Temperature & Entropy - Revision history

Ghood6: Undo revision 48072 by Ghood6 (talk)

2026-04-26T23:49:47Z

Undo revision 48072 by Ghood6 (talk)

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Ghood6 at 23:48, 26 April 2026

2026-04-26T23:48:31Z

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MaxwellH: /* Notable Scientists */

2024-12-04T20:59:24Z

Notable Scientists

← Older revision		Revision as of 16:59, 4 December 2024
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	*[[Nicolas Leonard Sadi Carnot]]		*[[Nicolas Leonard Sadi Carnot]]
	*[[Rudolf Clausius]]		*[[Rudolf Clausius]]
			*[[Ludwig Boltzmann]]

	== See also ==		== See also ==

MaxwellH: /* Examples */

2024-12-04T20:56:53Z

Examples

← Older revision		Revision as of 16:56, 4 December 2024
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	<math>{N+q-1 \choose q} = \frac{(N+q-1)!}{q!(N-1)!}=\Omega</math>		<math>{N+q-1 \choose q} = \frac{(N+q-1)!}{q!(N-1)!}=\Omega</math>

			===MATLAB Code===

			%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

			% You may change one of these 2 variables, keep in mind once q is over 100

			% It may be hard to compute

			q = 10; % Quanta of energy

			N = 3; % Oscillators


			%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

			f = @(q,N)((factorial(q + (N - 1)))/(factorial(q) * factorial(N-1)));

			y = [];



			for i = 1:q

			y1 = f(i,N);

			y = [y y1];

			end

			x = [1:q];

			plot(x,y);

			xlim([1 q]);

			ylim([0, y(end)]);

			xlabel('q (Quanta of Energy)');

			ylabel('\Omega (Multiplicity)');

			title('Multiplicity vs. Energy Quanta');

			box on

			grid on;

	==Connectedness==		==Connectedness==

Jbrandt35: /* Examples */

2022-04-24T16:23:04Z

Examples

← Older revision		Revision as of 12:23, 24 April 2022
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	==Examples==		==Examples==

	~~Be sure to show all steps in your solution and include diagrams whenever possible~~

	===Simple===		===Simple===

Jbrandt35: /* Simple */

2022-04-24T16:22:53Z

Simple

← Older revision		Revision as of 12:22, 24 April 2022
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	Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.		Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.

	Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil. Thats <math>\frac{6 \times 10^{23}}{12}</math> Carbon atoms. For each atom, there are three oscillating springs under the Einstein Model, so there is <math>\mathcal{O}(10^{23})</math> oscillators, <math>N</math>. Obviously, <math>\Omega</math> will quickly become unmanageably large. Taking <math>\ln{10^{23}}</math> we get only <math>52</math>		Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil. Thats <math>\frac{6 \times 10^{23}}{12}</math> Carbon atoms. For each atom, there are three oscillating springs under the Einstein Model, so there is <math>\mathcal{O}(10^{23})</math> oscillators, <math>N</math>. Obviously, <math>\Omega</math> will quickly become unmanageably large. Taking <math>\ln{10^{23}}</math> we get only <math>52</math>. To give appropriate units, we need a constant to multiply, which <math> k_B </math> gives. It should make sense why working with entropy on everyday scales is manageable under this definition.

	===Middling===		===Middling===

Jbrandt35: /* Simple */

2022-04-24T16:21:53Z

Simple

← Older revision		Revision as of 12:21, 24 April 2022
Line 48:		Line 48:
	Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.		Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.

	Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil. Thats <math>\frac{6 \times 10^{23}}{12}</math> Carbon atoms. For each atom, there are three oscillating springs under the Einstein Model, so there is <math>\mathcal{O}(10^{23})</math> oscillators, <math>N</math>.		Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil. Thats <math>\frac{6 \times 10^{23}}{12}</math> Carbon atoms. For each atom, there are three oscillating springs under the Einstein Model, so there is <math>\mathcal{O}(10^{23})</math> oscillators, <math>N</math>. Obviously, <math>\Omega</math> will quickly become unmanageably large. Taking <math>\ln{10^{23}}</math> we get only <math>52</math>

	===Middling===		===Middling===

Jbrandt35: /* Simple */

2022-04-24T16:20:05Z

Simple

← Older revision		Revision as of 12:20, 24 April 2022
Line 48:		Line 48:
	Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.		Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.

	Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil. Thats <math>\frac{6 \times 10^{23}}{12}</math> Carbon atoms.		Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil. Thats <math>\frac{6 \times 10^{23}}{12}</math> Carbon atoms. For each atom, there are three oscillating springs under the Einstein Model, so there is <math>\mathcal{O}(10^{23})</math> oscillators, <math>N</math>.

	===Middling===		===Middling===

Jbrandt35: /* Simple */

2022-04-24T16:18:04Z

Simple

← Older revision		Revision as of 12:18, 24 April 2022
Line 48:		Line 48:
	Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.		Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.

	Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil.		Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil. Thats <math>\frac{6 \times 10^{23}}{12}</math> Carbon atoms.

	===Middling===		===Middling===

Jbrandt35: /* Simple */

2022-04-24T16:17:05Z

Simple

← Older revision		Revision as of 12:17, 24 April 2022
Line 48:		Line 48:
	Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.		Question: Given that entropy is commonly applied to objects at everyday scales, explain why Boltzmann's definition of entropy is appropriate.

	Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>		Solution: A pencil has a mass <math>\mathcal{O}(1 \text{ gram})</math>. Pencils are made of wood and graphite, so let's say a pencil is 100% Carbon. Carbon's molar mass is <math>12 \frac{\text{gram}}{\text{mole}} </math>. So there's <math>\frac{1}{12}</math> mole of Carbon atoms in a pencil.

	===Middling===		===Middling===