Log in 𝐢ᴀɴᴅʏ_Qɪ 2 years agoPosted 2 years ago. Direct link to 𝐢ᴀɴᴅʏ_Qɪ's post “Are "q" and "w" defined b...” Are "q" and "w" defined based on how useful the energy transfer is to humans? For example, what sets work done by a system apart from heat transferred? Can heat transferred be considered work? • (11 votes) Sully 2 years agoPosted 2 years ago. Direct link to Sully's post “At 7:72, how can a system...” At 7:72, how can a system absorb heat/energy from the surroundings, yet feel cold? Does it have something to do with forming/breaking chemical bonds or... • (3 votes) Richard 2 years agoPosted 2 years ago. Direct link to Richard's post “If you're handling someth...” If you're handling something undergoing an endothermic reaction it would feel cold to yourself because the reaction is drawing heat from the surroundings; in this case you. A lack of heat is interpreted by our bodies as being cold. Hope that helps. (7 votes) connorJung10 a year agoPosted a year ago. Direct link to connorJung10's post “I'm not sure if this is t...” I'm not sure if this is the best place to ask, but what is enthalpy exactly? Thus far, I have come across two definitions: 1) the total heat content of a system 2) the change in heat content during a reaction. Are both of these correct? If so, going off of the former definition, since enthalpy is impossible to calculate, how do we know that certain systems have more than others? If we raise both a cube of ice and a glass of water in isolated systems by the same temperature, their change in enthalpies, or just enthalpies I suppose, would be the same, is this correct? • (1 vote) Richard a year agoPosted a year ago. Direct link to Richard's post “Enthalpy is formally defi...” Enthalpy is formally defined as the sum of a system’s internal energy and the product of its pressure and volume. Or: H = E + PV, where H is enthalpy, E is internal energy, P is pressure, and V is volume. Internal energy is the sum of heat and work from the first law of thermodynamics, while the pressure-volume term expresses the work required to establish the system’s physical surroundings, or to make room for it by displacing its surroundings. Like internal energy, it’s difficult to measure the absolute amount of enthalpy in a system so instead we measure the change in enthalpy from one state to another. Additionally by making the assumption that external pressure is constant we can do some algebra to find that the change in enthalpy is equal to the heat exchanged by the system and the surroundings. We can make the constant pressure assumption because most chemical reactions occur in open containers which do not significantly alter atmospheric pressure. So: ΔH = q, where q is heat. So in the vast majority of chemistry situations, enthalpy is measuring the amount of heat released or absorbed by the chemicals over the course of a reaction. Hope that helps. (4 votes) Angelina Dewar 2 years agoPosted 2 years ago. Direct link to Angelina Dewar's post “I swear I've seen the exa...” I swear I've seen the exact opposite convention for work used... • (1 vote) Nicole 2 years agoPosted 2 years ago. Direct link to Nicole's post “Yes, it is very much poss...” Yes, it is very much possible you have seen the exact opposite convention used. This is explained in further detail in the Khan Academy article on first law thermo: "We have to be very careful with the first law. About half of textbooks, teachers, and professors write the first law of thermodynamics as ΔU=Q+W and the other half write it as ΔU=Q-W. Both equations are correct, and they say the same thing. The reason for the difference is that in the formula ΔU=Q+W, we are assuming that W represents the work done on the system, and when we use ΔU=Q-W, we are assuming that W represents the work done by the system. The two different equations are equivalent since, https://www.khanacademy.org/science/ap-physics-2/ap-thermodynamics/ap-laws-of-thermodynamics/a/what-is-the-first-law-of-thermodynamics?modal=1 (3 votes) Lightning Driver, Esq. 10 months agoPosted 10 months ago. Direct link to Lightning Driver, Esq.'s post “So he just said E is the ...” So he just said E is the sum of the kinetic and potential energy in the system, then said that it's the sum of the heat gained/released + the work done. First of all, what work is he referring to, and second of all, how can it be the sum of both of them? Edit: also, how does the formula show that energy cannot be created or destroyed, as the first law of thermodynamics states? Second edit: I learned in another course that enthalpy is the internal energy + the product of the pressure and volume. The change in internal energy, as the video says, is q + w. If the product and volume are constant, then the change in enthalpy should be equal to the change in internal energy. Then, in • (1 vote) Lightning Driver, Esq. 10 months agoPosted 10 months ago. Direct link to Lightning Driver, Esq.'s post “Yes I'm answering my own ...” Yes I'm answering my own questions :P. Feel free to correct me if I'm wrong. 