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Ch 19: The First Law of Thermodynamics

Chapter 19, Problem 19

A monatomic ideal gas that is initially at 1.50 * 10^5 Pa and has a volume of 0.0800 m^3 is compressed adiabatically to a volume of 0.0400 m^3. (c) What is the ratio of the final temperature of the gas to its initial temperature? Is the gas heated or cooled by this compression?

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Hey everyone welcome back in this problem. We have a 0.12 m cube container filled with an ideal di atomic gas at a pressure of 1.8 times 10 to the five pascal's. The gas is compressed added automatically to a final volume of 50.6 m cubed. And were asked to determine the ratio final to the initial temperature. Okay, so T. F. Over T knot. And then we're asked does the process heat or cool the gas? Okay. Alright. So we have an an automatic process. The first thing we think of when we hear about it is that we have no heat transfer. Okay? Or Q is equal to zero. Now we're told that we have an ideal di atomic gas. We're looking for some information about temperatures. Now when we have an idiomatic process, let's recall that we can relate the temperature to the volumes which were given through the following equation. We have T. Not V. Not to the game of -1 is equal to t. Final The final to the game of -1. All right now, what about gamma? Okay. We want to find the ratio of tina too T. F. We don't know game. Let's first go ahead and isolate for the ratio we want to find and then see what quantities we need. Okay, so we want to find T. F. Over tina. So if we divide we would get TF divided by T. Naught is equal to me not to the gamma minus one, divided by the earth To the γ -1. Okay so we know them two volumes and we know gamma. Then we will be able to find this ratio TF over tina. Okay, well we know the volumes. They were given the size of the container 0.13 m container filled and we're told that it's compressed to 0.6 m scoop. So we have V not B. F. Let's find gamma. Now we are told that this is a di atomic gas. And what di atomic gas tells us is that C. V. The heat capacity, constant volume Is equal to 5/2. What? Okay. We're also told that this is an ideal gas which tells us that the heat capacity, constant pressure C. P is equal to cv plus R. So we can write this C P. A seven halves are okay. Now that we have C. V. And C P. We can find gamma which is the ratio of the two. Gamma is equal to C P over C V. Which is equal to 7/2 are Divided by 5/2 are Okay, which is equal to 7/5. We're putting this into a decimal 1.4. Okay so we have the two volumes. Now we found gamma. We can go ahead and find the ratio TF over tea. Not that we're looking for. Okay, so this is going to be equal to the not 0.1, two m cute to the exponents 1.4 - Divided by the final 0. meters cute Again. To the exponent gambling minus 11.4 minus one. And when we work this out, we're going to get our ratio T. F. Over tea. Not the units will cancel will be less. Left dimension less number of 1.32. Okay. And so the ratio is 1.32. Now we need to answer the question. Does this heat or cool the gas? Okay. We're looking at the ratio TF over T knot. Okay, T. F over T knot is bigger than one. Which means that T. F. Is bigger than tina. Okay, so the final temperature is bigger than the initial temperature. That means that the final temperature is hotter. And so the gas was heated. Okay. And so the answer here is going to be e. The ratio of T. Final two T initial is 1.32. And because T final is bigger, the gas is heated. That's it for this one. Thanks everyone for watching. See you in the next video.
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Textbook Question
A monatomic ideal gas that is initially at 1.50 * 10^5 Pa and has a volume of 0.0800 m^3 is compressed adiabatically to a volume of 0.0400 m^3. (a) What is the final pressure?
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