The volume of a gas is halved during an adiabatic compression ...

Question

The volume of a gas is halved during an adiabatic compression that increases the pressure by a factor of 2.5. By what factor does the temperature increase?

Accepted Answer

Hey, everyone in this problem, we have Alex who created a new air compressor design. OK. So we have an adiabatic compression where the gass volume decreases by one third while the pressure increased threefold. And Alex asks his coworkers, what the temperatures increase factor will be in this situation? OK. So we have four answer choices. Option A four, option B six, option C eight and option D one. So let's start by writing out the information we're given. OK? We're not given a lot of information. We're not given any raw values, but we are given some relationships between volumes and pressures. So in our initial situation and we're gonna use subscript one, OK? For this first state. So we have the volume V one which we're just gonna call V. Similarly, we have the pressure P one which we're just gonna call P and we have the temperature T one which we're just going to call T now in the second state. OK. After this compression, we have V two and we're told that the volume decreased by one third. So V two is going to be one third of V. We have P two, our pressure after this compression, we're told increased threefold. So P two is going to be equal to three P and our temperature T two. OK. Well, we're looking for the increase factor. OK. So T two, we're gonna say is equal to KT where K is the increased factor that we are interested in? All right. So is what we are looking for. Now, we're told that this is an adiabatic process. OK. So there are two kind of equations we can use in an antibiotic process. The first T one V one to the exponent gamma minus one is equal to T two V two to the exponent gamma one minus one. All right. And we're gonna call this equation one. OK. So we have this adiabatic process, we're assuming that we have this ideal gas. And so we have this equation and this equation allows us to relate these temperatures and we want those temperatures in our equation because we want to find K which is part of T two if we know that we're gonna want to come back to this equation. But first we need to find this gamma value. Now recall that gamma is our heat capacity ratio. Now, in this case, we're not told really any information about what we have here, whether it's mo atomic diatomic or anything else. So we are going to need a different equation to find gamma. OK? And the equation we're gonna use is a one relating pressure and volume. OK? Because we have information about both of those things. So similarly to equation one recall that we can write P one V one to the exponent gamma is equal to P two V two to the exponent gamma again because we have this adiabatic process. No filling in the information that we know. OK. P one is just P V one is just V. So we have PV to the exponent alpha is equal to two, which is three P multiplied by V two, which is one third B to the exponent. And sorry, that was to the exponent gamma, not to the exponent alpha. Same thing on the right hand side to the exponent gamma. All right. So on both sides, we have a P and A V to the exponent gamma term. So we can divide both sides by those things. And what we're gonna be left with is just one on the left hand side that's gonna be equal to three, multiplied by one third to the exponent gamma, dividing both sides by three, we get that one third is equal to one third to the exponent gamma. And this tells us that gamma must be equal to one. OK? If we have one third equal to one third to the exponent, something that exponent must just be one. So we have, and I've written it as a gamma, gamma is equal to one. All right. Now that we have our game of value, we can go back to equation one. OK. So let's go back to equation one and fill in what we know. Now T one is equal to T. So we have T multiplied by V one which is just V to the exponent gamma minus one. So that's gonna be one minus one. This is gonna be equal to T two, which we've said is going to be KT multiplied by V two which is one third V and this is all to the exponent gamma minus one. So one minus one. Now simplifying this expression, we get T multiplied by V to the exponent zero is equal to KT multiplied by one third of V to the exponent zero. OK. These terms to the exponent zero are just going to be one. So we're left with T is equal to KT dividing both sides by the temperature. T we get that K must be equal to one. And so the increase factor for our temperature is just going to be equal to one. OK. So that tells us that the temperature did not change during this process. Now, if we go back up to our answer choices and compare what we found, we can see that the correct answer here is going to be option D. Thanks everyone for watching. I hope this video helped see you in the next one.

The volume of a gas is halved during an adiabatic compression that increases the pressure by a factor of 2.5. By what factor does the temperature increase?

Key Concepts

Adiabatic Process

Ideal Gas Law

Temperature and Pressure Relationship in Adiabatic Processes

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