Is it possible to boil a liquid by just mixing many immiscible liquids together? Announcing...
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Is it possible to boil a liquid by just mixing many immiscible liquids together?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Vapor pressure of immiscible liquidsHow does the Freezing Point fall in a solution?What is the molecular interpretation of Raoult's law?Miscibility of pairwise miscible liquidsRaoult's Law and Mole FractionHeating of mixture of gases with one gas barely above its boiling pointIs the Ebullioscopic constant temperature-dependent?Boiling point elevation for a mixture of ethanol and waterReal boiling temperatureClausius–Clapeyron relation for states of water and vaporProperties of azeotropes
$begingroup$
In open air, when vapour pressure reaches 1 atm, boiling takes place.
I read that if we add two immiscible liquids together, the total vapour pressure of the 'mixture' is close to p=p*A+p*B, means that the vapour pressure of the 'mixture' is higher than the vapour pressure of both the constituent liquids, subsequently means that the 'mixture' boils at lower temperature than both of the constituent liquids.
If we have many liquids which are not immiscible with each other, and we add them together to form a 'mixture' such that p=p*A+p*B+...>1 atm, will boiling results at room temperature?
I suspect we have many liquids which are not miscible with each other, since we classify solvents as organic solvent and inorganic solvent only, but let's say we indeed have that (or we have two immiscible liquids volatile enough to make p>1atm), what will be observed? Boiling at room temperature? In that case, we can use the vapour to do work (eg rotate turbine), but when we mix immiscible liquid together there should be no energy transaction right? The particles do not react with each other. If all these are true, we are actually doing work without any input which violates the first law, so there must be something wrong with my reasoning above.
physical-chemistry thermodynamics mixtures
asked 2 hours ago
The99sLearnerThe99sLearner
156211
$endgroup$
add a comment |
$begingroup$
In open air, when vapour pressure reaches 1 atm, boiling takes place.
I read that if we add two immiscible liquids together, the total vapour pressure of the 'mixture' is close to p=p*A+p*B, means that the vapour pressure of the 'mixture' is higher than the vapour pressure of both the constituent liquids, subsequently means that the 'mixture' boils at lower temperature than both of the constituent liquids.
If we have many liquids which are not immiscible with each other, and we add them together to form a 'mixture' such that p=p*A+p*B+...>1 atm, will boiling results at room temperature?
I suspect we have many liquids which are not miscible with each other, since we classify solvents as organic solvent and inorganic solvent only, but let's say we indeed have that (or we have two immiscible liquids volatile enough to make p>1atm), what will be observed? Boiling at room temperature? In that case, we can use the vapour to do work (eg rotate turbine), but when we mix immiscible liquid together there should be no energy transaction right? The particles do not react with each other. If all these are true, we are actually doing work without any input which violates the first law, so there must be something wrong with my reasoning above.
physical-chemistry thermodynamics mixtures
asked 2 hours ago
The99sLearnerThe99sLearner
156211
$endgroup$
add a comment |
$begingroup$
In open air, when vapour pressure reaches 1 atm, boiling takes place.
I read that if we add two immiscible liquids together, the total vapour pressure of the 'mixture' is close to p=p*A+p*B, means that the vapour pressure of the 'mixture' is higher than the vapour pressure of both the constituent liquids, subsequently means that the 'mixture' boils at lower temperature than both of the constituent liquids.
If we have many liquids which are not immiscible with each other, and we add them together to form a 'mixture' such that p=p*A+p*B+...>1 atm, will boiling results at room temperature?
I suspect we have many liquids which are not miscible with each other, since we classify solvents as organic solvent and inorganic solvent only, but let's say we indeed have that (or we have two immiscible liquids volatile enough to make p>1atm), what will be observed? Boiling at room temperature? In that case, we can use the vapour to do work (eg rotate turbine), but when we mix immiscible liquid together there should be no energy transaction right? The particles do not react with each other. If all these are true, we are actually doing work without any input which violates the first law, so there must be something wrong with my reasoning above.
physical-chemistry thermodynamics mixtures
asked 2 hours ago
The99sLearnerThe99sLearner
156211
$endgroup$
In open air, when vapour pressure reaches 1 atm, boiling takes place.
