A 2.0 mol sample of oxygen gas in a rigid, 15 L container is s...

Question

A 2.0 mol sample of oxygen gas in a rigid, 15 L container is slowly cooled from 250℃ to 50℃ by being in thermal contact with a large bath of 50℃ water. What is the entropy change of (a) the gas and (b) the universe?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, determine the entropy change of eye the gas. And I I the whole system during a cooling process where a or where a chef is trying to create a refreshing culinary experience. The chef brings a 3.0 mole sample of nitrogen gas in a 4.0 liter rigid container down from 100 degrees Celsius to 25 degrees Celsius by putting it in a large basin of water at degrees Celsius. OK. So that's our end goal. We need to find two separate answers. We need to find the entropy and change of the gas and the whole system during the cooling process. OK? So we're given some multiple choice answers here and they're all for both parts I and parts I, I, they're both in the same units of jewels per Kelvin. So let's read them off to see what our final answer pair might be. A is 16 and 1.9 B is 14 and 1.7 C is 16 and 1.7 and D is 14 and 1.9. Ok. So first off, let us model nitrogen gas as an ideal gas. Let us also write down all of our known and variables. Ok. So we know that the initial temperature of the gas, let's call it T I is equal to 100 degrees Celsius. But let's convert that to Kelvin really fast. So all we have to do is just take 100 plus which is equal to 373. Kelvin, let us also like note that the final temperature of the gas, let's call it TF is equal to 25 degrees Celsius. Let's convert that to Kelvin. Same, same method, 25 plus 273 equals 298 Kelvin. OK. And then the temperature of the basin, let's call it TB is equal to degrees Celsius. So let's convert that to Kelvin. So 25 plus 273 equals 298 Kelvin. OK. And now we must recall and use the change in entropy for an ideal gas. So that states that delta S is equal to N where N is the number of moles multiplied by CV, where CV is the specific heat capacity at a consistent volume multiplied by the natural log of the final temperature divided by the initial temperature plus the number of moles multiplied by the universal gas constant multiplied by the natural log of the final volume divided by the initial volume. OK. So now if we consider the process to be iso Cork, meaning it has constant volume, the change in entropy or the gas will be delta S is equal to the number of moles multiplied by the specific heat capacity at a constant volume multiplied by the natural log of the final temperature divided by the initial temperature. OK. So note that an ideal, sorry note that the nitrogen gas is a diatomic gas. So we can say that the specific heat capacity at a constant volume CV is equal to five halves multiplied by the universal gas constant. Thus, we can write that delta S is equal to the number of moles multiplied by five halves multiplied by the universal gas constant, multiplied by the natural log of the final temperature divided by the initial temperature. So now at this stage, we could plug in all of our known and variables to solve or delta S which is the entropy. So let's do that. So delta S is equal to the number of malls which was 3.0 malls multiplied by five halves multiplied by the universal gass constant. And the numerical value for that is 8.31 joules per mole multiplied by Kelvin multiplied by the natural log of the final temperature which was 298 Kelvin divided by 373 Kelvin. So when we plug that into a calculator, you should get negative 13. jewels per Kelvin. But it has to be a positive number. So let us round and to 14 and make it positive. So it'll be 14 jewels per Kelvin. And that is our answer for part I fantastic. So to solve for part I I to find the entropy of the whole system, we must add the entropy of the gas and the entropy of the basin, we know the entropy of the gas. We found that for part I, so we now need to determine the entropy of the basin first. So let's call it delta SB is equal to the heat transfer into the basin divided by the temperature of the basin. So delta SB is equal to negative, the number of moles multiplied by CV, multiplied by delta T gas. So the change in the temperature of the gas divided by the temperature of the basin and CV, like we said is the specific heat capacity. So at this point, we could plug in all of our known variables to solve for delta SB. So let's do that. So delta SB is equal to negative 3. moles multiplied by five halves multiplied by the universal gas constant, which we forgot to squeeze that in there. So it's 8. J divided by jewels per I should say jewel per more multiplied by Calvin multiplied by the change in the temperature of the gas. So it's the final temperature minus the in the initial temperature. So 298 Kelvin minus 373 Kelvin divided by temperature of the basin which was 298 Kelvin. So when we plug that into a calculator, you should get that delta SB is equal to 15.69 joules per Kelvin. OK. So now we need to note that the entropy of the whole system, which we're gonna denote it as delta S subscript WS for the whole system is equal to delta SG, which is the answer we found in part I plus delta SB. OK. So let's plug in all of our known values to solve for the entropy of the whole system. So it's equal to and this time you have to use the negative value. So it's minus or negative, I should say negative 13.99 Jews per Kelvin plus 15.69 jewels per Calvin. So when we plug that into a calculator, the entropy of the whole system should be 1.7 joules per Calvin. So this is our final answer. For part I I Awesome. So let's go look at our multiple choice answers to see what the correct answer should be. The correct answer is the letter B 14 jewels per Kelvin. And for I and for I I, 1.7 jewels per Kelvin. Thank you so much for watching. Hopefully that helped and I can wait to see you in the next video. Bye.

A 2.0 mol sample of oxygen gas in a rigid, 15 L container is slowly cooled from 250℃ to 50℃ by being in thermal contact with a large bath of 50℃ water. What is the entropy change of (a) the gas and (b) the universe?

Key Concepts

Entropy

Thermodynamic Processes

The Second Law of Thermodynamics

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