Five moles of monatomic ideal gas have initial pressure 2.50 * 10...

Question

Five moles of monatomic ideal gas have initial pressure 2.50 * 10^3 Pa and initial volume 2.10 m^3 . While undergoing an adiabatic expansion, the gas does 1480 J of work. What is the final pressure of the gas after the expansion?

Accepted Answer

Hey everyone in this problem. We have a container fitted with a piston that contains 6.5 moles of an ideal mono atomic gas. The original pressure and volume are 3.2 times 10 to the four pascal's and 40.85 m cubed respectively. The gas is allowed to expand dramatically and it does 1640 jewels of work and were asked to calculate the final pressure following the expansion. Okay, so let's just start by writing out all of the information we're given. They were given a lot of information this problem. So let's just write out what we're given. So we have n the number of moles Is 6.5. We were given the initial pressure of 3.2 times 10 to the four. So p one is 3.2 times 10 to the four pascal's. We were also given the initial volume case of V 1.85 m cubed. And we were given the work, the work W is equal to 1640 jewels. Now we're told that this is an ideal mono atomic gas. So that's gonna come into play in a little bit. And we're also told that this is an idiomatic process. Okay, so when we're told information about what type of process we have, we want to start there because that's going to dictate which equations we use. Okay, so this is idiomatic. And what that means is we have no heat transfer. Okay, recall idea, biotic process. We have no heat transfer which tells us that Q. A zero and it also means that we can write delta U. Is equal to negative W Okay, because cuba zero. So now because we have W given in our equation, that means we also know delta U. Okay, so we can use equations with either of those because we have that value. Now, we're looking for the final pressure. We have a few ways to look at final pressure. Okay? We can look at final pressure. In terms of the ideal gas equation, we're talking about an ideal gas. In that case we're going to need to know the final volume and the final temperature were not given any information about the final volume or the final temperature. Okay, so let's start out by trying to find more information about the temperature and the volume and work her way up to information about the pressure. So, let's recall that we can write the following delta U. Is equal to N C V. Delta T. Okay, well, we have delta you because we know w And we know that it's an automatic process we have and the number of moles cv. We can find because we know what type of gas we have. Okay, this is going to give us information about the change in temperature. Okay, so that's a good start. Okay, so let's start with C V. Okay, we have an ideal gas and it's mono atomic. Okay, so because it's mono atomic. We know that C V. It's going to be equal to 3/2 are okay And then cp because this is an ideal gas, it's just going to be equal to cv Plus our So those heat capacities are related through this formula and that gives us 5/2ves april Right now. We just need C. V. So we have that. Let's go ahead. Okay, Delta U. Well that's negative. W. So we get negative 1640 jewels on the left hand side is equal to N 6.5 more Time. 3/2 are times delta T. All right now Filling in some more information. We know what our is our is our gas constant. So we get 6.5 more times. Three halves times are the gas constant. 8.314. And the unit there is jewel Permal Calvin times, delta T. And this is going to give us a delta T. Okay, we have jewel, we're going to divide to the unit of jewel will cancel the unit of mole cancels on the right hand side. We're gonna be left with the unit of Calvin which is perfect. We're looking at a temperature so that makes sense. Our units check out And we get Delta T. is equal to negative 20. caliber. Okay. Alright, well we know delta T. And again we're trying to find some information about the temperature and the volume after the expansion. That's going to allow us to find the pressure after the expansion. Okay, so we know delta T. But we don't know that final temperature. Okay. If we know delta T. We want to find the final temperature. Well we can do that if we have the initial temperature. Now let's recall. We have an ideal gas. So we can use the ideal gas equation. Okay, so let's give ourselves some more room. We're gonna work with the ideal gas equation which tells us that P. V. Is equal to N. R. T. Okay, so this relates the pressure, the volume and the temperature of our gas. Now we're looking at the first time point. Okay so we're gonna have P. One V. One And T. one. Okay, so before the expansion now we were given an initial pressure of 3.2 Times 10 to the four pascal's Okay. an initial volume of 0. meters cubed. We were given the number of moles. n. 6.5 R. S. R. Gas constant. 