I read that if we add two immiscible liquids together, the total vapour pressure of the 'mixture' is close to p=p*A+p*B, means that the vapour pressure of the 'mixture' is higher than the vapour pressure of both the constituent liquids, subsequently means that the 'mixture' boils at lower temperature than both of the constituent liquids.
If we have many liquids which are not immiscible with each other, and we add them together to form a 'mixture' such that p=p*A+p*B+...>1 atm, will boiling results at room temperature?
I suspect we have many liquids which are not miscible with each other, since we classify solvents as organic solvent and inorganic solvent only, but let's say we indeed have that (or we have two immiscible liquids volatile enough to make p>1atm), what will be observed? Boiling at room temperature? In that case, we can use the vapour to do work (eg rotate turbine), but when we mix immiscible liquid together there should be no energy transaction right? The particles do not react with each other. If all these are true, we are actually doing work without any input which violates the first law, so there must be something wrong with my reasoning above.
physical-chemistry thermodynamics mixtures
physical-chemistry thermodynamics mixtures
asked 2 hours ago
The99sLearnerThe99sLearner
156211
asked 2 hours ago
The99sLearnerThe99sLearner
156211
asked 2 hours ago
The99sLearnerThe99sLearner
156211
asked 2 hours ago
The99sLearnerThe99sLearner
156211
asked 2 hours ago
The99sLearnerThe99sLearner
156211
156211
add a comment |
add a comment |
2 Answers
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$begingroup$
Nothing special would happen, immiscible liquids would just form layers. As for the expression, $$p_T=Sigma p^o_i$$
I suggest you read this answer. Quoting Ivan Neretin:
This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.
It would take infinite time for that expression to prove itself correct.
In speaking within an average man's lifespan, the liquids would remain in separate layers, with the pressure above liquid surface would remain approximately the pure liquid vapour pressure ($p^o$) of the uppermost layer if left undisturbed (according to your question: no energy transactions).
Don't worry about the first law: it is rarely violated these days.
This "equilibrium" can be attained sooner by agitation (stirring) of the mixture, but mechanical stirring certainly counts as work done.
From a chemguide page:
Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapour pressure you measure will simply be the vapour pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapour. The top one is sealing it in.
Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapour pressures are bound to reach the external pressure before the vapour pressure of either of the individual components get there.
In other words, although less evident, even if you do get the mixture to boil and do work, it would be because you had done work on it earlier during agitation.
answered 1 hour ago
William R. EbenezerWilliam R. Ebenezer
82718
$endgroup$
add a comment |
$begingroup$
Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil. This is what steam distillation is all about.
As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.
So it goes.
answered 1 hour ago
Ivan NeretinIvan Neretin
23.9k34990
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
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$begingroup$
Nothing special would happen, immiscible liquids would just form layers. As for the expression, $$p_T=Sigma p^o_i$$
I suggest you read this answer. Quoting Ivan Neretin:
This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.
It would take infinite time for that expression to prove itself correct.
In speaking within an average man's lifespan, the liquids would remain in separate layers, with the pressure above liquid surface would remain approximately the pure liquid vapour pressure ($p^o$) of the uppermost layer if left undisturbed (according to your question: no energy transactions).
Don't worry about the first law: it is rarely violated these days.
This "equilibrium" can be attained sooner by agitation (stirring) of the mixture, but mechanical stirring certainly counts as work done.
From a chemguide page:
Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapour pressure you measure will simply be the vapour pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapour. The top one is sealing it in.
Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapour pressures are bound to reach the external pressure before the vapour pressure of either of the individual components get there.
In other words, although less evident, even if you do get the mixture to boil and do work, it would be because you had done work on it earlier during agitation.
answered 1 hour ago
William R. EbenezerWilliam R. Ebenezer
82718
$endgroup$
add a comment |
$begingroup$
Nothing special would happen, immiscible liquids would just form layers. As for the expression, $$p_T=Sigma p^o_i$$
I suggest you read this answer. Quoting Ivan Neretin:
This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.
It would take infinite time for that expression to prove itself correct.
