Hooray! The particles feel attractive forces, and the stored PE is ultimately given by the definition of potential energy: ΔPE = ∫F•dr, where dr is the displacement over a very small interval. We would apply this for each interaction between neighboring/bonded molecules. Calculus is needed because the force itself is a function of the distance.
Great video! The only detail is that the potential energy actually increases as the molecules (or atoms) get closer. For instance, in a system of two molecules with opposite charges, there is an attractive electrostatic force between the molecules. This force can pull the molecules together, and as they get closer, the potential energy stored in the system increases. The closer the molecules get, the greater the potential energy stored in the system.
@Sergio Miguel Espinoza Orellana, you've cited a common misconception. In fact, molecules *lose* potential energy when they move closer in the presence of an attractive field force (such as an attractive electrostatic force). The misconception usually arises because it's intuitive--but incorrect--to think, "the system has gained stability when it transitions to a solid, so there must be more PE stored in the system once it's a solid." The opposite is actually true: the increased stability exists because the system is at a *lower* energy state. To decrease in energy and reach that lower energy state, it must *lose* potential energy. This is precisely what energy / phase transition graphs show. The same concept also applies to binding energy, which suffers from similar misconceptions and is the energy that is *lost* when a system binds together. See, for example: "As the water cools, it will reach 32°F, at which point it will start to freeze. Heat still flows out, but it comes from the energy lost as the water freezes. You could say that's an exothermic process." van.physics.illinois.edu/ask/listing/1629
@@danielm9463 I think I misunderstood your explanation of PE in the video. In conclusion, I agree with you, but I would like to mention that the PE of two molecules when they are separated, is considered zero. When you said: "farther apart means more PE" it made me think of a positive quantity. Anyway, I think we both will agree with this idea: when two molecules are separated enough so that the intermolecular forces between them are negligible, the PE of the system is zero, and as the molecules get closer, the potential gets more negative (so, as you said decreases) until reaching a minimum when there is a net attraction, and phase changes may occur. So considering that the potential energy gets more negative as molecules get closer, yes, you can conclude that when they are farther apart PE increases but we must remember that it approaches zero. These would also clarify another comment in this video (see below): "the farther they are ... the greater is the PE" ? That would mean if they were on an infinite distance between... the PE would also be infinite ?" Considering this explanation, the PE would be just zero at an infinite distance.
@@sergiomiguelespinozaorella214, this is another extremely common misconception, which even appears in some texts. The PE is not 0 at infinity *because* the intermolecular force is 0 at infinity. Instead, the PE is 0 at infinity because we arbitrarily select infinity as our reference point. This statement: "the farther they are ... the greater is the PE? That would mean if they were on an infinite distance between... the PE would also be infinite?" is based on a mathematical misconception that PE is linearly related to distance, which is not true. The definition of PE is: PE_{at point A} = -∫dW_{done by field when moving from an arbitrary reference point to the point A} = -∫F*dr. Thus PE increases as distance increase in the presence of an attractive force. If we reach a distance where the intermolecular force is 0 (which never actually happens but does effectively occur), then no additional work is required to separate the molecules farther apart because there is no force. Hence, it's possible for both statements to be true: (a) the PE increases with distance in the presence of an attractive intermolecular force, and (b) the PE is not infinite at infinite distance. Statement (b) would still be true even if we arbitrarily decided that PE=1 keV at infinity rather than PE=0 at infinity. So, PE=0 is unrelated to the fact that F=0 at infinity. PE is a relative quantity. Absolute values of PE require an arbitrary decision wherein we choose a reference point at which we will set PE=0. We can choose any point as the reference. The reason we use a point infinitely far away is because this creates a convention where bound systems have negative PE and unbound systems have 0 PE.
Hello, How do you explain that "the farther they are ... the greater is the PE" ? That would mean if they were on a infinite distance between... the PE would also be infinite ?... Actualy PE between 2 molecules decrease with distance (and so with the accessible volume : see Van der Waals interaction for exemple).