1. So he did say that E is the sum of the kinetic and potential energy in the system, but he didn't say that E is also the sum of the heat gained/released and the work done. He said that ΔE = q + w. Delta means change. When the energy in a system changes, it is because it gained/lost heat and/or had work done on it/the surroundings. So the work referred to is the work done on it, which will be negative work, or on the surroundings, which will be positive work. 2. The equation ΔE = q + w doesn't prove the first law of thermodynamics, but is rather a derivation from the law. When a system does work, it releases energy to the surroundings. The energy released is the heat released and the work exerted. In other words, energy cannot be created or destroyed, therefore when a system changes it's energy, it's because it gained/lost heat and/or did work/had work done on it. Number 3 I still didn't figure out, but my best guess is that the enthalpy change referred to is only for chemical reactions, and any work done is accidental. Therefore, the work done will be about 0, so it can be neglected. Again, if I am wrong, feel free to correct me, especially for question 3. (2 votes) connorJung10 a year agoPosted a year ago. Direct link to connorJung10's post “At 4:39, Jay states how t...” At • (1 vote) Richard a year agoPosted a year ago. Direct link to Richard's post “With an open container, t...” With an open container, the reaction is mostly going to doing work to the air surrounding the container (the air still being part of the surroundings). This will affect the air pressure of the room very slightly. But the pressure change of a chemical reaction on atmospheric pressure is so insignificant that we can still assume a constant air pressure in the surroundings. Hope that helps. (2 votes) alexfontainesstatus a year agoPosted a year ago. Direct link to alexfontainesstatus's post “kind of a confusing expla...” kind of a confusing explanation. so does q represent only heat? When you say 6000 J are flowing into the system do you mean that the piston is being force down compressing the He gas? how do you know the reaction is carried out by constant external pressure at • (1 vote) Richard a year agoPosted a year ago. Direct link to Richard's post “Yes, q is the usual varia...” Yes, q is the usual variable used in these kind of thermodynamic problems. Heat is the transfer of energy, specifically thermal energy or the energy associated with temperature, from one object to another. When 6000 J of heat is being added to the container, we are supplying 6000 J of thermal energy to the container. Pushing down on the piston would be us doing work on the container (positive work), which isn’t what is happening. Instead the 6000 J of heat is expanding the volume of the gas causing it to push back on the piston and do work on the surroundings (negative work). Jay says the reaction is performed in an open container at Hope that helps. (1 vote) jaehyung.jason a month agoPosted a month ago. Direct link to jaehyung.jason's post “at 3:05. I can not unde...” at • (1 vote) Richard a month agoPosted a month ago. Direct link to Richard's post “w is work, which is defin...” w is work, which is defined, in a physics sense, as a force acting through a distance. Such as if you push a box across a floor, you perform work on the box. Essentially you transform some of your potential energy into kinetic energy for the box. Work is a separate type of energy transfer than heat. In this example, a gas is inside a container with a moveable piston which can expand or contract. The type of work such a set up can result in is pressure-volume work, or the force which is caused by a volume change against an external pressure. If the gas expands, its volume increases, then the resulting increased pressure inside the container will cause the piston to expand and move against the external pressure outside the container. This would be negative work since the gas is doing work on something else, in essence it is losing energy to the surroundings. If the external pressure pushes back harder on the piston, it will cause the gas to contract, a decrease in volume, which is interpreted as work being done the gas, or positive work. In essence, the surroundings are giving energy to the gas. This type of pressure-volume work with a piston is the basis for an internal combustion engine in an automobile. Hope that helps. (1 vote) issahkadiallo8 2 years agoPosted 2 years ago. Direct link to issahkadiallo8's post “And I wonder How the tuto...” And I wonder How the tutor got 2044? • (0 votes) Richard 2 years agoPosted 2 years ago. Direct link to Richard's post “For your first question, ...” For your first question, that -2044 kJ value was provided in the question. Your second question is too vague to answer, you need to specify more so. (2 votes) chicken nugget 2 months agoPosted 2 months ago. Direct link to chicken nugget's post “This guy's voice is very ...” This guy's voice is very annoying and smug. Detracts from the learning. Bring Sal back • (0 votes)Want to join the conversation?