In speaking within an average man's lifespan, the liquids would remain in separate layers, with the pressure above liquid surface would remain approximately the pure liquid vapour pressure ($p^o$) of the uppermost layer if left undisturbed (according to your question: no energy transactions).
Don't worry about the first law: it is rarely violated these days.
This "equilibrium" can be attained sooner by agitation (stirring) of the mixture, but mechanical stirring certainly counts as work done.
From a chemguide page:
Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapour pressure you measure will simply be the vapour pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapour. The top one is sealing it in.
Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapour pressures are bound to reach the external pressure before the vapour pressure of either of the individual components get there.
In other words, although less evident, even if you do get the mixture to boil and do work, it would be because you had done work on it earlier during agitation.
answered 1 hour ago
William R. EbenezerWilliam R. Ebenezer
82718
$endgroup$
add a comment |
$begingroup$
Nothing special would happen, immiscible liquids would just form layers. As for the expression, $$p_T=Sigma p^o_i$$
I suggest you read this answer. Quoting Ivan Neretin:
This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.
It would take infinite time for that expression to prove itself correct.
In speaking within an average man's lifespan, the liquids would remain in separate layers, with the pressure above liquid surface would remain approximately the pure liquid vapour pressure ($p^o$) of the uppermost layer if left undisturbed (according to your question: no energy transactions).
Don't worry about the first law: it is rarely violated these days.
This "equilibrium" can be attained sooner by agitation (stirring) of the mixture, but mechanical stirring certainly counts as work done.
From a chemguide page:
Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapour pressure you measure will simply be the vapour pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapour. The top one is sealing it in.
Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapour pressures are bound to reach the external pressure before the vapour pressure of either of the individual components get there.
In other words, although less evident, even if you do get the mixture to boil and do work, it would be because you had done work on it earlier during agitation.
answered 1 hour ago
William R. EbenezerWilliam R. Ebenezer
82718
$endgroup$
Nothing special would happen, immiscible liquids would just form layers. As for the expression, $$p_T=Sigma p^o_i$$
I suggest you read this answer. Quoting Ivan Neretin:
This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.
It would take infinite time for that expression to prove itself correct.
In speaking within an average man's lifespan, the liquids would remain in separate layers, with the pressure above liquid surface would remain approximately the pure liquid vapour pressure ($p^o$) of the uppermost layer if left undisturbed (according to your question: no energy transactions).
Don't worry about the first law: it is rarely violated these days.
This "equilibrium" can be attained sooner by agitation (stirring) of the mixture, but mechanical stirring certainly counts as work done.
From a chemguide page:
Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapour pressure you measure will simply be the vapour pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapour. The top one is sealing it in.
Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapour pressures are bound to reach the external pressure before the vapour pressure of either of the individual components get there.
In other words, although less evident, even if you do get the mixture to boil and do work, it would be because you had done work on it earlier during agitation.
answered 1 hour ago
William R. EbenezerWilliam R. Ebenezer
82718
edited 1 hour ago
answered 1 hour ago
William R. EbenezerWilliam R. Ebenezer
82718
answered 1 hour ago
William R. EbenezerWilliam R. Ebenezer
82718
answered 1 hour ago
William R. EbenezerWilliam R. Ebenezer
82718
82718
add a comment |
add a comment |
$begingroup$
Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil. This is what steam distillation is all about.
As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.
So it goes.
answered 1 hour ago
Ivan NeretinIvan Neretin
23.9k34990
$endgroup$
add a comment |
$begingroup$
Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil. This is what steam distillation is all about.
As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.
So it goes.
answered 1 hour ago
Ivan NeretinIvan Neretin
23.9k34990
$endgroup$
add a comment |
$begingroup$
Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil. This is what steam distillation is all about.
As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.
So it goes.
answered 1 hour ago
Ivan NeretinIvan Neretin
23.9k34990
$endgroup$
Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil. This is what steam distillation is all about.
As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.
So it goes.
answered 1 hour ago
Ivan NeretinIvan Neretin
23.9k34990
answered 1 hour ago
Ivan NeretinIvan Neretin
23.9k34990
answered 1 hour ago
Ivan NeretinIvan Neretin
23.9k34990
answered 1 hour ago
Ivan NeretinIvan Neretin
23.9k34990
23.9k34990
add a comment |
add a comment |
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