Thanks for the message and the questions. Your first statement: "the farther they are ... the greater is the PE? That would mean if they were on an infinite distance between... the PE would also be infinite?" is based on a mathematical misconception that PE is linearly related to distance, which is not true. The definition of PE is: PE_{at point A} = -∫dW_{done by field when moving from an arbitrary reference point to the point A} = -∫F*dr. Thus PE increases as distance increase in the presence of an attractive force. If we reach a distance where the intermolecular force is 0 (which never actually happens but does effectively occur), then no additional work is required to separate the molecules farther apart because there is no force. Hence, it's possible for both statements to be true: (a) the PE increases with distance in the presence of an attractive intermolecular force, and (b) the PE is not infinite at infinite distance. Regarding the second statement about Van der Waals forces, that level of complexity is beyond the scope of this video (which is simply an introduction to temperature change and phase change for a high school physics class), but also is a reasonable thing to ignore when simplifying and focusing on net long-range interactions associated with phase transitions. The simplification that the net intermolecular is always attractive explains why the temperature doesn't rise when changing phase (because we're adding PE to the system) and explains why the increase in temperature is sometimes slightly less than what we'd calculate if all the heat converted to KE of the particles. The true picture is more complex, of course, as you suggest. Even the Van der Waals interaction you cite can be attractive or repulsive depending on the distance. However, to return to your first question, it is always true that increasing separation causes the PE to increase in the presence of an attractive force. This does not imply that infinite separation produces infinite PE, because the net attractive force decreases with distance. As long as there is any attractive force at all, no matter how small, additional work is required to move the molecules farther apart. This work constitutes a tiny additional bit of PE that gets added to the system. If we reach a point so far away that the force is 0, then additional separation does not add to the PE because no additional work is required to move the molecules farther apart--in other words, there is no longer an attractive force, and so the statement "PE increases with distance in the presence of an attractive intermolecular force" does not apply. At some point, the increase in PE caused by additional separation is negligible because the force becomes negligible. In fact, because of the way the net intermolecular force drops off, almost all of the increase in PE occurs when changing phase, and this is exactly how introductory physics classes explain the latent heat that gets absorbed when changing phase.
@@davidbuckingham6594 This is not correct. PE is lower in the solid state than in the liquid state. If this were true, then we would have to ADD energy to water in order to change it to ice. The fact that we must REMOVE energy from water indicates that we're reducing the PE, and thus the ice (the solid) has less PE than the water (the liquid).
@@davidbuckingham6594 That is incorrect, but it's a very common misconception. The PE is higher in the liquid state than in the solid state. This is why you must add heat to melt ice. The heat energy you add to ice (a solid) becomes PE in the liquid.
Sir does this statement PE increase with distance Always true As this is only true if the electric field is constant and its area of application is infinite ? And one more doubt sir We generally see in our daily life as well as in classical physics that PE AND KE ARE INVERESELY proportional But in terms of molecules its not why ?? Well at the last I loved your explanation thanks sir Hope I will get a reply 😀
These are great questions!! PE increases with distance whenever we have an attractive force. This is true even when the field is *not* constant. Most force fields are infinite (they stretch out forever) but they decrease in strength as you move farther away from the source. Imagine you're holding an electron that is 1 km away from a proton. The attractive force is nearly 0. You take one more step away from the proton, increasing your distance to 1.001 km. The electron in hand *does* experience an increase in PE. But because the force is negligible (F is almost 0), the added PE will also be negligible (nearly 0). For a field that decreases with distance (or decreases with distance squared), when we're far enough away, we can imagine that the field is 0 even though it's infinite. This is what we do when we model gases: we ignore any additional PE that the molecules might gain as the volume of the gas increases. However, it's more complicated in the real world with real molecules because sometimes the force can switch between attractive and repulsive. Everything in the previous paragraph comes straight from the definition of potential energy: (∆PE between point A and B) = -(work done by the field in moving from point A to B). If the force is attractive, then moving opposite that force will produce negative work by the field. Then applying a negative sign to that work will create positive ∆PE. I really love your observation that, in most cases we've studied as high school/university physics students, PE and KE increase/decrease in opposite directions. Yet with thermal physics that doesn't seem to be the case. Here's the explanation for that. The other scenarios we've studied are things like: "you drop a ball from a height of 2 m. As it falls, it gains KE and loses PE." In these types of mechanics situations, there's *no energy being inputted into the system*. Thus, what we're studying are transfers of energy between different types (e.g., KE → PE or PE → KE). Phase changes are completely different, because they *require* external energy to be input into the system or removed from the system. This is very different from many of the other physics situations we've studied. With phase changes, because we're putting energy into the system, that added energy can go into PE or KE, and they can increase together.
