Allen M need help to clarify doubt about: : what heigh must an oxygen molecule fall in a vacuum so KE=average energy of a molecule at 312K?
From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of an oxygen molecule at 312 K?
Trying to do this one, but unsure of the equation? Any help is appreciated!
Try this:
Answer by joex444
You need to know what the average energy of a molecule is at 312K.
I’m pretty sure that E = 3/2 K * T, where K is Boltzmann’s constant, and T is temperature in Kelvin. (I looked this up, it is true. A molecule has 3 directions which it can vibrate in. Each direction has KT of energy, but 50% of it is potential and 50% is kinetic so it’s 3/2 KT for a molecule’s kinetic energy).
What you really want to relate, though, is potential energy to 3/2KT. So, you want to solve mgh = 3/2 * KT. You know m, g, K, T so this should be pretty simple. The only thing you need to look up is the mass of oxygen (or you could try to derive it… Oxygen is essentially 8 protons and 8 neutrons, and the mass of a neutron is basically the mass of a proton, so this is 16 * m_p. But if you don’t know the mass of a proton, then you could remember than it is 1836x the mass of an electron, which we all know is 9.11*10^-31 kg. Oh… well, I know that, whatever.)
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