What is the entropy change when the volume of 1.6 g of O2 increases from 2.5 L to 3.5 L at a constant temperature of 75 °C? Assume that O2 behaves as an ideal gas.

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Hello. Everyone in this video, we're assuming that methane behaves in an ideal gas. We want to determine the change of entropy when methane decreases from 1.95 liters to one point oh five liters in volume at a constant temperature of 209 eight kelvin's. So first let's go ahead and determine the number of moles of that thing that we have. So we see here that we have a mass here having kind of change our mass into the number of moles or just moles is by using the molecular weight. So here again, we're talking about anything which has a molecular formula of C2H6. We can calculate the molecular weight by hand. We're gonna go ahead and just get it from Our own sources. So you can either find it in your textbook using the pr table, you find it online or you can even be given to you by your professor. So the molecular weight value that I have is equal to 30.069 units being g per mole. So now we're gonna go and use this information to convert the g that we have into moles. So again, we'll just denote the number of moles of that. They need to just end. So again, we're starting off with the math has given that's 3.90 g of things of C two H six. Again, we said we're gonna go ahead and use the molecular weight. So that is for every one mole of methane we have 30.69 g. You can see here that the units of grams will cancel leaving us with just moles. So once I put the numbers into my calculator, I see that the number of moles of methane that we get is 1.2970 times 10 to the negative one moles of methane. So C two H six. So the entropy change at a constant temperature and were given the number of moles on ideal gas. And we have the volume as such. We can go ahead and relate this all with the formula that is delta S. Equals to the number of moles multiplied by our deal gas law constant, multiplied by the natural log of our final volume over our initial volume. So V T r V two which is the final volume and V one is our initial volume. Alright, so of course we are given some of this information. So we are given that the V one is going to be 1.95 liters since we're saying we're having this, going to this. So this is our final volume. Right? So this means that our V two then we just said is 1. liters. So here we do see that we have basically all of our parts of our equation. So we have the number of moles already. We already have course know the constant here and we have identified V two, N. V one. So we go ahead and directly use this formula. So now just putting it in, then we'll have delta s equaling two n. So we used that in blue and that's 1. times 10 to the negative one moles. We're going to multiply that with the ideal gas law constant, that being 8.314 units being jewels, per mole, kelvin or times kelvin's. And then of course the natural log, we said B two is equal to one point oh eight liters over B one is 1. liters receive. Once you put everything here into the calculate for the numerical value that we get, that's equal to negative 6.37 times 10 to the negative one. And of course our units are just going to be equal to jewels per kelvin. So this right here, my delta S. Is my final value for this problem. Thank you so much for watching.

What is the entropy change when the volume of 1.6 g of O2 increases from 2.5 L to 3.5 L at a constant temperature of 75 °C? Assume that O2 behaves as an ideal gas.

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Key Concepts

Entropy

Ideal Gas Law

Change in Entropy for an Ideal Gas