Great animation and explanation !! Many thanks !
i have been searching all over google to understand how PE works in particles and now i understand
Hooray! The particles feel attractive forces, and the stored PE is ultimately given by the definition of potential energy: ΔPE = ∫F•dr, where dr is the displacement over a very small interval. We would apply this for each interaction between neighboring/bonded molecules. Calculus is needed because the force itself is a function of the distance.
Great video! The only detail is that the potential energy actually increases as the molecules (or atoms) get closer. For instance, in a system of two molecules with opposite charges, there is an attractive electrostatic force between the molecules. This force can pull the molecules together, and as they get closer, the potential energy stored in the system increases. The closer the molecules get, the greater the potential energy stored in the system.
@Sergio Miguel Espinoza Orellana, you've cited a common misconception. In fact, molecules *lose* potential energy when they move closer in the presence of an attractive field force (such as an attractive electrostatic force). The misconception usually arises because it's intuitive--but incorrect--to think, "the system has gained stability when it transitions to a solid, so there must be more PE stored in the system once it's a solid." The opposite is actually true: the increased stability exists because the system is at a *lower* energy state. To decrease in energy and reach that lower energy state, it must *lose* potential energy. This is precisely what energy / phase transition graphs show. The same concept also applies to binding energy, which suffers from similar misconceptions and is the energy that is *lost* when a system binds together.
See, for example:
"As the water cools, it will reach 32°F, at which point it will start to freeze. Heat still flows out, but it comes from the energy lost as the water freezes. You could say that's an exothermic process."
van.physics.illinois.edu/ask/listing/1629
@@danielm9463 I think I misunderstood your explanation of PE in the video. In conclusion, I agree with you, but I would like to mention that the PE of two molecules when they are separated, is considered zero. When you said: "farther apart means more PE" it made me think of a positive quantity. Anyway, I think we both will agree with this idea: when two molecules are separated enough so that the intermolecular forces between them are negligible, the PE of the system is zero, and as the molecules get closer, the potential gets more negative (so, as you said decreases) until reaching a minimum when there is a net attraction, and phase changes may occur. So considering that the potential energy gets more negative as molecules get closer, yes, you can conclude that when they are farther apart PE increases but we must remember that it approaches zero. These would also clarify another comment in this video (see below): "the farther they are ... the greater is the PE" ? That would mean if they were on an infinite distance between... the PE would also be infinite ?" Considering this explanation, the PE would be just zero at an infinite distance.
@@sergiomiguelespinozaorella214, this is another extremely common misconception, which even appears in some texts. The PE is not 0 at infinity *because* the intermolecular force is 0 at infinity. Instead, the PE is 0 at infinity because we arbitrarily select infinity as our reference point.
This statement: "the farther they are ... the greater is the PE? That would mean if they were on an infinite distance between... the PE would also be infinite?" is based on a mathematical misconception that PE is linearly related to distance, which is not true. The definition of PE is: PE_{at point A} = -∫dW_{done by field when moving from an arbitrary reference point to the point A} = -∫F*dr. Thus PE increases as distance increase in the presence of an attractive force. If we reach a distance where the intermolecular force is 0 (which never actually happens but does effectively occur), then no additional work is required to separate the molecules farther apart because there is no force. Hence, it's possible for both statements to be true: (a) the PE increases with distance in the presence of an attractive intermolecular force, and (b) the PE is not infinite at infinite distance.
Statement (b) would still be true even if we arbitrarily decided that PE=1 keV at infinity rather than PE=0 at infinity.
So, PE=0 is unrelated to the fact that F=0 at infinity. PE is a relative quantity. Absolute values of PE require an arbitrary decision wherein we choose a reference point at which we will set PE=0. We can choose any point as the reference. The reason we use a point infinitely far away is because this creates a convention where bound systems have negative PE and unbound systems have 0 PE.
great animation
Hello, How do you explain that "the farther they are ... the greater is the PE" ? That would mean if they were on a infinite distance between... the PE would also be infinite ?...
Actualy PE between 2 molecules decrease with distance (and so with the accessible volume : see Van der Waals interaction for exemple).
Thanks for the message and the questions.
Your first statement: "the farther they are ... the greater is the PE? That would mean if they were on an infinite distance between... the PE would also be infinite?" is based on a mathematical misconception that PE is linearly related to distance, which is not true. The definition of PE is: PE_{at point A} = -∫dW_{done by field when moving from an arbitrary reference point to the point A} = -∫F*dr. Thus PE increases as distance increase in the presence of an attractive force. If we reach a distance where the intermolecular force is 0 (which never actually happens but does effectively occur), then no additional work is required to separate the molecules farther apart because there is no force. Hence, it's possible for both statements to be true: (a) the PE increases with distance in the presence of an attractive intermolecular force, and (b) the PE is not infinite at infinite distance.
