ago
1
Due to the room's size you are not considering the difference in potential energy between particles at differeht heights.
– Wolphram jonny
2 hours ago
@Wolphramjonny The height distribution does not change much with temperature. Potential energy does not matter. What changes is the rotational kinetic energy of air molecules.
– Pieter
24 mins ago