8.314 joules per mole Calvin. Okay. And then we have our temperature T. one. Alright, if we divide so we're gonna divide um by 6.5 times 8.314. The unit of mole will cancel here. And then when we divide the pascal meters, cute. That's going to be a jewel. Okay, so that unit will cancel. We're gonna be left with the unit of Calvin which is what we want for the temperature. We get a temperature T. One. You go to 503.3 to Calvin. So we have delta T. We have T. One. So now we can find T. Two like we wanted. Okay. So let's just do that over here. On the right hand side, delta T. Is equal to T two minus T. One. Okay, so negative 20.232 Calvin. K. A. Negative delta T. Means that our temperature decreased. T 2 -503 Calvin gives us a temperature T two of 483 point 088 Calvin. Okay. Alright. So we're getting closer we have a final temperature. Okay and again, we're trying to find the final pressure. We're going to do that through the ideal gas equation but in order to find the final pressure through that equation, we need the final temperature and the final volume. Okay. We found the final temperature. Now let's do the volume. Okay. And let's recall because we have an idiomatic process, we have the following relationship between temperature and volume two. U 1. V 1 to the gamma -1 Is equal to T two. V 2 to the game of -1. Okay. All right, what is this? Gamma gamma is going to be The ratio of the heat capacities. Now we found our heat capacities above C. V. And c p. And so gamma is just gonna be the ratio of the two. Gamma is equal to C. P over C. V. And in this case this is just gonna be 5/3. Yes we have our game of value. We have T. One and V. One and we have T. Two. Okay, so this is going to allow us to find V. Two. Alright, so T one we found to be 0.32 Calvin kv one was given in the problem. 0.85 m cubed. Okay, gamma. We found by knowing that this was an ideal mono atomic gas. So we get five thirds minus one K. T two. We found 483. Calvin. And finally the two that value we're looking for to the exponent given -15/3 -1. Alright, so if we divide by 483.88 Calvin and work out that left hand side. We're going to get V two to the exponent two thirds. Okay five third minus one is two thirds Is equal to 0.9349. All right now, in order to get rid of this exponent two thirds, what we need to do is take the exponent three halves. Okay, when you take something to the exponents, you multiply the exponents. So we get two thirds times three halves. We get an exponent one on V two Which is what we want. And on the right hand side we get 0. 3953 K. And r. Units are going to be meters cute when we divided the unit of Calvin cancel. And so we had unit meters cubed to the two thirds and then we end up with meters cubed when we take that. Okay, we're getting closer, we have our final temperature, we have our final volume. Now we can switch over to the ideal gas equation to find the final pressure we're looking for. Okay, one more time. We'll give ourselves a little bit more room to work here. Now recall the ideal gas equation. P V is equal to N R T K. We used it above to find one of the temperatures. Now in this case we're looking at the final state. So after the expansion and so we're going to consider P. Two V two and T. Two. So we end up with P. two, which is what we're trying to find times V two which we just found.  aN:aN:000NaN 0. m cubed is equal to end 6.5 more times Are the gas constant? 8.314 joules per mole Calvin Okay, Times T two. Which we found to be 483.088. Alright. And if we work this out we're gonna divide by the 0.903. The unit of meters cubed. And sorry, this should be meters cubed. This unit on this final volume Calvin. Sorry, temperature Calvin. There we go. Okay so the unit of Calvin will cancel. The unit of mold will cancel. And we're gonna have jewel divided by meters. Cute. Which is gonna give us a unit of pascal. Okay so we end up with pressure P. 2 28, 20.44 pascal's. We write this in scientific notation. We get 2.89 Times 10 to the four Pascal's. And so this is the final pressure. The pressure after the expansion that we were looking for. And if we go back up to our answer choices, this is going to correspond with answer. A thanks everyone for watching. I hope this video helped see you in the next one.

Five moles of monatomic ideal gas have initial pressure 2.50 * 10^3 Pa and initial volume 2.10 m^3 . While undergoing an adiabatic expansion, the gas does 1480 J of work. What is the final pressure of the gas after the expansion?

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