Regarding the second statement about Van der Waals forces, that level of complexity is beyond the scope of this video (which is simply an introduction to temperature change and phase change for a high school physics class), but also is a reasonable thing to ignore when simplifying and focusing on net long-range interactions associated with phase transitions. The simplification that the net intermolecular is always attractive explains why the temperature doesn't rise when changing phase (because we're adding PE to the system) and explains why the increase in temperature is sometimes slightly less than what we'd calculate if all the heat converted to KE of the particles.
The true picture is more complex, of course, as you suggest. Even the Van der Waals interaction you cite can be attractive or repulsive depending on the distance.
However, to return to your first question, it is always true that increasing separation causes the PE to increase in the presence of an attractive force. This does not imply that infinite separation produces infinite PE, because the net attractive force decreases with distance. As long as there is any attractive force at all, no matter how small, additional work is required to move the molecules farther apart. This work constitutes a tiny additional bit of PE that gets added to the system. If we reach a point so far away that the force is 0, then additional separation does not add to the PE because no additional work is required to move the molecules farther apart--in other words, there is no longer an attractive force, and so the statement "PE increases with distance in the presence of an attractive intermolecular force" does not apply. At some point, the increase in PE caused by additional separation is negligible because the force becomes negligible. In fact, because of the way the net intermolecular force drops off, almost all of the increase in PE occurs when changing phase, and this is exactly how introductory physics classes explain the latent heat that gets absorbed when changing phase.
PE is higher in the solid state than the liquid state.
@@davidbuckingham6594 This is not correct. PE is lower in the solid state than in the liquid state. If this were true, then we would have to ADD energy to water in order to change it to ice. The fact that we must REMOVE energy from water indicates that we're reducing the PE, and thus the ice (the solid) has less PE than the water (the liquid).
@@davidbuckingham6594 That is incorrect, but it's a very common misconception. The PE is higher in the liquid state than in the solid state. This is why you must add heat to melt ice. The heat energy you add to ice (a solid) becomes PE in the liquid.
Sir does this statement
PE increase with distance
Always true
As this is only true if the electric field is constant and its area of application is infinite ?
And one more doubt sir
We generally see in our daily life as well as in classical physics that PE AND KE ARE INVERESELY proportional
But in terms of molecules its not why ??
Well at the last
I loved your explanation thanks sir
Hope I will get a reply 😀
These are great questions!!
PE increases with distance whenever we have an attractive force. This is true even when the field is *not* constant. Most force fields are infinite (they stretch out forever) but they decrease in strength as you move farther away from the source. Imagine you're holding an electron that is 1 km away from a proton. The attractive force is nearly 0. You take one more step away from the proton, increasing your distance to 1.001 km. The electron in hand *does* experience an increase in PE. But because the force is negligible (F is almost 0), the added PE will also be negligible (nearly 0). For a field that decreases with distance (or decreases with distance squared), when we're far enough away, we can imagine that the field is 0 even though it's infinite. This is what we do when we model gases: we ignore any additional PE that the molecules might gain as the volume of the gas increases. However, it's more complicated in the real world with real molecules because sometimes the force can switch between attractive and repulsive.
Everything in the previous paragraph comes straight from the definition of potential energy: (∆PE between point A and B) = -(work done by the field in moving from point A to B). If the force is attractive, then moving opposite that force will produce negative work by the field. Then applying a negative sign to that work will create positive ∆PE.
I really love your observation that, in most cases we've studied as high school/university physics students, PE and KE increase/decrease in opposite directions. Yet with thermal physics that doesn't seem to be the case. Here's the explanation for that. The other scenarios we've studied are things like: "you drop a ball from a height of 2 m. As it falls, it gains KE and loses PE." In these types of mechanics situations, there's *no energy being inputted into the system*. Thus, what we're studying are transfers of energy between different types (e.g., KE → PE or PE → KE). Phase changes are completely different, because they *require* external energy to be input into the system or removed from the system. This is very different from many of the other physics situations we've studied. With phase changes, because we're putting energy into the system, that added energy can go into PE or KE, and they can increase together.
@@danielm9463 thanks a lot sir 😀
Why do the particles have more internal potential energy when they are farther apart?
Refer 2:20
Why they vibrate
Due to close packing