A cylinder contains 0.0100 mol of helium at T = 27.0°C. (a) How m...

Question

A cylinder contains 0.0100 mol of helium at T = 27.0°C. (a) How much heat is needed to raise the temperature to 67.0°C while keeping the volume constant? Draw a pV-diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0°C to 67.0°C? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

Accepted Answer

Hey everyone in this problem. We have 0.013 moles of neon heated under conditions of constant volume and constant pressure from 17°C to 80°C. We're told that the heat required is different. Were asked why are the two values different? Which process requires more heat? And how does the process use the extra heat? Okay. All right, well let's start with the constant volume process. Okay, so if we have a constant volume process, this is ice, a cork. And what we know in a constant volume process is that the work is equal to zero? Okay. There's no work done on the surrounding, On the surroundings. So the work is zero which also tells us that the change in internal energy delta U. Is equal. Okay, now we're asked about the heat. Okay, so the heat is the quantity Q. How can we find? Well, we know that neon is mono atomic. Hey nan is mono atomic, which tells us that the heat capacity, constant volume is 3/2 are alright. And we can write Q. And we're gonna write Q V. To indicate Q. The heat at constant volume is equal to N. C V delta T. Okay, so now we have an equation that allows us to find the heat at constant volume. And we know and the number of moles were given, we know C V. We've just written it because we know that neon is mono atomic and we know the change in temperature delta T. So QV. Let's give ourselves some more space and we have 0.013 moles. Okay so we're going to use that in just a minute. So we have 0.013 mole Times 3/2 are And then delta T. Well the first temperature is 80 degrees or sorry? The second temperature. Okay, we're going from 17 degrees Celsius to 80 degrees Celsius. So the final temperature is going to be 80 degrees Celsius. We're gonna convert this into Calvin. Okay, so 273.15 Calvin. Okay. And we're going to subtract the initial temperature which we were given us 17 degrees Celsius. So same thing 17 plus to 73.15 Calvin. Well what about our okay, our is our gas constant so we can fill that in 0. more eight times 3/2ves Times are 8.314. And the unit here is jewel Permal Calvin. Okay and then with delta T. We are left with a Delta t. here of 63 Calvin. Okay, so we have unit jewel per mole kilogram times mole. Sorry, mole Calvin times more times Calvin. Okay so the mole and the Calvin will cancel. And we're gonna be left with the unit of jewels which is a unit of heat. We want. Okay so our units check out and we get 10.21375 jewel. Alright so this is our heat at constant volume. Now let's do the same under constant pressure. Okay we want to figure out which one requires more heat. So let's figure out that constant pressure, constant pressure. We have an ice a barrick process. Okay. And we have some information about um the work of an isometric process. But in this problem we're looking for heat. So we don't need to go into that. Okay, so again we're going to use the same equation Q. In this case it's QP constant pressure is N. See and instead of using the or the heat capacity for constant volume, we're going to use the heat capacity at constant pressure. Cp many times DELTA T. And we can treat this as an ideal gas which tells us that C. P. Is equal to Cv. Plus our Okay Cv We found above was three halves are so C. P. Is going to be five halves times are Okay. Alright. So plugging in our values. Let's give ourselves a little bit more room And we get n 0.013 more time. Cp five halves are times delta T. And it's the same temperature change. Okay so we're gonna have 63 Calvin again. Okay, are is oops not five thirds five halves are is 8.314 jewels per mole Calvin. Okay, 63 Calvin. And again the unit of mole and Calvin will cancer. We're going to be left with jewels. We get the heat under constant pressure is 17. to 3 jewels. All right, so comparing the two. What we see is that QP is bigger than QV. Okay, so the constant pressure process requires more heat than the constant volume process. Why are they different? Okay, well let's compare our equations. We use the same end in both and we use the same delta T in both. The only thing that changed was this heat capacity CV or C. P. Okay. And so these are this is bigger because the heat capacities are different. Alright, so let's go back up to our possible answer choices and look at question three as well. Okay. Alright. So we found that the constant pressure process requires more heat. Okay, and it requires more heat because these values are different because heat capacities are different. Okay, so we're looking at answer either B or C. The only thing we have left to do is answer question three. Which is how does the process use the extra heat? Okay, now A and B. We say it says that the heat is used to do work and in three it says one process has greater change in internal energy. Okay, well recall that delta U. The change in internal energy is path independent. Okay, so it doesn't matter whether we're in a constant volume or constant pressure process if we start at the same place and end at the same place, we're going to have the same change in internal energy. Okay, so we don't we don't have um see okay. We don't have options. See this process is going to use the extra heat to do work. Okay. And so we have be the key capacities are different, which causes the key to be different. Okay. The process of constant pressure requires more heat and that extra heat is used to do work. Okay. Thanks everyone for watching. I hope this video helped see you in the next one.

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