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If increase in temperature and volume of an ideal gas is two times, then the initial pressure $mathrm{P}$ changes to(a) $4 mathrm{P}$(b) $2 mathrm{P}$(c) $mathrm{P}...

Question

If increase in temperature and volume of an ideal gas is two times, then the initial pressure $mathrm{P}$ changes to(a) $4 mathrm{P}$(b) $2 mathrm{P}$(c) $mathrm{P}$(d) $3 mathrm{P}$

If increase in temperature and volume of an ideal gas is two times, then the initial pressure $mathrm{P}$ changes to (a) $4 mathrm{P}$ (b) $2 mathrm{P}$ (c) $mathrm{P}$ (d) $3 mathrm{P}$



Answers

$\bullet$$\cdot$ An ideal gas at 4.00 atm and 350 $\mathrm{K}$ is permitted to expand adiabatically to 1.50 times its initial volume. Find the final pressure and temperature if the gas is (a) monatomic with $C_{p} / C_{V}=\frac{5}{3},$ (b) diatomic with $C_{p} / C_{V}=\frac{7}{5} .$

This is Jumper 15. Problem number 60 were given an ideal guess. Up initial pressure off four atmospheres and the initial temperature is 350. Calvin, um, this guess is permitted to expand of you pathetically to its one point type five times of its initial value. So if you call the initial volume beasts of I, then fine was gonna be 1.5 times this initial volume were asked to find the pea final final pressure. Uh, and final temperature of the guests. So grass playing fresher and final temperature of the gas if the, um ratio C p over CV is five over three. If this gas is another attempt, guests. So there couple equations that we need to remember for an independent process. So the initial pressure times the initial volume times gamma being, um, this ratio right equals final pressure times The final volume to power. Gamma. Since we're after final pressure, I divide both of these equations by the final call. You gamma power of it. Right? So these eggs cancel out. So the final pressure's gonna be written. And is the initial believe the final time scam a sedan? We have our equation complained the numbers for him from spheres initial divided by one and half times even shoulders they're gonna cancel out. And his gamble. We have five over three. From here we find out the final pressure to be 2.4 atmospheres. Now, when it comes to the final temperature, we have to remember our outer our our other equation. That goes as, um, initial temperature turns the initial volume. Camel minus one equals final temperature time supplying a ball game a nice one. Now, since we're after final temperature, if you divide both sides by this turn a room and sweet Lizzie isolating the final temperature. Right. Initial times being initial over the final power gamma lines. One initial temperature has given us a scream from over 50 Calvin's and initial volume divided by 15 times the initial volume to the power off five over three minus one. So official volumes they're gonna cancel out we're gonna find out the temperature to the 267 Calvin now in part, a thistles party party we're gonna do, huh? A similar process on Lee. For now, the ratio is gonna be different. That is given to us a spiral over three. This time we're dealing in the bad, Tommy. Guess so. It's gonna be seven over five. But apart from that, we will be using the same explosions. You know, in order to get to our answers. The being these two equations Self party given no CP over. See me to be seven over five. Then the final pressure beaming stole, right. I'm just copying the vision that we had initial pressure stiffer after spears and initial volume. One thing back times initial warrior, am I This time is seven over five. So we find a final pressure to be 2.2. Sad, fabulous fear. Now, the final temperature is the initial temperature Times initial volume finals volume getting minus one initial temperatures. 350 Calvin's, um, tradition volume over looking and for the final on 1.5 times the initial values. So I have been over five like this one, the initials they're gonna cancel of it. So the final temperature is gonna be 298

