Thermodynamics
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Thermodynamics (part 1)
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Thermodynamics (part 2)
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Thermodynamics (part 3)
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Thermodynamics (part 4)
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Thermodynamics (part 5)
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Macrostates and Microstates
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Quasistatic and Reversible Processes
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First Law of Thermodynamics/ Internal Energy
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More on Internal Energy
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Work from Expansion
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PV-diagrams and Expansion Work
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Proof: U=(3/2)PV or U=(3/2)nRT
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Work Done by Isothermic Process
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Carnot Cycle and Carnot Engine
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Proof: Volume Ratios in a Carnot Cycle
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Proof: S (or Entropy) is a valid state variable
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Thermodynamic Entropy Definition Clarification
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Reconciling Thermodynamic and State Definitions of Entropy
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Entropy Intuition
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Maxwell's Demon
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More on Entropy
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Efficiency of a Carnot Engine
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Carnot Efficiency 2: Reversing the Cycle
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Carnot Efficiency 3: Proving that it is the most efficient
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Enthalpy
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Heat of Formation
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Hess's Law and Reaction Enthalpy Change
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Gibbs Free Energy and Spontaneity
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Gibbs Free Energy Example
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More rigorous Gibbs Free Energy/ Spontaneity Relationship
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A look at a seductive but wrong Gibbs/Spontaneity Proof
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Stoichiometry Example Problem 1
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Stoichiometry Example Problem 2
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Limiting Reactant Example Problem 1
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Empirical and Molecular Formulas from Stoichiometry
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Example of Finding Reactant Empirical Formula
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Stoichiometry of a Reaction in Solution
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Another Stoichiometry Example in a Solution
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Molecular and Empirical Forumlas from Percent Composition
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Hess's Law Example
Limiting Reactant Example Problem 1 Limiting Reactant Example Problem 1
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- We're told methanol, which is used as a fuel in racing cars
- and fuel cells, can be made by the reaction of carbon
- monoxide and hydrogen.
- So this is the methanol right there.
- They're giving us 356 grams of carbon monoxide.
- So carbon monoxide we have 356 grams of it.
- And they're giving us 65 grams of hydrogen,
- of molecular hydrogen.
- They're mixed and allowed to react.
- They say what mass of methanol can be produced?
- And then they say what mass of the excess reactant remains
- after the limiting reactant has been consumed?
- So that tells you this is a limiting reactant problem,
- that we have too much or too little of one
- of these two reactants.
- These are the two reactants there.
- The one that we have less of is the limiting reactant and
- that'll dictate how much of the product we can produce.
- And the one that we have more of is the excess reactant.
- But first we have to figure out which is the limiting and
- which is the excess.
- And before we even do that, we should always check that our
- equation is actually balanced.
- So let's just check that.
- On the left-hand side of this equation we have 1 carbon
- right there.
- On the right-hand side we also have 1 carbon, so that looks
- good so far.
- On the left-hand side of the equation we have 1 oxygen, and
- on the right-hand side we have 1 oxygen.
- Looks good so far.
- Left-hand side we have 4 hydrogens, 2 times 2.
- On the right-hand side we have 4 hydrogens, 3 plus 1.
- So this is balanced so we can proceed to try to figure out
- what the limiting reactant is.
- So the way we want to do it is to figure out how many moles
- of each of these we're given, and then figure out how many
- moles the stoichiometric ratio that's
- required by this reaction.
- We already know what it is.
- We know that for every two moles of hydrogen required, we
- require-- you can see it right here from the equation-- 1
- mole of carbon monoxide.
- This is what the equation tells us, so let's see how
- many moles of hydrogen and how many moles of carbon monoxide
- we have.
- The first place to start-- we've done this several
- times-- is what is the molecular
- weight of carbon monoxide?
- So carbon monoxide molecular weight-- let me write it
- here-- molecular weight for carbon monoxide.
- It's a carbon, has a molecular weight of 12-- it's good to
- memorize that, carbons and oxygens and hydrogens show up
- so frequently.
- And then oxygen is 16.
- So it's equal to 28-- a molecular weight of 28, which
- tells us that carbon monoxide, if we want to view it this
- way-- so let's do the carbon monoxide first. So we're
- dealing with the carbon monoxide.
- So we have 356 grams of carbon monoxide, and we want to write
- this in terms of how many moles of carbon monoxide do we
- have. So what we want is how many grams per mole-- but I
- actually write-- 1 mole has how many grams?
- So 1 mole of carbon monoxide has how many grams?
- Well it's molecular weight is 28 so if we
- have a mole of them.
- If we have that 6.02 times 10 to the 23rd molecules of
- carbon monoxide.
- That's going to weigh 28 grams. So 1 mole of carbon
- monoxide is 28 grams, or I guess you have 1 mole per
- every 28 grams. The reason why I wanted to put the grams in
- the denominator so it cancels out over here.
