2.0 mol of helium at 280℃ undergo an isobaric process in which the helium entropy increases by 35 J/K. What is the final temperature of the gas?

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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 at an initial temperature of 300 degrees Celsius. A 50.0 g sample of a recently identified mono atomic gas nova undergoes an isobaric transformation. The molar mass of nova is 25. g per mole. During this process, the entropy of nova increases by 40.0 jules per Kelvin calculate the final temperature of the gas in the in Kelvin K. OK. So that's our end goal. So we're given some multiple choice answers in degrees Calvin. So let's read them off to see what our final answer might be. A is 1425 B is 1210 C is 1327 and D is 1500. So first off, let us recall and use the equation for the entropy change at constant pressure which states that delta S is equal to the number of moles multiplied by the specific heat capacity at constant pressure multiplied by the natural log of the final temperature divided by the initial temperature. OK. Also recall that the equation for the specific heat capacity at a constant pressure for a mono atomic gas is CP is equal to five halves multiplied by R where R is the universal gas constant. So we can then write entropy which is delta S is equal to the number of moles multiplied by five halves multiplied by R multiplied by the natural log of the final temperature divided by the initial temperature. So note our end goal is to solve for the final temperature. So we must rearrange this equation to solve for the final temperature using a little bit of algebra we get, so we can start isolating, we can start this whole process by isolating the natural log of the final temperature divided by the initial temperature and then take the exponential of both sides and then multiply the initial temperature over to the other side. So then when we get in the end, we should get that the final temperature all by itself is equal to the initial temperature multiplied by E to the power of two multiplied by delta S divided by five multiplied by N multiplied by R. And then we'll call this equation one. So now we need to find the number of moles. And let's remember and recall that the number of moles is equal to the mass divided by the molar mass or the molecular mass. So now we need to plug in our known variables to sulfur. And so the mass is given to us as 50 g. And then the molar mass was given to us in the problem as 25.0 g per mole, the grams cancel out leaving moles. So when we plug that into a calculator, we should get two moles. Now, we can substitute this value back into equation one. But before we do that, let's quickly convert our initial temperature in our final temperature from degrees Celsius to degrees. Now, from degrees Celsius to fair like from degrees Celsius to Kelvin. So we have our initial temperature was 300 degrees Celsius. So to do the conversion, all we have to do is just take 300 add 273. So when we do that, when we add it, we get 573 Alvin and then the final temperature same, that's the value that we're trying to find. So now that we know that our initial temperature we can solve for our final temperature. So we take our initial temperature which was Kelvin. And we multiply it by e to the power of and now we need to plug in our known variables. So two multiplied by 40 jewels per Kelvin divided by I multiplied by two moles multiplied by the universal gas constant, which the numerical value for that is 8. jewels per mole multiplied by Calvin. So when we plug that into a calculator, we should get Kelvin and that is our final answer. So the correct answer is the letter D 1500 Kelvin. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

2.0 mol of helium at 280℃ undergo an isobaric process in which the helium entropy increases by 35 J/K. What is the final temperature of the gas?

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Key Concepts

Isobaric Process

Entropy

Ideal Gas Law