Second Law of Thermodynamics and Entropy - Revision history

Laurence12799: /* Difficult */

2019-08-05T03:15:41Z

Difficult

← Older revision		Revision as of 23:15, 4 August 2019
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	===Difficult===		===Difficult===
	:'''Insert Difficult Example ~~her~~'''		:'''Insert Difficult Example here'''

	==Connectedness==		==Connectedness==

Laurence12799: /* Difficult */

2019-08-05T03:15:20Z

Difficult

← Older revision		Revision as of 23:15, 4 August 2019
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	===Difficult===		===Difficult===
			:'''Insert Difficult Example her'''

	==Connectedness==		==Connectedness==

Laurence12799: /* Middling */

2019-08-05T03:14:26Z

Middling

← Older revision		Revision as of 23:14, 4 August 2019
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	===Middling===		===Middling===
	~~:give number of different micro-states of a system, ask for entropy~~
	~~:change number of micro-states, ask for entropy again~~
	~~:ask for change in entropy, is this possible naturally?~~

	A small glass jar contains <math>18.06 \times 10^{23}</math> molecules of a gas, or 3 moles of the gas. Each micro-state is equally probable, and the entropy of the gas is <math>112 \ \frac{J}{mol \cdot K}</math> initially.		A small glass jar contains <math>18.06 \times 10^{23}</math> molecules of a gas, or 3 moles of the gas. Each micro-state is equally probable, and the entropy of the gas is <math>112 \ \frac{J}{mol \cdot K}</math> initially.

Laurence12799: /* Middling */

2019-08-05T03:13:52Z

Middling

← Older revision		Revision as of 23:13, 4 August 2019
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	:::<math>\Omega = e^P = e^{40.4449} = 3.673 \times 10^{17}</math> micro-states		:::<math>\Omega = e^P = e^{40.4449} = 3.673 \times 10^{17}</math> micro-states

			:'''b) Now, the number of micro-states is reduced to <math>\Omega = 1.54 \times 10^{11}</math>. What is the change in entropy of the system?'''

			::The initial entropy can be calculated as:

			:::<math>S_{i} = k_{B} \ \text{ln}(\Omega_{i}) = (1.38 \times 10^{-23})\ \text{ln}(3.673 \times 10^{17}) = 7.488 \times 10^{-22}</math>

			::The final entropy can be calculated as:

			:::<math>S_{f} = k_{B} \ \text{ln}(\Omega_{f}) = (1.38 \times 10^{-23}) \ \text{ln}(1.54 \times 10^{11}) = 3.555 \times 10^{-22}</math>

			::Therefore, the change in entropy is:

			:::<math>\Delta S = S_f - S_i = 3.555 \times 10^{-22} - 7.488 \times 10^{-22} = (3.555 - 7.488) \times 10^{-22} = -3.933 \times 10^{-22} \ \frac{J}{K}</math>

			:'''c) Would the change in entropy calculated in the previous part normally be possible in a natural system?'''

			::No, a negative change in entropy corresponds to an increase in order of the system. This would not normally happen in a natural system.

	===Difficult===		===Difficult===

Laurence12799: /* Middling */

2019-08-05T02:52:02Z

Middling

← Older revision		Revision as of 22:52, 4 August 2019
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	:::<math> \frac{nS_{mole}A}{k_{B}} = \text{ln}(\Omega)</math>		:::<math> \frac{nS_{mole}A}{k_{B}} = \text{ln}(\Omega)</math>

	:::<math>\Omega = e^{\frac{nS_{mole}A}{k_{B}}}</math>		:::<math>\Omega = e^{\frac{nS_{mole}A}{k_{B}}} = e^{P}</math>

	:::<math>\frac{nS_{mole}A}{k_{B}} = \frac{3 \times 112 \times \frac{1}{6.02 \times 10^{23}}}{1.38 \times 10^{-23}} = \frac{3 \times 112}{(1.38 \times 10^{-23}) \times (6.02 \times 10^{23})} = \frac{3 \times 112}{1.38 \times 6.02} = 40.4449</math>		:::<math>P = \frac{nS_{mole}A}{k_{B}} = \frac{3 \times 112 \times \frac{1}{6.02 \times 10^{23}}}{1.38 \times 10^{-23}} = \frac{3 \times 112}{(1.38 \times 10^{-23}) \times (6.02 \times 10^{23})} = \frac{3 \times 112}{1.38 \times 6.02} = 40.4449</math>

			:::<math>\Omega = e^P = e^{40.4449} = 3.673 \times 10^{17}</math> micro-states

	===Difficult===		===Difficult===

Laurence12799: /* Middling */

2019-08-05T02:49:32Z

Middling

← Older revision		Revision as of 22:49, 4 August 2019
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	:::<math>\Omega = e^{\frac{nS_{mole}A}{k_{B}}}</math>		:::<math>\Omega = e^{\frac{nS_{mole}A}{k_{B}}}</math>

