First Law of Thermodynamics

Concept: Changes in Internal Energy

9m
Video Transcript

Hey guys, in this video we want to talk about changes in internal energy and what type of energy exchange will produce changes in internal energy let's get to it. Remember guys the internal energy is the energy of the particles within a system. This includes individual particle kinetic energy and individual particle interaction energy right the potential energy due to interactions between particles within a system, remember guys for an ideal gas specifically there is no interaction energy there's only kinetic energy so the temperature sorry the internal energy depends entirely on that kinetic energy which depends only on the temperature of the gas so the internal energy depends only on the temperature of the ideal gas and this is nothing new we've seen this a bunch of times but it's important to always have this on the forefront of our minds in order for there to be a change in internal energy in order for there to be a delta U there must be a change in temperature or some delta T. The question we want to answer is what energy transfers what type of energy transfers I mean by that will lead to a change in the temperature because if there's a change in temperature there will be a change in internal energy we know from calorimetry which we did a while ago that heat will change the temperature right we talked before that heat added at a constant volume will change the temperature of a system. We also saw in discussing the concept of mechanical equivalence that work too can change the temperature of a substance alright what this means is the internal energy of a substance of any system can be changed by two energy transfers by either heat or work. So far we've only specifically talked about how heat causes a change in temperature we haven't discussed how work causes a change in temperature right heat leading into a change in temperature is calorimetry and that we spent a while discussing. Consider the scenario in the figure above me at the top if we have an ideal gas contained in a thermally insulated piston that means heat can't go in, heat can't go out there's no heat transfer with the environment, if the piston moves down that means that you're decreasing the volume this was some initial volume this is some final volume and clearly that volume is less if you decrease the volume the temperature is going to rise on the gas and so the internal energy is also going to rise. This gas we know is putting a pressure on the piston head, it's putting pressure on all surfaces of the cylinder because gas molecules are colliding with the piston head we've talked before in kinetic theory of gases how to describe the pressure a gas puts on a surface due to these collisions but either way there is a pressure on that piston head so that means that there is some force F on that Piston head. Now when that piston moves right that piston was originally here and it moved some delta X against that force there had to be work done work is being done. You talk about we talk about a force against a displacement this force this displacement that means that there has to be work done and the work happens to increase the internal energy of the gas because we know that delta U increases so by pushing on the piston head you're pushing against the pressure of the gas so you're doing some work because you're pushing against the force that the gas puts on the piston head you're doing work on the gas or increasing that gases temperature you are adding energy into that system and that causes the internal energy to rise due soley to work being done. Now if we consider a second scenario sort of a counter example. If we consider the piston head being free to move in this figure right here to the right. Now we're not controlling the piston head the piston head is moving on its own and the walls are not insulated so heat can go in heat can go out if we add heat the heat goes in then we know that we have an increased kinetic energy each particle is going to absorb that heat and increase its individual kinetic energy so there's an increased average kinetic energy we know that leads to more collisions per second so there's an increased pressure of the gas. So what that means is this force that was originally on the piston head gets larger because the gas is now putting a larger pressure there's an increased force on the piston head so that force gets larger when that force gets larger it puts a net force on the piston head and moves it up originally before this heat was added in the piston was an equilibrium the pressure of the gas molecules inside equaled the pressure of the air but now these gas molecules are moving faster they're colliding more they increase the pressure so they increase the force now this force is larger than the force that the atmospheric air puts on the piston and it moves up so there is some force the piston was originally here and it moved some delta X so clearly some force some displacement there was work being done. But the temperature didn't have to rise there didn't have to be of an increase in temperature here nothing was said ensures the temperature is going to go up but we know that some of this heat had to be converted into work not all of that heat could be converted into change in internal energy which would lead to a change in temperature that only occurs when the volume is constant. So Q going to internal energy that's what leads to a change in temperature but now Q also produces work and that leads to a change in volume so both of these things are competing with one another alright and it's important to realize that so we know that work can change the internal energy of a system it doesn't have to but it can change also a change in internal energy can also lead to work being done exactly like I said here when heat is added the internal energy does change but not as much as it normally would because some work is being done right this goes both ways work can produce a change in internal energy a change internal energy can produce a work either way work clearly influences the internal energy of a system so the internal energy is not only inextricably linked to heat which we've seen repeatedly before but it's also linked to work and that's the important thing to take away from this video. Alright guys that wraps up this discussion into changes in internal energy and what types of energy exchanges can lead to a change. Thanks for watching us.

