Why can $\beta $ not be linearly proportional to T, that is $\beta =constant\times T$

Almintas2l
2022-07-20
Answered

Why can $\beta $ not be linearly proportional to T, that is $\beta =constant\times T$

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salumeqi

Answered 2022-07-21
Author has **15** answers

Because by convention, we want to write $\beta $ as the coefficient in front of energy E in the exponent $\mathrm{exp}(-\beta E)$. Exponents have to be dimensionless so $\beta $ has to have units of inverse energy. That's why it has to be objects such as $\beta =1/kT$ because $kT$ has units of energy. The latter statement holds because the energy per degree of freedom increases with the temperature. At the end, that's the fundamental answer to your question. The exponential is $\mathrm{exp}(-E/kT)$ because $E\sim kT$ and "hot" means "highly energetic". It's linked to the fact that the temperature has an absolute lower bound, the absolute zero, much like energy is bounded from below.

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Is the loudness of a ball that you drop a measure of the amount of energy you put in to lift it?

When you lift a heavy object off the floor and raise it then you've exerted an amount of work/energy on that body. When you release the heavy object gravity exerts the same work/energy. But when you lift the ball nothing dramatic happens. You lift it and that's it. When you drop the ball there is a loud bang when the ball hits the floor. What accounts for this difference? It seems that it must be that there is an equivalent amount of energy in your body as in the sound (assuming that the floor doesn't change so that most of the energy goes into sound). Is this accurate reasoning? Can you do sound calorimetry?

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3. Combine your previous two answers to generate a simplified expression (involving no derivatives) for the pressure of the ideal monatomic gas

1. What is the Helmholtz energy of this system? (Be sure to substitute the form of any relevant partition function(s) from your formula sheet).

2. Base don the Thermo dynamic square, what is there lation ship between Helmholtz energy and pressure. (Give a general thermodynamic expression not involving the par- tition function.)

3. Combine your previous two answers to generate a simplified expression (involving no derivatives) for the pressure of the ideal monatomic gas

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