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03:09

Aditya P.

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

05:29

Kathleen T.

(I) What is the weight of a 68-kg astronaut ($a$) on Earth, ($b$) on the Moon ($g =1.7 m/s^2$) ($c$) on Mars ($g = 3.7 \,m/s^2$) ($d$) in outer space traveling with constant velocity?

0:00

Suman Saurav T.

(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?

00:39

Averell H.

I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?

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welcome to the next section in our conversation about the kinetic theory of love thermodynamics, and we consider a constant volume process. Then, for a constant volume process. We know that work is zero because there's no Delta V and we can use N C v Delta T, where CV is our Mueller specific heat for the gas. Because we're talking about a gas process solving servi for C V. Then in this case, and substituting in what we just learned about the energy in a the thermal energy inside of gas weaken, say, three halfs and our delta T over and Delta T is going to be three halfs are so CV, then is some constant and we've seen this before, and we found that it was true for Mon atomic gas is. But if you look it back at your table in your textbook, you're going to find that for Die Atomic and Polly Atomic, CASS is it's not true. In fact, the Moller specific heat for a constant volume processes closer to these numbers, which correspond to approximately five halfs are, and three are. So what's going on here? Why is it that the diatonic and Polly atomic gasses seem tohave mawr energy on average, for their molecules than the monitor Amit gasses do. It comes down to a question of what's known as degrees of freedom. Degrees of freedom are the ways in which anatomy or molecule can move. So, for example, a single Adam gas where you have single atoms that are unbound to each other, this Adam can move in the X direction it can move in. The white can move in the Z, but if you were to rotate it in any direction, you wouldn't notice any difference, because it's going to look the same on all sides. It's approximately symmetric for any access that you would try to rotate it about. So this means that really it can Onley have energy in these three directions because spinning would be in material on the other hand, for a dye atomic gas. If you it can move in those three translational directions. But then you can also rotate it this way because that would cause a difference in how the Adam the molecule is oriented. And you can also rotate it this way because that would change how the molecules oriented, meaning it has two extra degrees of freedom notice. So we had three degrees of freedom. And here we have five degrees of freedom. And for a poly atomic gas, we can also rotate around this third axis where we couldn't with the dye atomic gas for the same reason that we couldn't rotate with the Mon atomic, that it doesn't actually change how the atoms look. But here it would, because we have this extra atom. So for a poly atomic gas, we have six degrees of freedom. So a succinct statement of the ECU a partition theme would be that the energy is in a in a gas is split evenly between each degree of freedom of the gas molecules. That is to say, each degree of freedom has, on average and energy of one half the gas, constant times the temperature of the gas. Now that means that we can then write out the total energy is three halfs and R T and three halfs capital and K B T for a mon atomic gas. And we could replace this constant three halfs with the five halfs for a dye atomic gas, or with the three for a poly Atomic gas. So this is kind of an interesting situation, and it allows us to analyze something like this where we have a box with two different gasses. They have different temperatures, and they're separated by a thin membrane that the gasses can't penetrate through. But the membrane that Kim will deflect, meaning that the molecules in gas A can pass energy to the molecules in gas be because of this over time, they will eventually come to some final temperature that is the same, so setting this up mathematically, we have initial temperatures, and then we have initial energies associated with those temperatures, and those two different gasses will have two different energies. But they should add up to some total amount of energy. Meanwhile, after some amount of time, the temperatures will come to be the same, meaning that the final temperature of gas A will be equal to the final temperature of gas be. And we still have this condition over here that the two energies must add up to be equal to the total energy. So in order for this to be true, what we're looking for is that each molecule in a in each molecule and be should have the same average energy, meaning that e a f over the total number of moles would be equal to E B F. Over the total number of moles which is equal to eat total divided by the sum of the moles in gas A and gas be weaken, right? Right. This in terms of number of atoms or molecules if you prefer notice here that what we find explicitly is that this is energy for inside each individual, atom or molecule, depending with your dealing with a monotone, Mick or die atomic gas. So, having looked at this, then we can do we can continue to look at the solution in the solution. We would find that the final energy would be equal to this ratio multiplied by the total energy. And here the final energy of B would be equal to this other ratio multiplied by a total notice. They're the same, except for what's in the numerator. Here we have the number of moles and gas A and here we have the number of moles in gas Be notice. If you add these two together, you come up explicitly with E total now moving on we confined another result we can ask ourselves. So what's the total amount of heat exchange? Here we have a heat that leaves gas a and heat that goes into gas be. And if we add those together, notice that we have e a final plus e B final, and then we have minus E a initial and minus E B initial. But this could be rewritten, rewritten as e total minus e total, giving us that Q A plus. Q B is equal to zero jewels or that Q. A is equal to negative Q B. So they're the same amount, just going in opposite directions. One is heating, the other is cooling. So we're going to take a couple of looks at examples off how this works in Applied World.

The Second Law of Thermodynamics

03:19

03:30

03:27

04:40