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Boyle'8 Law posits thaf when the temperature ofa confined gas i8 congtant, Kaen tbe pre!g ute (in pound: per 8q uare inch varie8 inve mely a6 the volume V (in ...

Question

Boyle'8 Law posits thaf when the temperature ofa confined gas i8 congtant, Kaen tbe pre!g ute (in pound: per 8q uare inch varie8 inve mely a6 the volume V (in cubic inches For a certain gas s"ppose800P(v)Find {le instanfaneous rte Ol change Iunction lor P(V ):

Boyle'8 Law posits thaf when the temperature ofa confined gas i8 congtant, Kaen tbe pre!g ute (in pound: per 8q uare inch varie8 inve mely a6 the volume V (in cubic inches For a certain gas s"ppose 800 P(v) Find {le instanfaneous rte Ol change Iunction lor P(V ):



Answers

One cubic foot of gas under a pressure of 80 pounds per square inch expands adiabatically to 4 cubic feet according to the law $p v^{1.4}=c .$ Find the work done by the gas.

Yes. Law here. I've got a constant chemicals 10 of these other conditions that are right here. So basically we're trying to find the partial of T with respect to time and that's equal to perch. Enroll partial T perspective. V times T V t T plus partial of tea and respect to the pressure times dP DT. Okay. So before I even proceed with this, we want to make sure that we got everything in terms of T. Multiplying by V dividing by case. So I get T is equal to PV over. Okay. All right. This little little area. All right. So then taking that into consideration here. Now we can go ahead and take our partials. So the first part would be P over K times four plus fee over K times negative one. So, since we have PS five K's 10, It's going to give us 5/10 times four Plus 200 over 10 times negative one. That's 2 -20. So, your answer is negative 18 kelvin per second. Okay.

Hi, this is Human Fatima. Next question question number 21 is one more of an ideal gas at 300 Kelvin is expanded as a family from an initial volume of one leader to 10 liter. The change in internal energy Delta you for the gas in this process is option E 1 63.7 Gallery B zero C1381 0.1 Gallery D nine. Uh and uh and it must freak pressure. So the answer is answer is option B. That is zero. What can be the day 0? Yeah. Mhm. For I'd guess for ideal guess delta U. That is changing Internal energy is equal to and see we D. D. For eyes. Atomic presses for isO terminal process D. D. Because it will do zero. The physical 20. Hence changing internal energy is zero during ice. Ottoman expansion of a guest. So option B is the correct answer. That is zero. Thank you.

Here in this problem given that we physical two k so physical too. Okay, bye. We solving it for the very differentiated with respect to time so deeply by DT is equal to mine escape by these but d v by DT going forward here, give me by DT is given as minus five. We is given as 1000 p is given as 40. So I can write. K is equal to b p is called to 1000 multiplication 40 is equal to 40,000. So going back to the question we have dp by DT is equal to two multiplication three so it can be further return it dp by DT is equal to minus 40,000. Bye 1000 is choir multiplication minus five, which is equal to 0.2. So pressure change at £0.2 was squired inch for Howard

Hey, guys. So in this question were given three scenarios with different volumes and temperatures for gases and were asked to use. Charles is Lalit itself for the unknown parts of each question. So starting with part A were given starting temperature and starting volume and a final temperature and final volume and need to find a final ball, You excuse re. So the first thing I'm gonna do is I'm gonna convert our temperatures to Kelvin since this Gasol equation, and we need our temperatures of the gases to beating Kelvin. So at 273 to each of these values and for the 1st 1 I get to 97 Kelvin. And for the 2nd 1 I get 3 21 Kelvin. Now, with these values, I can go ahead and use Charles his law and self review to. So when I do that I get V two is equal to be one times t two over t one plugging in our values that were given. We have 9.14 leaders want to play that by 3 21 Calvin. And that's all divided by 297 Kelvin plugging all of this in tor calculator. We find that V two is equal to 9.89 Leaders master answer for part A moving on to part B. We have volume and starting temperature as well as a final volume. You need to find the final temperature in degrees Celsius. So the first thing I'm gonna do against convert this to Kelvin by adding to 73 in this yields a temperature of 2 61 Kelvin for t one. Now that we have that, I can go ahead and plug in our values into our Charles equation. So that is t two is equal to be to times t one over you one plaguing our values, we have B two which is 49.9 mils multiplied by 2 61 Calvin all divided by 24.9. When we plug this into our calculator, we find that our temperature is to this 523 Calvin to convert this back to Celsius. I just subtract 2 73 and our Celsius temperature is 250 degrees Celsius. This is our answer for part B. And lastly we could move on to part C where were given the same values or similar values as we had in part A. And we're already given temperatures and Kelvin so we can go ahead and just use our equation from last time. So V two is equal to be one times t two over t one, and that is equal to 9 25 mills multiplied by 2 73 Calvin all over 25 Calvin plugging all this into our calculator, we find that our final volume for this was 1.1 times 10 to the fourth mills. And this answer can, of course, be converted into a more suitable unit that it's not middle leaders.


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