- So those cancel out when we multiply.
- And so we are left with 356 times 1 divided by 28 moles of
- carbon monoxide.
- So let's figure what that is.
- That's 356 divided by 28-- right, that's essentially what
- we're doing-- divided by 28 is equal to 12.71.
- Let's just say it's 12.7 moles of carbon monoxide.
- Now let's do the same thing for the hydrogen.
- We have 65 grams of molecular hydrogen.
- What's the molecular weight of hydrogen?
- I'll do this in green over here.
- Molecular weight of hydrogen is-- well each hydrogen atom
- has a molecular weight of 1 times 2, which is equal to 2.
- So we have 65 grams of a molecular hydrogen, and the
- same way, we want to write it in moles.
- So we're going to multiply it times-- 1 mole of hydrogen is
- equal to how many grams?
- Well we just figured it out.
- 1 mole is equal to its molecular weight is 2.
- So a mole of it is going to have a mass of 2 grams. You
- can view this as 2 grams per mole or 1 mole per 2 grams.
- And we want the grams in the denominator so it
- cancels out over here.
- So let's do the math.
- That cancels out with that.
- And we have 65 times 1 divided by 2.
- 65 divided by 2 is what-- 32.5 moles of hydrogen.
- Now we know exactly how many moles of carbon monoxide and
- how many moles of hydrogen they've given us.
- Let's figure out what the ratio is.
- And we'll do it in the exact same way as we wrote up here.
- So we have 32.5-- this is what we're given-- moles of
- hydrogen-- let me write that a little bit
- neater-- moles of hydrogen.
- And we're given 12.7 moles-- I'll do that in the same
- color-- 12.7 moles of carbon monoxide.
- So if we were to just divide this, what is this?
- I guess you could imagine, divide the numerator and the
- denominator by 12.7, or just divide 32.5 divided by 12.7,
- what do we get?
- 32.5 divided by 12.7-- no, that's not-- let
- me write that again.
- 32.5-- man, these buttons are-- let me do it with the
- keyboard-- 32.5 divided by 12.7-- that's a lot better--
- is 2.5-- I'll just say 2.56.
- So this can be re-written as-- so if I just re-write this-- I
- should have written it here to begin with-- I could write
- this as we have 2.56 moles of molecular hydrogen for every 1
- mole of carbon monoxide.
- So this is what we need for a reaction to occur.
- This is what the balanced equation tells us.
- It tells us that we need 2 moles of hydrogen for every
- mole of carbon dioxide.
- Based on what they've given us, we just figured out that
- we have 2.56 moles of hydrogen for every
- mole of carbon dioxide.
- So we have more than enough hydrogen.
- We only need 2 for every mole of carbon dioxide.
- We have 2.56.
- So we have an excessive amount of hydrogen.
- So the excess reactant is the hydrogen.
- Hydrogen is excess reactant.
- And the other one's going to be the limited reactant.
- We don't have enough carbon monoxide to react all of the
- hydrogen, right?
- We only have 1 for every 2.56.
- We would actually, for this, says you need 1.25 or whatever
- for every, or 1.28 for every 2.56.
- It should be a 1:2 ratio.
- So we don't have enough carbon monoxide to
- react all of the hydrogen.
- So carbon monoxide is the limiting reactant.
- Now given that this is the excess reactant, we can use
- the stoichiometric ratios to figure out how much methanol's
- going to be produced.
- It's all going to be limited by our carbon monoxide.
- Did I just say-- hydrogen's not the limiting reactant,
- carbon monoxide is the limiting reactant.
- We have more than enough hydrogen.
- So how much carbon monoxide do we have?
- We already figured it out.
- We have 12.7 moles of carbon monoxide.
- And we look at our-- let me write this over here-- so we
- have 12.7 moles of carbon monoxide.
- And looking at our original equation, we see for every
- mole of carbon monoxide, we produce 1 mole of methanol.
- So let's write that down.
- Times-- and we want the carbon monoxide in the denominator.
- So for every 1 mole of carbon monoxide used we have 1 mole
- of methanol, which is, what, CH3OH-- did I get that right,
- yup-- produced.
- We get that straight from the balanced equation.
- And this math is pretty easy, but it gives
- us the right unit.
- Remember, we're using the carbon monoxide, not the
- hydrogen because the carbon monoxide's
- the limiting reagent.
- That's what's telling us what's going-- if we used
- hydrogen as the limiting reactant, then we wouldn't
- have enough carbon monoxide for the reaction to occur.
- So this is what's kind of capping off on how much this
- reaction can move forward.
- But the whole point of this was to cancel that and that.
- So, obviously, if we're using 12.7 moles of carbon monoxide,
- we're going to produce 12.7 moles of
- methanol will be produced.
- And now we just have to figure what is the mass of 12.7 moles
- of methanol.
- And we just think about what the atomic weight.
