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$\bullet$$\bullet($ a) One-third of a mole of He gas is taken along the path $a b c$ shown as the solid line in Figure $15.42 .$ Assume that the gas may be treated as ideal. How much heat is transferred into or out of the gas? (b) If the gas instead went from state $a$ to state $c$ along the horizontal dashed line in Fig. $15.42,$ how much heat would be transferred into or out of the gas? (c) How does $Q$ in part (b) compare against $Q$ in part (a)? Explain.

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a) $3000 . J$b) $2000 . J$

Physics 101 Mechanics

Chapter 15

Thermal Properties of Matter

Temperature and Heat

The First Law of Thermodynamics

The Second Law of Thermodynamics

Cornell University

University of Michigan - Ann Arbor

University of Winnipeg

McMaster University

Lectures

03:15

In physics, the second law of thermodynamics states that the total entropy of an isolated system can only increase over time. The total entropy of a system can never decrease, and the entropy of a system approaches a constant value as the temperature approaches zero.

03:25

The First Law of Thermodynamics is an expression of the principle of conservation of energy. The law states that the change in the internal energy of a closed system is equal to the amount of heat energy added to the system, minus the work done by the system on its surroundings. The total energy of a system can be subdivided and classified in various ways.

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(a) One-third of a mole of…

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One mole of an ideal diat…

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One-half mole of an ideal …

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One mole of an ideal diato…

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One mole of an ideal monoa…

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One mole of an ideal gas i…

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In Fig. 12.16, the amount …

This is Chapter 15. Problem number 76 are given to peopie diagram off a guess of, um, number of moments of one over three. And then, in fact, they were asked how much he was transferred into or out of this guest. So we're looking for total Cure. Um, let's remember, between human Terminal ever Joe, the system equals to the heat given to the system, minus the work done by the system. Um, so from here, Q then is gonna be cool to Delta. You plus the work done. Right. So we're gonna have to determine what Bell to you is to change internal energy. And then we're gonna determine well what the work done by the system is to the work. That portion is actually pretty straight forward is you know, the area under the PV and curve gives us the work done. So it's probably smart to start back with us. They were talking about this entire area, right? It isn't supposed Bye. Um, the curve that has given to us. So if you pay attention to consist of a rectangle like this and then, uh, too right triangles that are actually basically identical to each other. So we If we add the errors of these three regions, then we're gonna get to the total work done in order to do Let the work done that is gonna be Let's start with the triangle. This hype is one thing. Santa Barbara five times 0.1 minus 0.2 Right. This is a ridiculous this trash portion. Plus, Now let's try to determine what this length is. So assuming that this is 3.5 times 0.2 3.5 10 to 4 or five, right, so 3.5 times five minus one times 5 1 10 to 5. This length times this length 0.6 minus 0.2 over to write divided by house. Whatever. We have two identical. This area is also identical to this area, so we might as well move plight by two. Sit him from here. We're gonna figure out that the work down his 1800 jewels now as far as did out of you, is concerned for an ideal gases. You know, we're looking at the point A and point C, which is the initial and final off points for this for the process is so they'll to use the vehicle 10 cv times Delta T Delta t being here in Stevie. I doubt the TV and the temperature had seen minus final temperature minus Dangel temperature. We need to compute that now. How are we going to compute that? What's going on in page? We can see clearly the pressure at point A in pressure point see their equal to each other, right? A big fools PC. So then if we write the ideal gas law frees his P B and r A T. P s constant B is changing. So can rewrite the Delta V and R Down city. Since, uh, nothing else was changing about except for the temperature. Right. So then you're free. Oh, right. While if you did by both sides by end times are sometimes are deserving can slots adult. The tea that we're looking for is actually pressure times. Delta bi go from hate to see delighted by end times are right. So here, um P his constant greasy lens We a then and are right So we're giving all this information. P is warm town sent to gratify the pressure at point a right passports and B C is also given to us this 0.1 meter. Cute B A is going zeros, your meter tube divided by how many moles? 1/3. And then we had a 0.315 joules per more cover. So Delta T from here is gonna found out to be 289 Calvin. So now, since we know the delta, um t going back to the internal energy change equation that we've previously previously introduced here and CV and see you delta t we can get works with is from the table on her textbook. It's 12.47 jewels. Mole Calvin Fright. We're going to use this information here. So one over three moles times 12.47 jewels for more cal. Been times 289 Calvin men. So the internal change of internal energy use we're gonna find a TV 1200 jewels. All right, Now, all we need to do is to go back and calculate to heat them right down to lost work. From here, Huey cools down through. You must work. Datta, you is well 100 jewels. Plus the work. Done. Um, we calculated to be 1800 jewels here 1800. Jule search gives us 3000. Jules, this is positive, right? This is positive. It means, um the he is transferred into us. Transferred into. Okay, let me put here. This means cute is transferred even two. The gas, Right? Because it's positive. All right, Now, let's warm with part B in part B, we are asked to assume if the process was directly going from eight to see skip in, um, be here. Then what will be the queue? So if correctly we're going from here to see, then what is cute, right? That's that's what we're asked. So you'd be drifting out from a to see the whole process would be an isil Barrett process, because the pressure is not changing. That the pressure point a equal suit pressure points in any anywhere along the road. Then you'd simply I still battery process. If it's a massive Eric process than the heat is gonna be calculated. His n c p times Delta t again see, be something we can look up on its 20.7 here. Jules. Thermal. Calvin, You know, we have calculated what Belle Titi is already and it's one over three morals or gas. 27 e Jews per mole cabin to 80,000,000. Calvin. Um so from here, we find it to be 2000 Jules again. It's positively trees. Heat is transfer into Yes, right now in proxy were asked to compare the heat of that that is transferred into guests. So if you look at the amount in Party of the Troll, this cube be okay is 2000 jewels and of you a thank you. The heat that we calculated in parties 3006 to 8 is better than to be obviously. Because so when we get the the inter changing internal energy for both cases, we're going to see that it is the same because we used the same initial and final points when we're calculating Delta, you right, save because initial and final states are the same. So the difference comes from the work. Actually, that's why we have a reader of heat transferred into dancing party. Then we have a party because work done well is greater in a and that's why we have the heat transferred into guests greater in part

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