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Ch 19: The First Law of Thermodynamics

Chapter 19, Problem 19

The pV-diagram in Fig. E19.13 shows a process abc involving 0.450 mol of an ideal gas.

(c) How much heat had to be added during the process to increase the internal energy of the gas by 15,000 J?

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Everyone welcome back in this problem. We have a PV diagram that shows a process followed by 0.82 moles of an ideal gas. If the internal energy rises by jewels were asked to determine the heat that must be transferred into the gas during that process. Okay, Alright. We're looking for heat. We're told some information about the internal energy. So, let's recall the relationship between the internal energy and the heat. Okay, we know that delta U. The change in internal energy is equal to Q. The heat minus the work. W hmm. Now, we know delta. You were told the change in internal energy. We're trying to find the heat Q. So, if we can find W. We'll be able to plug in those values and use this equation to find Q. All right, well, what information are we given that can help us find? W Well, we're given a PV diagram. Okay. And let's recall that work. W is equal to the area under the curve of PV diagram. So, under the PV curve. Okay, So, what we need to find is the area under this P V curve. And that area is going to be this whole area below the curve to the X axis. Okay, well, let's break this into two parts. We have this triangle here, and then we have this rectangle below. Okay. And that's gonna give us that total area under the curve. And what you'll notice is that the process from X to Y. It contributes to this total area, but the process from White Jersey does not. Okay now let's think about that. The process from white is it is a vertical line that means it's constant volume and we know in a constant volume process the work is equal to zero. Okay. And so that makes sense that that does not contribute because the work is zero. Um So that's not gonna contribute to our total work. Alright, so let's start the work w again the area under the curve which is going to be the area of the triangle, which is that blue triangle? We drew A Plus the area of the rectangle which we drew and green. Okay. Alright. Well the area of a triangle we know is 1/2. Base times height, area of a rectangle is going to be length times width. This is gonna be one half. Okay, the base of this triangle and we're going from 0.25 Up to 0.625, sorry zero point 0.875. Okay we have four notches between here which gives us the base of 0.625. I got a little bit ahead of myself there. Alright, so the base of our triangle, 0.625. The height of our triangle, 0.1. The length of our rectangle. Well it's the same as the base of the triangle. 0.625. Okay. Times the width 011. All right. So on the left hand side we are going to get 0. plus 0.625 Were a total of 0.09375. Okay. And what about units? Okay, we didn't include the units above. Let's think about units. Well we're multiplying the base or the length of the rectangle. We're multiplying a particular volume case, we're in meters cubed. And when we're doing the height or the width we're in mega pascal's. Okay, so multiply mega pascal by a meter. Cute. You get mega pascal meter. Cute. Let's go ahead and put this into pascal meters. Cute because the answer is in jewels and we know that pascal meter cube is equal to a jewel. And so we get 93, jewels. Alright, so we found the work. But let's remember what we're trying to find. Okay, we're trying to find Q. The heat, we can relate the tube by this equation we wrote above. Change in internal energy is equal to Q minus the work. W So let's move down and find that final answer. We're looking for that heat. Okay, so again, the change in internal energy is equal to the heat Q minus the work. W The internal energy, we're told rises by 2850 jewels. Okay, so 2850 jewels is equal to the heat Q minus the work 93, jewels. Moving this over to the left hand side and adding we get the heat Q. Is equal to 96,600 jewels. So that is our final answer and that is going to correspond with answer D. Thanks everyone for watching. I hope this video helped see you in the next one.
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A cylinder contains 0.0100 mol of helium at T = 27.0°C. (a) How much heat is needed to raise the temperature to 67.0°C while keeping the volume constant? Draw a pV-diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0°C to 67.0°C? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?
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Textbook Question
A cylinder contains 0.0100 mol of helium at T = 27.0°C. (a) How much heat is needed to raise the temperature to 67.0°C while keeping the volume constant? Draw a pV-diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0°C to 67.0°C? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?
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