A cylinder contains 0.100 mol of an ideal monatomic gas. Initiall...

Question

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 * 10^5 Pa and occupies a volume of 2.50 * 10^-3 m^3. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.

Accepted Answer

Hey everyone in this problem. We have a tube filled with 0.15 moles of an ideal mon atomic gas. Okay. With the pressure initial of 1.1 times 10 to the five pascal's the initial volume of gas were given is 0.42 m cubed. The tube expands to a final volume, that is three times the original volume. And the process is idiomatic. Okay. And were asked to determine the final temperature in Calvin and the final pressure. Okay. Alright. So this is an idiomatic process when we think about it, the first thing that comes to mind is no heat transfer. Okay, so Q is equal to zero. Alright, so again, robotic process, we're looking for a temperature and a pressure. So let's recall the equations that we have in an automatic process. Okay, we're looking for temperature. We're looking for pressure and were given information about volume. Okay, so let's recall that we can write the following two equations. We can relate temperature and volume by T one. V one to the gamma minus one is equal to T two. V two to the gamma minus one. And similarly P one V one to the gamma is equal to P two, V two. To the gamma. Okay, and again, the final temperature T two and the final pressure. P two is what we are looking for. Okay. Were given the volume initial and final. Okay. As well as the initial pressure. So let's go ahead and start by finding gamma. We're going to need that in both equations. Okay, now we have a mono atomic gas. Okay, Which means that C. V. It's going to be equal to 3/2 times are all right. And because it's an ideal gas, we have a relationship between Cp and CV where C P is equal to CV. Plus our Okay. Which in this case is going to be 5/2 are. Now let's recall that gamma is the ratio between these values. So gamma cp over CV Just going to be 5/2 are Divided by 3/2 are which gives us a gamma value of 5/3. Okay. Alright, so we have gamma we can use that in our equation Now if we're looking for the temperature, recall that we don't know T1. Okay, so we don't know T one but we do know the volumes. We know gamma. Now, how can we find T. One. Okay, well let's recall that. We can write the ideal gas law. Okay, P V is equal to N. R. T. Okay, so we can relate the pressure, the volume and the temperature and again, we're looking for T one to be able to use in the equation to find T. Two. Okay, so that means we're gonna be worrying about P one and V one as well. Okay, all of the values from that same initial point. Okay, let's give ourselves some more room here And go ahead and find T1 and actually let's go back first. Okay, P one P one. We are given in the problem. 1.1 times 10 to the five pascal's that initial pressure and the initial volume. 0.42 m cubed. Okay so we're going to use those values in our equation. And so we get 1. Times 10 to the five Pascal's Times. m. cute Is equal to end. The number of moles which were told is 0. Are the gas constant? 8.314. The unit here is jewel per mole Calvin Times the temperature T. one that we're looking for in order to find our final temperature. Alright so working this out on the left hand side we get 424.2 we have pascal meters cube. So that's unit jewel on the right hand side. The unit of mobile cancel. We're left with 1. jewels per Calvin times T. one. Okay we divide by the 1.2471. The unit of jewel will cancel. And we're left with T. One is equal to 340. Calvin. Okay so now we have our T. One and we have our gamma value. So we can go ahead And go back up here. We can go ahead and use this relationship between the temperature and volume of an idiomatic process to find T. two. Okay. Alright so let's do that. So we have T. One V. One. The gamma minus one is equal to T two. V two. To the gamma minus one Again. T. one. We just found 340.15 Calvin V. One. We were given in the problem. 0.42 m cubed. Okay. And gamma we found to be five thirds. So we're gonna have five thirds minus one. Okay, this is going to give us two thirds five thirds minus three thirds. That's gonna be two thirds the exponent. Okay. And this is equal to T two times V two. Now we're not given the exact number for V two but we're told that it's three times the initial volume. Okay, So we're gonna have three times 0.42 m. Cute. And that's going to be again to the exponents 5/3 -1. Okay, now we can divide and we're going to end up with T to the unit of meters cube to the exponent. Two thirds will cancel. And we're going to get T. Two is equal to 163. kelvin. Okay. And if we round to the nearest Calvin, this is going to give us 160 for Calvin. Okay, we do that. Rounding that's the format that the possible answers are given in. Okay. And so we have our value of T. two. Okay. no Let's go back up for a second. We found T. two. But the problem is also asking us to find the final pressure. Okay now let's recall we had this relationship between temperature and volume for an idiomatic process. We have the same thing for that pressure and volume relationship. So let's go ahead and use that in order to find our final answer. Okay. And again this is our final temperature that we were looking for. Now we have the relationship P one V one to the exponent gamma is equal to P two V two to the exponent gamma, K. P one. The initial pressure were given 1.1 times 10 to the five pascal's. The initial volume were given 0. m cubed to the exponent gamma which we found to be five thirds is equal to p to that pressure. We're looking for times the final volume which again is three times the initial volume. So three times 0.42 m. Cute. Okay and this is again all to the exponent five thirds. Now if we divide Okay, the units of meters cube to the exponent five thirds is going to cancel. And we're gonna be left with pascal's and we get a pressure P two of 1.62 times 10 to the four pascal's. Okay, so that is the final pressure we were looking for. So if we go back up to our possible answers, we see that we found a temperature of 164 kelvin and a pressure of 1.62 times 10 to the four pascal's. And so we have answer. C. Thanks everyone for watching. I hope this video helped see you in the next one.

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