			:::<math>\frac{nS_{mole}A}{k_{B}} = \frac{3 \times 112 \times \frac{1}{6.02 \times 10^{23}}}{1.38 \times 10^{-23}} = \frac{3 \times 112}{(1.38 \times 10^{-23}) \times (6.02 \times 10^{23})} = \frac{3 \times 112}{1.38 \times 6.02} = 40.4449</math>

	===Difficult===		===Difficult===

Laurence12799: /* Middling */

2019-08-05T02:44:37Z

Middling

← Older revision		Revision as of 22:44, 4 August 2019
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	:ask for change in entropy, is this possible naturally?		:ask for change in entropy, is this possible naturally?

	A small glass jar contains <math>18.06 \times 10^{23}</math> molecules of a gas, or 3 moles of the gas. Each micro-state is equally probable, and the entropy of the gas is <math>~~2,530~~ \ \frac{J}{K}</math> initially.		A small glass jar contains <math>18.06 \times 10^{23}</math> molecules of a gas, or 3 moles of the gas. Each micro-state is equally probable, and the entropy of the gas is <math>112 \ \frac{J}{mol \cdot K}</math> initially.

	:'''a) How many micro-states are there?'''		:'''a) How many micro-states are there?'''
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	::Since each micro-state is equally probable, we can use Boltzmann's simplified equation for entropy to solve for the number of micro-states:		::Since each micro-state is equally probable, we can use Boltzmann's simplified equation for entropy to solve for the number of micro-states:

	:::<math>S = k_{B} \ \text{ln}(\Omega)</math>		:::<math>nS_{mole}A = k_{B} \ \text{ln}(\Omega)</math>, where <math>n = 3</math>, <math>A = \frac{1 \ mol}{6.02 \times 10^{23} \ \text{molecules}}</math>, and <math>S_{mole} = 112 \ \frac{J}{mole \cdot K}</math>

	:::<math> \frac{S}{k_{B}} = \text{ln}(\Omega)</math>		:::<math> \frac{nS_{mole}A}{k_{B}} = \text{ln}(\Omega)</math>

	:::<math>\Omega = e^{~~S/k_~~{B}} ~~= e^~~{ ~~120 \ / \ 1.38 \times 10^~~{~~-23~~}} = </math>		:::<math>\Omega = e^{\frac{nS_{mole}A}{k_{B}}}</math>

	===Difficult===		===Difficult===

Laurence12799: /* Mathematical Model */

2019-08-05T01:27:55Z

Mathematical Model

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	'''''System 3: The Universe'''''		'''''System 3: The Universe'''''
	:Consider the Universe as the system. Energy flows to the freezer (despite insulation, the Temperature of the freezer will inevitably increase). Therefore, ΔQ is positive. Meaning that the entropy must increase. Consequently, we find that the energy principle is still satisfied, as the energy is transformed, and not created nor destroyed.		:Consider the Universe as the system. Energy flows to the freezer (despite insulation, the Temperature of the freezer will inevitably increase). Therefore, <math>\Delta Q</math> is positive. Meaning that the entropy must increase. Consequently, we find that the energy principle is still satisfied, as the energy is transformed, and not created nor destroyed.

	In general if:		In general if:

Laurence12799: /* Mathematical Model */

2019-08-05T01:26:33Z

Mathematical Model

← Older revision		Revision as of 21:26, 4 August 2019
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	If each possible micro-state is assumed to be equally probable, then:		If each possible micro-state is assumed to be equally probable, then:

	:<math>S = -k_{B}\sum_{i} p_{i}\ \text{log}(p_{I}) = -k_{B}\sum_{I}^{\Omega} \frac{1}{\Omega} \ \text{log}\left(\frac{1}{\Omega}\right) = -k_{B} \ \text{ln}(\Omega)</math>		:<math>S = -k_{B}\sum_{i} p_{i}\ \text{log}(p_{I}) = -k_{B}\sum_{I}^{\Omega} \frac{1}{\Omega} \ \text{log}\left(\frac{1}{\Omega}\right) = k_{B} \ \text{ln}(\Omega)</math>

	Also, The Second Law of Thermodynamics states the entropy of an isolated system will always stay the same (if a reversible process occurs) or increase (if an irreversible process occurs), as a system tends towards thermodynamic equilibrium.		Also, The Second Law of Thermodynamics states the entropy of an isolated system will always stay the same (if a reversible process occurs) or increase (if an irreversible process occurs), as a system tends towards thermodynamic equilibrium.

Laurence12799: /* Middling */

2019-08-05T01:23:34Z

Middling

@@ Line 139: / Line 139: @@
 :change number of micro-states, ask for entropy again
 :ask for change in entropy, is this possible naturally?
 ===Difficult===