Hi, guys. I'm doing problem 13 in chapter 18. It says that the volume of Amon Atomic ideal gas triples and I so thermal expansion. By what factor does this pressure change? So whenever we're dealing with ideal gases, we can use the ideal gas law PV equals N R t. Or, if you prefer P vehicles and Katie doesn't matter, Um, course, in the second case, you have big end. So you're talking about number of particles of stuff moles, But here we'll use P vehicle on Artie and we're told that an ice with thermal expansion is going on, okay, and then the volume is tripling. Okay, So what we noticed about this equation is that if we're talking about one gas and doing an isotope thermal expansion, the entire right side is going to remain the same. And so this this equation is always true for an ideal gas. So if it's true before, has to be true. After and so, if you're going to change one of these variables by a factor, some other variables must change by a factor in order to keep this equation true. Okay, so if we undergo a nicer thermal expansion the entire right side remains the same. Okay, so this really means that therefore, that p p one V one equals p to V two. Okay. And so this is a separate equation that sometimes people memorize. But you don't really need to memorize it. Because if you just have the the ideal gas law memorized already, then it's very simple to derive this just by noticing in this first equation that in nicely thermal expansion, the entire right side is the same. So then the stuff that's changing must be on the left side. Okay, so then it says the volume is tripling, so V is giving up by a factor of three. So what must happen to pee in order to make this the same? Well, P as you should immediately notice P is going to go down by a factor of three or P p going 2 to 1/3. But it wasn't originally eso another way to see this. If if you didn't immediately see it from above, you can say V two equals three V one. Okay, this is what it means when the volumes tripling right? This just means the volumes tripling. And then put that into the second equation, right? And so you'll get something like, um, so you you'll get something like P one V one equals three p to V one V of fee one. Right. And then, since you have the one on each side, you can cancel out the V ones dividing through and you'll see that P one equals three p two or pee, too. P two equals 1/3 p one, which is the same thing as we said before that the pressure is changing is going down by a factor of three or being scaled by a factor of 1/3.

Continuing on with our work related to gas. So what we're taking a look at is Anton's law that the pressure of a given amount of gas is directly proportional to its temperature. So P directly proportional to temperature. So the equation we need today is P two over P one is equal to t two over t one we're solving for peak too. So we have he won t two over tp one. So if the temperature is doubled, that is t two is equal to two t than the equation becomes Pete two is equal to P one two t one over Tijuana. So therefore we have p two is equal to two p one. So therefore I see temperatures double the pressure will also double.

In this example, we're going to determine what the ratio of the final pressure to the initial pressure is when we know what the initial and final temperatures of an ideal gas are. And also we're assuming that the volume of this ideal gases constant during this whole change, which is very important. So we're going to start that. Okay, so if we hear ideal gas, we should think of the ideal gas law, which is PV equals and times are times T. All right. So if the volume of our gas is constant, notice if we solved for volume here by dividing both sides by P, if our volume is constant, that means that this whole right side of this equation is also constant. Right? So what that means is that at any one point in time, we can equate this value to any other Gwen time for different temperatures and pressures. So that means that our times a t at one moment I was a P at that same moment is going to be equal. Teoh and our times t it's another moment over P. It's a mother moment or this leaves us with tea. One over p one is equal to t to overpay to or since we're looking for the ratio of P final two p initial or P two p one we can solve for that. So if we cross multiply that equation, we'll have p to overpay one equals t two over t one. So for this set up, all we need is the initial and final temperatures to solve for this ratio here. So for this 1st 1 we have 70 for the final temperature and 35 for the initial temperature. So if we plug those in, we'll have 70/35 Calvin, which is just too. So the ratio of the final pressure to the initial pressure for this 1st 1 is just too. Now, let's look at a slightly difference in Are you? Where are initial temperature is 35 degrees Celsius in our final temperature is 70 degrees Celsius. So first we're gonna want to convert these two. Kelvin, So remember to get killed. Then we're gonna add 273.15 to our Celsius number here. So these and Kelvin would be 308 0.15 Kelvin. And over here we have 343.15 Children. And what we can see is that this change here is actually a much smaller change in till then, right? We only went up about 40 degrees Kelvin, whereas in the previous example, we're sorry we went up about 40 degrees in both, but we're going to see that overall, this ratio is gonna be much smaller. So if we do, the exact same thing is before say that the ratio of P two to P one is the same as teacher T one. Just the same as 343.15 killed it over 108.15 killed them. We're going to get a value of one point 11 All right, so what do we just show? We showed that if we know we're dealing with a constant volume process for an ideal gas that the initial and final pressures are related to the initial and final final temperatures. Just by this equation here, p two over p one is equal to teach you oversee one. So if we have t two and t one, we could solve for the ratio of the pressures


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