Gas density and PV=nRT, the ideal gas law

[@aarishhanif1837]

Finally, someone talks about the relationships of density. Thanks!

[@adnonalumbert9008]

agreed

[@adnonalumbert9008]

LUV this vid, and seems to be filling a gap on youtube. Perfectly done, perfectly explained. THANKS!

[@marigeo24]

@3:14 how the value for R differed based on wether the pressure was measured in atm vs. kPa, how can we convert between units of measurement? In other words, how did you know that “under these conditions, the pressure would be 101.3 kPa”? Or better, what can I search on KZbin to find videos explaining this? To add on to this: If you’re given the values for R, V, T, and n, which give you a value of 101.3 for P, but you don’t know the unit of measurement, how do you know it’s kPa rather than atm?

[@CrashChemistryAcademy]

I think this video should give you your answers kzbin.info/www/bejne/f5Tbp2uolql6eLM

[@marigeo24]

@@CrashChemistryAcademy So from what I understand, no matter how high or low the atm/kPa for a given ideal gas, the rest of the variables including R will be proportional to it (the given pressure), in the same ratios. In other words, you either have an R of 0.082 when using atm, or 8.314 when using kPa, no matter what the exact values for P, V, N, and T happen to be for that problem. So we can just plug in either of the two Rs depending on which measurement of pressure is being used. Did I understand correctly?

[@CrashChemistryAcademy]

Yes that is correct!

[@smartscience5305]

Hola!!! I am twelve years and i am admired by your lesson toooooo much but which class we'll gonna take the lesson☺

[@CrashChemistryAcademy]

This is for a chemistry class. It could easily be given in a physics class as well. It depends on your school/teacher, but usually it is chemistry.

[@smartscience5305]

@@CrashChemistryAcademy thank you, great vedios!!!😄😄

[@bubbahotep6316]

Another great video. You have a really great channel.

[@uveselic200]

Excellent! May I ask which "blackboard" software do you use for explanations. Looking for suggestions for me for my students

[@CrashChemistryAcademy]

I use microsoft powerpoint. It has a lot of drawing and animation features.

[@smartscience5305]

Hi again, we'll gonna take this lesson in grade 10 or 11 or 12 or college?

[@CrashChemistryAcademy]

usually grade 10, although not all teachers would teach it--it depends on the level of the introductory chem class. The higher the level, the more likely it will be taught.

[@smartscience5305]

@@CrashChemistryAcademy Thank you 😁😁

[@pauldolinko9905]

great explanation!! Thanks!!!. Luv yer channell!!!!

[@Galileosays]

Personally I prefer the expression: P = rho.RT as the ideal gas law, because it is more intuitive. When the molar density (rho) goes up, more molecules hit the wall and thus pressure should be higher. Likewise, when the temperature goes up, the speed of the molecules goes up and so the pressure increases. In your presentation it would be better to call "d" the specific density, which follows from the molar density: d=M.rho=MP/RT.

[@CrashChemistryAcademy]

I agree, P = ρRT gives a more intuitive (and mathematically) direct relationship between number of particles and pressure, and so gives a thinking student a direct connection to Avogadro's Law of V1/n1 = V2/n2. It would have been a good addition to the vid, I did not think of it.

[@Galileosays]

@@CrashChemistryAcademy Thanks for the reply.

[@bobbleheadgary]

beautiful, thanks for posting

[@chemistfatimasmartscientis3923]

Great Vedios👍

[@KingSkkrt]

Thank you!

[@Sunita73-cx5vb]

Si unit of density is kg/m³ by MP/RT how plzz tell

[@CrashChemistryAcademy]

It depends on the unit used for R. I use units for R that result in density expressed in g/L, which is a common way to express gas density. If you use R with m³ instead of L, then you would have to express molar mass as kg/mol in order to get to kg/m³. If you notice that there are 1000 g in 1 kg, and there are 1000 L in 1 m³, that means 1 g/L = 1 kg/m³. So whatever density you get from the math that is expressed as g/L, you can use the same value to express it as kg/m³. So for example 3.07 g/L = 3.07 kg/m³, etcetera.

[@Sunita73-cx5vb]

@@CrashChemistryAcademy when I solved the equation I got kg/L so how can i make it kg/m³

[@CrashChemistryAcademy]

What unit for R are you using?

[@Sunita73-cx5vb]

@@CrashChemistryAcademy kpa/mol.k

[@CrashChemistryAcademy]

Your unit is missing volume. The R you gave must have a volume unit (usually liters) in the numerator.

[@minniemouse7348]

Thank you sooo much!!

[@হিরোআলম-ণ৮ভ]

Sir, you are too great . Could you please recommend me a good CHEMISTRY textbook for HiGhScHOoL

[@MutoniRitah18]

Thnx so much

[@CrashChemistryAcademy]

You're welcome! Thanks for watching.

[@TechnicalGamerDino99]

Thanks .

[@Sunita73-cx5vb]

Is kg/L = kg/m³

[@CrashChemistryAcademy]

No. There are 1000 g in 1 kg, there 1000 L in 1 m^3. So 1 g/L = 1 kg/m^3. And 1 kg/L = 1000 kg/m^3.

[@Sunita73-cx5vb]

@@CrashChemistryAcademy sir i used D = PM/RT D= kpa*kg/mol whole divided by Lkpa/mol*k * k After cancelling kpa , mol, K I got kg/l But si unit is kg/m³ so what do I do Sir can u help me by Commenting or making a video on it Sir I am very thankful to you because you replied me on every comment ❤

[@CrashChemistryAcademy]

the relationship between kg/L and kg/m^3 is 1:1000, so you multiply your amount of kg/L by 1000 to get the amount of kg/m^3

[@ааа-п4н5ч]

Thanks👓🐦💚

[@returnofthemomos]

First

[@CrashChemistryAcademy]

Thanks Owen, wherever you are...