W(on gas) = -W(by gas)"
5:07 4:39 5:00 4:11 3:05
And also how to deal this with equations
Video transcript
- [Educator] Before we get into the terms, endo and exothermic, We need to look at someother thermodynamics terms that are used. For example, system. The system refers tothe part of the universe that we are studying. For our example, we're goingto consider a monatomic gas. Let's say we have some heliumparticles in a container. And the helium gas represents our system. The surroundings are everythingelse in the universe. So that would include this piston here and the cylinder in which the gas is in. And the universe consistsof both the system and the surroundings. Next, let's look at thefirst law of thermodynamics which can be summarized by writing delta E is equal to Q plus W. Delta E is the changein the internal energy of the system. And the internal energy refers to the sum of all the kinetic and potential energies of the components of the system. Since we have a monatomicgas for our system, we only have kinetic energy. So if you could imagineadding up the kinetic energy for each particle, the sum of those kineticenergies for this example would be equal to the internal energy of the system. Q refers to the heat that's transferred from the system. So it's either transferredto or from the system. And W refers to the work done on or by the system. Let's look at the sign conventions for the first law of thermodynamics. When we think about Q; when heat flows into thesystem from the surroundings, we say that Q is positive. When heat flows out of the system and into the surroundings, we say that Q is negative. For work, when work is done on the system by the surroundings, the work is positive. But if work is done by the system on the surroundings, work is negative. It's very useful to thinkabout internal energy like a bank account. So if Q is positive and work is positive, that's like money cominginto your bank account. But if Q is negative or the work is negative, that's like money leavingyour bank account. Let's look at an example ofthe first law of thermodynamics using our sample of helium that's in a cylinder with a movable piston. So let's say that 6,000 joules of energy. So 6,000 joules flows from thesurroundings into the system. And that heats up our helium particles, which then expand and push the piston up. So the piston gets pushed up, which is the system doingwork on the surroundings. And let's say that, that's 2000 joules of work done by the system on the surroundings. Using our assigned conventions, since heat flowed into thesystem from the surroundings, that's a positive value for Q. And since the work was done by the system on the surroundings, that's a negative for the work done. So we can go ahead and plug in to the first law of thermodynamics. The heat transferred ispositive 6,000 joules and the work done as negative 2000 joules, therefore delta E or the change in theinternal energy is equal to positive 4,000 joules. Thinking about internalenergy like our bank account, we've gained 4,000 joules. So that could be like gaining $4,000. Since the system has gained 4,000 joules, that must mean the surroundingshas lost 4,000 joules. But since energy is conserved, the total energy of theuniverse remains constant. Let's apply the firstlaw of thermodynamics to the combustion of propane and an open containerat constant pressure. For the combustion of propane, the reactants and products for the combustion reaction are considered to be the system, and everything else is the surroundings. So this combustion reactiongives off 2,044 kilojoules of energy. So that's the heat that'stransferred from the system to the surroundings. The system also doestwo kilojoules of work on the surroundings. So by convention, we make that negative. To find the change in theinternal energy of the system, we add Q plus W and get negative 2,046 kilojoules. Since this reaction was carried out under constant external pressure, we can write Q sub P here. So Q sub P is the heat that's transferred at constant pressure, and that is equal to thechange in the enthalpy, which is symbolized by delta H. So the change in enthalpy is the heat that's transferred at constant pressure. So the change in enthalpy for the combustion of propane is equal to negative 2,044 kilojoules. And notice how the changein enthalpy is almost the same value as the changein the internal energy. So the work done by thesystem is a very small amount in this case, and that's usually the case. And since most chemical reactions are done under constant pressure, chemists care more aboutthe change in the enthalpy than they do about the changein the internal energy. When delta H is negative, we call that an exothermic process. So the combustion of propaneis an exothermic reaction. An endothermic process iswhere heat is transferred from the surroundings to the systems. So the system has gainedheat from the surroundings. The change in enthalpy, delta H is positive foran endothermic process. An example could be melting an ice cube. If heat flows from thesystem to the surroundings, the system has releasedheat to the surroundings, and the change in enthalpy, delta H is negative. And we call this an exothermic process. And we already saw an example of that. The combustion of propaneis an exothermic reaction. So a phase change could be an endo or an exothermic process. A chemical reaction could also be an endo or an exothermic process. And even the formation of asolution could be classified as an endo or an exothermic process. Let's think about making a solution. So let's say that we have a beaker and we have it full of water, and we take a solid and we dissolve the solid in the water to form a solution. If the dissolution processis an exothermic process, that means the system releases heat to the surroundings. And since the beaker ispart of the surroundings, if we were to put our hand on the beaker and the beaker feels hot, we know that the disillusion of this particular solidis an exothermic process. If we do the same thingwith another solid; so we dissolve this solid in water to form a solution. And we put our hand on the beaker, and this time the beaker feels cold. The reason the beaker feels cold is because energy was transferredfrom their surroundings to the system. And so, the surroundings lost energy. And so, therefore wecan say the dissolution of this particular solidwas an endothermic process.