- So if you look at methanol, CH3-- let me
- put the H3-- H3OH.
- Its atomic weight is 12 plus 3 times 1 plus 3 plus 16 plus 1.
- So what is this?
- This is 20 plus 2 is equal to 32, or if we think in moler
- terms-- or not moler terms, I should just say 32 moles--
- this is its atomic weight.
- So that tells us that if we have a mole of it we're going
- to have 32 grams of methanol per 1 mole of methanol.
- And once again, we got that by figuring
- out its atomic weight.
- Now to convert the number of moles that we have of methanol
- to the number of grams, we just
- multiplied that times that.
- That units work out.
- This is in the numerator, this is in the denominator.
- Let me just copy and paste it.
- So we have that, copy and paste, and then we can
- multiply it times that.
- Let me copy and paste it.
- You get that right there, and I'll pick a different color,
- maybe a blue.
- Just like that, and then let's multiply these two.
- We have moles of methanol, and we're left with 12.7 times 32
- is equal to-- so let me just keyboard again, let me clear
- it out-- 12.7 times 32, 406.4.
- I'm kind of adding an extra significant digit there, but
- let's just go with it.
- 406.4-- actually, I should just stay
- with significant digits.
- So we'll say 406 grams-- I'm rounding down-- 406 grams of
- methanol-- let me write it with the units-- so grams of
- methanol-- I went off the screen-- of methanol produced.
- I could write the produced down here
- since we're all done.
- So I think that was the first part of our question.
- How many grams of methanol.
- What mass of methanol, that's 400-- I have a horrible
- memory-- 406 grams. And then they say what mass of excess
- reactant remains after the limiting
- reactant has been consumed?
- Now there's a couple of ways you can do this.
- The easiest way to think about it is that the
- mass has to be conserved.
- So we started off over here with 65 grams of-- let me be
- careful-- we started off here with 356 grams of carbon
- monoxide, and 65 grams of hydrogen, and we were able to
- produce 406 grams of methanol.
- And we figured out that carbon monoxide was
- the limiting reactant.
- So all of this gets consumed, and only some
- of this gets consumed.
- So if you do the math, what gets consumed has to be equal
- to 406 grams because that's what gets produced.
- So let's think about it a little bit.
- The left-hand side, how many total grams do we have?
- So if we have 356 plus 65, we're starting off with 421
- grams of reactant.
- So we're starting with 421 grams of reactant, and then we
- end up with 406 grams of product, of our methanol.
- So that means that 421 minus 406 grams of
- reactant was not used.
- So that means that 421 minus 406, which is equal to what?
- 21 minus 6 is 15 grams of reactant not used.
- Now the reactant that's not going to be used is the one
- that you have an excess of.
- And we figured out that we have an excess of hydrogen.
- So all of that 15 grams must have been 15 grams of hydrogen
- that was not used.
- So 15 grams of hydrogen left over.
- Now the other way you could have done this exact same
- problem is you could have said look, we're starting with 12.7
- moles of carbon monoxide-- that's the limiting reactant.
- It's a 2:1 ratio-- you need 2 moles of hydrogen for every
- mole of carbon monoxide.
- So you said, OK, if we have 12.7 moles of this, I need
- twice that many moles of hydrogen.
- So you would say well, what's twice that?
- That's 25.4.
- You need 25.4 moles, you have 32.5, so you subtract the
- difference-- you subtract 32.5 minus 25.4-- And that number
- of moles of hydrogen is left over.
- You'd multiply it times the grams per mole, which is 2,
- and then you, once again, would get the same thing.
- You would get 15 grams.
- Let's do that.
- I just want to make sure you understand.
- So we're starting off with 12.7 moles of carbon monoxide.
- And we know that we are required that 2 moles of
- hydrogen are required for every 1 mole of CO, of carbon
- monoxide, that is required.
- So you know that this cancels with that, you know 12.7 times
- 2 is 25.4 moles of H2 required.
- And what is the mass of 25.4 moles of H2.
- So let's write that.
- 25.4 moles of hydrogen required.
- Times-- well we know the molecular weight of H2 is 2,
- so 1 mole, we're going to have 2 grams of H2
- per 1 mole of H2.
- This cancels out, mole of H2, mole of H2.
- So we're going to have 25.4 times 2 is what?
- That is 50.8.
- That is going to be 50.8 grams of H2 required.
- That's how much we're going to consume in the reaction.
- Now they wanted to know how much is left over.
- So we're going to consume 50.8.
- We started with 65.
- So if you subtract 50.8 from 65-- 65 minus 50.8-- you're
- going to get what?
- You're going to get 14.2 grams left over.
- And that compares to the 16 we got left over when we just
- took the 421 minus the 406.
- And the difference between the 14.2 and the 15 is really just
- a little rounding here and there with our digits.
- Anyway, hopefully you found that helpful.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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