Concept: The First Law of Thermodynamics

7m
Video Transcript

Hey guys, in this video we're going to talk about the first law of thermodynamics. We're not really going to talk about applications here so much as just the first law itself but from now on until we discuss the second law of thermodynamics we're mainly going to be focusing on applications of the first law. Alright let's get to it. Remember both heat and work can act to change the internal energy we focus for a long time on how heat can change the temperature we discussed calorimetry extensively we saw recently that work can also change the temperature so work can also change the internal energy both heat and work can change the internal energy now the first law of thermodynamics is a mathematical law and it states that the change in internal energy is equal to the heat transfer plus the amount of work done alright now the sign for each of these is very very important we've seen before the sign of the heat positive Q is heat entering negative Q is heat leaving this is nothing new we saw in calorimetry if you calculate a positive Q heat was entering the substance if yu calculated a negative Q heat was leaving the substance but work is something that we need to establish a sign convention for now if we use a positive sign in between the two and some of your textbooks might not use a positive sign they might use a negative sign instead but I'm going to use a positive sign because conceptually that makes the most sense if we use a positive sign between Q and W in the equation what we're saying is when work is positive work is done on the gas work that increases the gas's energy. If we're saying that W is negative that work is negative this is work done by the gas this is work that decreases. This should say increases not increasing, this is work that decreases the gas's energy and this is very important we use the exact same sign convention when we discussed mechanical work and this is why I like to use a positive sign here between Q and W because it allows me to use these sign conventions that make the most sense conceptually in mechanical work when work was positive the kinetic energy increased the energy of the substance increased work was done on the substance, substance gained energy when negative work decreased the kinetic energy of the substance so the substance lost energy right if you're going over a surface with a lot of friction when you cross that surface on delta X your work is negative always because you have to expand work to move across a surface with friction on it alright so this sign convention allows us to keep the same conceptual idea that we had in mechanical work with what a positive work means and what a negative word means. 3 moles of an ideal monoatomic gas is stored in a cylinder right with a movable piston simultaneously the piston moves as 300 joules of heat is added to the gas. Raising the temperature from 300 Kelvin to 320 Kelvin how much work was done is this work done by the gas or work done on the gas so what we're told is heat added was 300 joules this is positive because this is heat going into our system now normally we would just say that this heat goes into our system to change the internal energy of our system this is what we used to say but now we can't say that because some of the heat could produce a change in internal energy but some of the heat could produce work as well. We need to use the first law of thermodynamics which says that the change in internal energy is the heat plus the work done and so we can say that the work done is the change internal energy minus the heat so this will tell us how much work is done we know the heat we can calculate the change in internal energy for an ideal gas. Now remember the internal energy for an ideal gas is F over 2 N, R, T since this is a monoatomic gas there are 3 degrees of freedom the gas only has translational motion so this is three halves N, R sorry T. Now if we want to talk about a change in internal energy. The number of moles doesn't change because this gas is stored in the cylinder whose volume can change but the cylinder is closed there's nothing that implies in this problem that gas can leave or enter the cylinder so if we talk about a change in internal energy the only thing that can change is the temperature so a change in internal energy produces a change in temperature. Now what is that change in internal energy well we're talking about 3 moles of ideal gas the ideal gas constant is 8.314 what's the change in temperature we're going from 300 Kelvin to 320 Kelvin so clearly the change is an increase by 20 Kelvin and this is going to be 748 joules that's the amount of energy that the internal energy changes by. So given our first law the work is going to be delta U minus Q which is going to be 748 joules minus the 300 joules that we added. If this problem had said 300 joules was removed that 300 would be negative 300 and this negative sign will become a positive sign. So we'd actually get a number larger than 748 but since heat was added that heat has to stay positive so this becomes 448 joules that's the amount of work done and because this work was positive this is work done on the gas that's what a positive work means a positive work always means work is being done on the gas. Alright guys this wraps up our introduction into the first law of thermodynamics. Thanks for watching.

Problem: 2.5 moles of an atomic hydrogen gas are stored in a cylinder with a moveable piston, sitting in a bucket of ice. The piston is moved SLOWLY, decreasing the volume from 0.03 m3 to 0.01 m3 . How much ice melts during this movement? Work done at constant pressure is given by 𝑊 = −𝑃Δ𝑉, and work done at constant temperature by 𝑊 = −𝑛𝑅𝑇𝑙𝑛 ( 𝑉𝑓/𝑉𝑖 ). Note that the latent heat of fusion for water is 334 kJ/kg.

8m

First Law of Thermodynamics Additional Practice Problems

An athlete doing push-ups performs 650 kJ of work and loses 425 kJ of heat. What is the change in the internal energy of the athlete?

a) -225 KJ

b) -1075 KJ

c) 1075 KJ

d) 225 KJ

 

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Consider the head-on collision of two masses, m1 = 3 kg moving to the right with speed 6 m/s and m2 = 6 kg moving to the left with speed 3 m/s. Given that they stick together after the collision, what is the increase in internal energy of the system during the collision? Assume no energy is lost to the surroundings and neglect any external forces acting on the two masses.

1. 81.0 J

2. 56.25 J

3. 40.5 J

4. 216.0 J

5. 303.75 J

6. 112.5 J

7. 144.0 J

8. 375.0 J

9. 525.0 J

10. 72.0 J

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A skater pushes straight away from a wall. She pushes on the wall with a force whose magnitude is F, so the wall pushes on her with a force F (in the direction of her motion). As she moves away from the wall, her center of mass moves a distance d. Consider the following statements regarding energy.

I ΔKtrans + ΔEinternal = Fd
II. ΔKtrans + ΔEinternal = −Fd
III. ΔKtrans + ΔEinternal = 0
IV. ΔKtrans = Fd
V. ΔKtrans = −Fd

What is the correct form of the energy principle for the skater as a real system and as a point particle (PP) system?

1. Real: II, PP: V
2. Real: I, PP: V
3. Real: II, PP: IV
4. Real: V, PP: I
5. Real: IV, PP: III
6. Real: III, PP: IV
7. Real: V, PP: IV
8. Real: I, PP: IV
9. Real: III, PP: V
10. Real: IV, PP: IV

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The temperature of 5.0 moles of an ideal gas is increased from 100°C to 300°C. If this is done at constant volume, 3.0 x 104 J of heat energy flows into the gas. If the same temperature change is carried out at constant pressure, the heat flow into the gas is

(a) 3.0 x 104 J

(b) less than 3.0 x 104 J

(c) greater than 3.0 x 104 J

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Six moles of a monatomic ideal gas undergo the process shown in the figure. State 1 has pressure p= 4.00 x 105 Pa and volume V1 = 2.00 x 10-3 m3. State 2 has pressure p= 3.00 x 105 Pa and volume V2 = 6.00 x 10-3 m3

a) In this process, what is ΔU, the change in the internal energy of the gas?

b) What is the heat flow Q for this process? 

c) Does heat flow into the gas or out of the gas?

 

 

 

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