0.10 mol of nitrogen gas follow the two processes shown in FIGURE P19.58. How much heat is required for each?

Video transcript

Hey, everyone in this problem, equal samples of 0.25 mole of an ideal diatomic gas are treated through the processes represented below. And we're asked to determine the amount of heat required in each process. We're given a diagram, we have the volume in liters on the X axis, the pressure and atmospheres on the Y axis. And we have three points if and a second F OK. And we have two processes. One is ISOC and where the pressure is decreasing and the volumes remain the same. And another is adopta where both the pressure is decreasing and the volume is increasing. Now, we have four answer choices each containing different values for that heat or adiabatic process and our Isoc Cork process and we're gonna come back to those answer choices as we work through the problem. So let's get started and we're gonna write out some of the information we're given. And the first thing we're told is that we have 0.25 mole. So Anna is gonna be 0.25 mole. And then we're gonna look at each of these processes. So let's start with the adiabatic process. And in an adiabatic process, recall that there is no heat transferred and if there is no heat transferred from the surrounding area, then Q the heat required is just zero. OK. So for audio process Q is just going to be zero. OK. So let's take a look at our answer choices. We're already done. Part of this problem. Option A and option D have that Q for the antibiotic process is 1010 jewels, which is not what we found. So we can eliminate those. Option B and C both have it at zero jewels. So either of those could be the correct answer. We're gonna move on with the second process. Now, in order to figure out which one it is. So now we're looking at the other process and the other process is an iso coric process. OK? So we have an ISOC process. Remember that that means that we have a constant volume and we can see that from the graph as well. This is a vertical line, the volume is not changing. So we have a constant volume. And if we have an ISOC Corp process, the work done is just equal to zero, which tells us that the change in internal energy delta U is gonna be equal to that key Q. So if we want to calculate Q, we can just calculate delta U change in internal energy which we know can be written as N multiplied by CV multiplied by delta T, OK. So the number of moles heat capacity specific or at constant volume and the change in temperature. So there are a couple of things we need to figure out to use this equation. And one of them is this CV value. OK. So we're gonna have to calculate that and the other is the temperature. So we need to know the change in temperature. So let's start as an aside with CV And CV is simple. In this case, we're told this is a diatomic gas because this is a diatomic gas. And it's ideal CV is just going to be five halves multiplied by R and we are the ideal gas constant. OK. So we have CV. Now what about delta T? OK. So for delta T, we know that this is gonna be equal to the final time point T two minus the initial time T one. OK. So temperature at the final time minus temperature initially, how can we do this? Well, from our graph, we know the pressure and the volume uh before the process and after the process, so we can use our ideal gas law. OK. So for T one, what do we have? Well, we know that the pressure P one. OK. Let's go back up to our diagram. The pressure before our process at point I is three atmospheres, the volume is 2 L. After we're done our isoc cort process, the pro the pressure at point F is going to be one atmosphere and the volume is going to be 2 L still. So P one is equal to three atmospheres converting this to our standard unit. This is gonna be three multiplied by 1.01325 multiplied by 10 to the exponent five pascal and the volume is 2 L or two multiplied by 10 to the exponent negative 3 m street. All right. Our I ideal gas law tells us that P one V one is equal to N RT one substituting in our values. Actually, let's rearrange before we substitute in. We want to find the temperature T one and that's gonna be equal to P one V one divided by. And uh and so now if we substitute in our values, we have three multiplied by 1.01325 multiplied by 10 to the exponent five pascals multiplied by two tend to the exponent negative 3 m cubed all divided by 0.25 mil multiplied by 8.31 joules per mole Calvin. And when we work this out, we get a temperature T one of 292 0.6 three 538 health. But so we have T one, we're gonna do the exact same thing for T two. OK. So for T two, what do we have? Well, P two is gonna be one atmosphere or 1.013 25 multiplied by 10 to the exponent five pascals and the volume V two is the exact same as the volume V 1 2 L or two multiplied by 10 to the exponent negative 3 m. Cute. And we're gonna use the exact same equation here. P two V two is equal to it. N RT two for solving for T two which is gonna be P two V two divided by NR and we can substitute in our values 1.01325 multiplied by 10 to the exponent five pascals multiplied by two, multiplied by 10 to the exponent negative 3 m cubed all divided by 0.25 mole multiplied by 8.31. Jules, Verma Alvin. OK. Working this one out on our calculator, we get a value for T two of 97.545. How H our final temperature is smaller than our initial temperature? And the temperature has decreased for this process. So now that we have CV, we have our T values, we can get back to calculating Q and in this case, we're gonna have Q equal to 0.25 mole. OK? Multiplied by that CV value. So five halves multiplied by our ideal gas constant, 8.31 joules per mole Calvin multiplied by delta T and delta T. We're gonna take T two subtract T one and this is gonna give us a value of negative 195.09 Kelvin. When we work all of this out, we get a Q value of negative 1013.25 jewels. Maybe if this negative value, our temperature is decreasing, we're losing heat. And in this question, we really want the magnitude, we want the amount of heat. So we're talking about a magnitude here. So if we take the absolute value of this, we get about 1013 joules of heat and that's it. That's the second part of this um problem. So we take a look at our answer choices. We had narrowed it down to B or C. OK. Option C has that Q should be 1420 which does not match what we found. But option B looks good. It had that for the antibiotic process. Q is zero joules and for the Isoc Cork process, it's 1013 jewels, which is exactly what we found. And so B is the correct answer. Thanks everyone for watching. I hope this video helped see you in the next one.