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I disagree. You can grasp the meaning of pv=nrt without memorizing the value of r. You can understand that electrons exist in probability shells without rememb

I disagree. You can grasp the meaning of pv=nrt without memorizing the value of r. You can understand that electrons exist in probability shells without remembering what atomic number the d orbital starts at. You can understand that e=mv^2 is only an approximation that works at low velocities without being able to do a Lorentz transform.

laretluval 7y ago
Solving problems doesn’t necessarily mean getting numerical answers. If you don’t know how to apply the equations you list algebraically then you don’t really understand them.
pducks32 7y ago
In all my undergrad and graduate level Physics, I don't think I ever used the value of R. The answer has nothing to do with the numerical value--that you can google. It has to do with knowing how to solve the problem. Though having a grasp of the numbers can help with scale and the often forgotten "does that make any reasonable sense" check.
cambalache 7y ago
> You can grasp the meaning of pv=nrt without memorizing the value of r. Do you actually think that memorizing physical constant values is solving problems?And ,yes, anybody can grasp what PV= nRT means,but what about stuff like: "The average speed of the molecules doubled, what is the temperature now"."Derive the relation between pressure and height for an ideal gas and use it to estimate at which height the atmospheric pressure halves".Interesting stuff like that and not plugging numbers in a formula. > you can understand that electrons exist in probability shells without remembering what atomic number the d orbital starts at The d orbital "starts" at n=1. Hydrogen has d orbitals.You are confusing orbital existence with the occupation of them by electrons. > You can understand that e=mv^2 is only an approximation that works at low velocities without being able to do a Lorentz transform. The e = mv^2 you propose here is not an actual formula in physics. It is either Ek = (1/2)mv^2 in classical physics, so just kinetic energy, or E =mc^2+mv^2/(1-gamma^2)^(1/2) in special relativity. Which of course at low velocities is approximated by E= mc^2 + 0.5mv^2 the second term being the Ek as calculated in the classical theory. I think your comment shows beautifully what I meant with my original assertion.

laretluval 7y ago

Solving problems doesn’t necessarily mean getting numerical answers. If you don’t know how to apply the equations you list algebraically then you don’t really understand them.

pducks32 7y ago

In all my undergrad and graduate level Physics, I don't think I ever used the value of R. The answer has nothing to do with the numerical value--that you can google. It has to do with knowing how to solve the problem. Though having a grasp of the numbers can help with scale and the often forgotten "does that make any reasonable sense" check.

cambalache 7y ago

> You can grasp the meaning of pv=nrt without memorizing the value of r. Do you actually think that memorizing physical constant values is solving problems?And ,yes, anybody can grasp what PV= nRT means,but what about stuff like: "The average speed of the molecules doubled, what is the temperature now"."Derive the relation between pressure and height for an ideal gas and use it to estimate at which height the atmospheric pressure halves".Interesting stuff like that and not plugging numbers in a formula. > you can understand that electrons exist in probability shells without remembering what atomic number the d orbital starts at The d orbital "starts" at n=1. Hydrogen has d orbitals.You are confusing orbital existence with the occupation of them by electrons. > You can understand that e=mv^2 is only an approximation that works at low velocities without being able to do a Lorentz transform. The e = mv^2 you propose here is not an actual formula in physics. It is either Ek = (1/2)mv^2 in classical physics, so just kinetic energy, or E =mc^2+mv^2/(1-gamma^2)^(1/2) in special relativity. Which of course at low velocities is approximated by E= mc^2 + 0.5mv^2 the second term being the Ek as calculated in the classical theory. I think your comment shows beautifully what I meant with my original assertion.