For the diver question have I missed something or shouldn't it increase to 4 atmospheres because for every 10m it increases by 1atm and the question states this is at 30m below sea level, you did it for 10m?? Thanks though your videos are really helpful
for the diver question, I did it in a different way. It says it increases 1atm every 10m but we do not know where it started so we start at p and then 30 meters below where it has increased by 3, 3P, which in the calculations cancel each other. I found out it does not matter the initial pressure but the proportionality of both. Thank you
Aqmal Fiqri I’m quite sure he made a mistake, check his reply to Matthew Clisby. But it should not be 3 atm but rather 4 atm since the pressure increases by 1atm every 10 metres, and the pressure is 1atm before it goes below water. Therefore 1 + 3 x 1 = 4atm
If take N molecules of gas, and increase the volume in which they can travel, there will be fewer molecules, (less mass) per unit volume, and mass per unit volume is density. If we add molecules in a fixed volume, pressure increases since pressure is created by the molecules hitting the container walls, but so does density since here is more mass per unit volume.
Hello sir, I'm just wondering what the difference is between (3/2)(R)(T) and (3/2)(K)(T) where R is the gas constant and K is the Boltzmann constant? when would (3/2)(R)(T) ever be applied?
+Muhammad H Basically if you are working with moles use KEave=3/2RT /Na, and if you are working with number of particles use KE=3/2kT. The internal energy of an ideal gas is the sum over all molecules of the the average KE, so U=N3/2kT or U=n 3/2RT.
In mathematics, the symbols are more precisely defined than in the sciences and engineering. It is very common to use a straight and one or two wavy lines together to mean "approximately equal to". See en.wikipedia.org/wiki/List_of_mathematical_symbols I am not saying its right to do this, but I'm not offended by it either.
For the diver question have I missed something or shouldn't it increase to 4 atmospheres because for every 10m it increases by 1atm and the question states this is at 30m below sea level, you did it for 10m?? Thanks though your videos are really helpful
Thank you Matthew, right you are. Too many 30's and 10's in the problem. I've added an annotation. It should have been 10 below the surface not 30 m.
saying thank you for the third time today lol, just finished the whole series and all my concepts are clear!
Thank you again.
Thanks for these videos they really help
bless your sweet soul
for the diver question, I did it in a different way. It says it increases 1atm every 10m but we do not know where it started so we start at p and then 30 meters below where it has increased by 3, 3P, which in the calculations cancel each other. I found out it does not matter the initial pressure but the proportionality of both. Thank you
just found out I am wrong, If I do 1atm as the initial pressure, the final should be 4atm as matthew said
Thank you for answering my questions throughout the year, I'm sure it must have gotten annoying. I’ll need your videos for ib sl year two next year 😂
Enjoyed the questions and happy you are benefitting from the channel.
at 15:00, in IB exams would I need to memorise the mass number of all the elements?! or will it be given to me? same goes to molar mass
God bless your soul
hello! I wanted to ask why is the pressure 1 atm when for the first scenario the balloon is AT the surface not below?
Air pressure is 1 atmosphere.
@@donerphysics For the second scenario, wouldn't pressure be 3 atm since it is 30m below the surface?
Aqmal Fiqri I’m quite sure he made a mistake, check his reply to Matthew Clisby. But it should not be 3 atm but rather 4 atm since the pressure increases by 1atm every 10 metres, and the pressure is 1atm before it goes below water. Therefore 1 + 3 x 1 = 4atm
Hi chris, are u sure the answer of question 2 isn't A? ratio of 1/(3/2) = 2/3 could u explain please
X has 3X the temperature but half the number of moles, so its pressure is higher by a factor of 1.5
This is so helpful! subscibed.
Thank you. Continue to spread the word.
How to find pressure if name of gas, mean square speed and density are given?
Name of gas gives grams per mole and grams per particle. Speed and mass give KE per particle which will give T. Density will give volume and moles.
Why does low density correspond to low pressure and high volume?
If take N molecules of gas, and increase the volume in which they can travel, there will be fewer molecules, (less mass) per unit volume, and mass per unit volume is density. If we add molecules in a fixed volume, pressure increases since pressure is created by the molecules hitting the container walls, but so does density since here is more mass per unit volume.
Thanks so much !!
Hello sir,
I'm just wondering what the difference is between (3/2)(R)(T) and (3/2)(K)(T) where R is the gas constant and K is the Boltzmann constant?
when would (3/2)(R)(T) ever be applied?
+Muhammad H
Basically if you are working with moles use KEave=3/2RT /Na, and if you are working with number of particles use KE=3/2kT. The internal energy of an ideal gas is the sum over all molecules of the the average KE, so U=N3/2kT or U=n 3/2RT.
+Chris Doner (C. Doner's IB Physics) Thanks so much!
That symbol is "congruent to," not "approximately equal to."
In mathematics, the symbols are more precisely defined than in the sciences and engineering. It is very common to use a straight and one or two wavy lines together to mean "approximately equal to".
See en.wikipedia.org/wiki/List_of_mathematical_symbols
I am not saying its right to do this, but I'm not offended by it either.
I was referring to the ~ + = symbol, but it makes sense now, thanks.
Is n always constant in PV=nRT?
It is really just the amount of gas. So it is constant unless there is gas going in or out of the system. This is usually the case.
@@donerphysics Thank you
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Sir, you spelt physics wrong haha
Thanks ... I edited.