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Ch 19: Work, Heat, and the First Law of Thermodynamics

Chapter 19, Problem 19

The volume of a gas is halved during an adiabatic compression that increases the pressure by a factor of 2.5. a. What is the specific heat ratio γ?

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Hey, everyone in this problem, we have a high altitude weather balloon, the strong insulated container that holds a sample of gas as the balloon descends to a lower altitude, the external pressure changes causing the volume of the gas to decrease to one third of its initial value and the pressure to increase by a factor of three, we're asked to determine the specific heat ratio of the gas in the container. Hey, we're giving this note that the process is a about it. We're given four answer choices here. Option A gamma is equal to 1.4. Option B, it's equal to one. Option C is equal to 1.2 and option D it is equal to 1.6. Hm. Ok. So let's just start by writing out what we know. So what we know is that we have some initial volume, we're just gonna call that the and we have some initial pressure which we're gonna call P hm. Now, after the balloon descends, the pressure changes the volume changes. Ok. We're told that the volume which is V two decreases to one third of its initial value. So V two is just going to be one third of that initial volume. V hm. Similarly, the pressure P two increases by a factor of three. So P two is going to be equal to three P. OK. So we don't have these exact values, but we have the relationship between the two. And what we are looking for is that specific heat ratio gamma. Now, we're told that this is an adiabatic process. OK. So recall that we have an equation for an adiabatic process that relates all of these variables that we have information. And that's gonna be P one multiplied by V one to the exponent gamma is equal to P two multiplied by V two to the experiment gamma. OK. Now, don't get overwhelmed by not having numerical values. We're gonna substitute in what we know about the relationship between P one V one and P two V two and see where that gets us. OK? All right. So let's substitute in what we know. So P one, we're just calling P V one, we're just calling V. So on the left hand side, we have P multiplied by V to the exponent G. On the right hand side, we have P two, which is gonna be three, P multiplied by V two to the exponent gamma. So we get one third V to the exponent. Yeah. All right. Now you can see that we just have this MP multiplied on both sides, we can divide both sides by that pressure. P. And what we're left with is that V to the exponent gamma is equal to three. Mhm We're gonna break this exponent up. We have one third of the all to the exponent gamma. OK. That exponent can be applied to each term individually because they're multiplied together. So we have one third to the exponent gamma multiplied by V to the exponent gamma. OK. Now we have V to the exponent gamma on both sides. Again, let's go ahead and divide by that V to the exponent gamma. So we get V to the exponent gamma divided by V to the exponent gamma is equal to three multiplied by one third to the exponent. Yeah. Well, V to the exponent gamma divided by V to the exponent gamma is just one. OK? So we have one is equal to three multiplied by one third to the exponent gamma. And you'll notice something really neat. Now, the only unknown we have is that value of gamma that we're looking for. And we've eliminated the pressure, we've eliminated the volume. So we didn't need numerical values. We just needed the relationship between the initial and final volumes and pressures. That's really neat. It doesn't matter what the actual value is just what the relationship is. We're getting close, we have one third to some exponent on the right hand side. Let's go ahead and divide both sides by three. And when we do that, we get one third on the left hand side and one third to the exponent gamma on the right hand side. Now, if we have fraction raised to an exponent equaling that same fraction race to some other exponent, the exponents must be equal and which tells us that gamma must be equal to what? And that's exactly what we wanted. We wanted to find that specific heat ratio gamma and now we've got it, let's compare this to our answer choices. We can see that the correct answer is going to be option B, the specific heat ratio gamma is equal to one in this case. Thanks everyone for watching. I hope this video helped see you in the next one.