Third Law of Thermodynamics - Video Tutorials & Practice Problems
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1
concept
Entropy of Perfect Crystals
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2m
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Now the 3rd law of thermodynamics states that the entropy of a perfect crystal is 0 at absolute 0. Now absolute 0 is 0 kelvin. Here we're going to say a perfect crystal is just a solid with a regular and ideal internal atomic arrangement. So here we have a perfect crystal where all of its components perfectly aligned with one another. And we're gonna say here, this happens again at a temperature of 0 Kelvin. It's frozen perfectly in place because the temperature is as low as it can get. There's no such thing as negative Kelvin. So this is the bottom of the temperature scale. And because it's frozen in place, it can't move around. So there's only gonna be one type of arrangement it can have, the one pictured. So here we're going to say it has 1 microstate. Now if the temperature happens to be above 0 Kelvin, then the particles are not frozen perfectly in place, they're gonna wiggle around a little bit, move around a little bit. Here we have this one in red, could maybe move around a little bit, and now it's over here. And, again, this happens when the temperature is above 0 Kelvin. Because it can arrange itself in more than one way, you say it has more than 1 microstate. Now microstates are just the number of possible energetic ways to arrange components, whether they be atoms, molecules, or ions of a system, which represents our chemical reaction. Now the 3rd law of thermodynamics is pie theoretical. We can't achieve absolute 0 here on earth, and on average, the universe has a temperature around 2 Kelvin. K? But so this is saying that if we could attain this absolute zero value, this was what would happen. It'd be like freezing a structure perfectly in place. The truth is, even solids around us don't stay perfectly still. If you have a powerful enough microscope and you look at it, you would see that the molecules of even your cell phone would be vibrating in place. That's because the temperature around us is higher than 0 Kelvin. So keep this in mind when we talk about microstates and the third law of thermodynamics.
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example
Third Law of Thermodynamics Example
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1m
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Now here in this example question it says, all the statements are correct except the greater number of molecular motion, the greater number of possible microstates. That's true. A perfectly ordered system has more than 1 microstate. Remember, if you're a perfectly ordered system, you're a perfect crystal. This happens at absolute 0, 0 Kelvin. You're frozen perfectly in place, so you'd only have 1 microstate. So this statement here is false. Any system at a temperature above 0 Kelvin has a positive change in entropy value. Yes. Because above 0, you can arrange yourself in different ways. You can move around. Greater motion, greater movement means there'll be greater change in the change of your entropy. So this is true. Perfect crystal exhibits no molecular motion. That is also true. A perfect crystal is frozen perfectly in place at 0 Kelvin, can't move around, can't rearrange itself, so it only has one type of arrangement, one microstate. So out of all the options, only option b is false.
3
concept
The Boltzmann Equation
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1m
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So the Boltzmann equation is credited to the Austrian physicist, Ludwig Boltzmann, where he related entropy to the number of microstates. Now, entropy uses the variable s, microstates uses the variable capital w. Here, the Boltzmann equation says that the entropy of my system, which is my chemical reaction, equals k times your natural log l n, times your number of microstates. Here k equals your Boltzmann constant and it is 1.38 times 10 to the negative 23 joules per Kelvin. And again, capital w just represents the number of microstates. Here, we would say that the greater number of the microstates, the greater the entropy. That's because more microstates means more freedom of motion, more movement, more arrangements, more chaos, more disorder. So this would mean an increase in our entropy. Now, a microstate state equal to 1 would mean I would plug in 1 here, 4 w, Allen of 1 is 0, which would mean at the end that my entropy is equal to 0. Right? So keep this in mind. This is just a mathematical way to calculate the entropy of my system if the number of microstates are known.
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example
Third Law of Thermodynamics Example
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49s
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Now, consider a system with a total of 3 times 10 to the 26 number of microstates. What is the entropy of such a system? Alright. So, here we utilize the Boltzmann equation, so we'd say that the entropy of my system, the change in entropy of my system equals k times ln of w. K represents my Boltzmann constant, which is 1 point 38 times 10 to the minus 23 joules per Kelvin. And then we'd say ln of the number of microstates, which is 3 times 10 to the 26. When we plug this in, this will give me 8.41 times 10 to the negative 22 joules per Kelvin. This will represent the change in the entropy of my given system.
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Problem
Problem
A brand new deck of cards which hasn't been shuffled yet, possesses only one arrangement. Another, older deck has been shuffled and possesses 8 × 1067 arrangements. Calculate and compare entropies of each deck.
A
deck 1 = 0, deck 2 = 2.158 × 10−21 J/K
B
deck 1 = 0, deck 2 = 9.371 × 10−22 J/K
C
deck 1 = 0, deck 2 = 9.371 × 1024 J/K
D
deck 1 = 0, deck 2 = 2.158 × 1025 J/K
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