FIGURE P19.62 shows a thermodynamic process followed by 120 mg of helium.

c. How much heat energy is transferred to or from the gas during each of the three segments?

Question

FIGURE P19.62 shows a thermodynamic process followed by 120 mg of helium.

c. How much heat energy is transferred to or from the gas during each of the three segments?

Accepted Answer

Hey, everyone in this problem, 0.2 g of neon is treated through the thermodynamic paths that are shown below. And we're asked to calculate the net heat energy that flows into or out of the gas in the three step process. So the diagram we're given has the volume in liters on the X axis, the pressure on the Y axis and we have three points. OK. So going from A to B, we have constant pressure but the volume increases, going from point B to C, the volume is decreasing, the pressure is increasing. And we get this curve and we're told this isothermal and then going from C to A, we have constant volume but the pressure is decreasing. We're told that the temperature at point A is 688 Kelvin and at point C is 1720. Kelvin. We're given four answer choices all in jewels. Option A negative 45 option B 570 option C 45 and option D negative 308. So what this problem is asking us to find is this net heat energy Q net and that's just gonna be made up of the heat energy from each of these processes on their own. So it's gonna be Q from A to B, post Q from B to C plus Q from C to A. Now, we're told that we have 0.2 g of neon. We know that in all of our calculations, we're gonna need to know the number of moles. So let's get that out of the way, the number of moles and it's just gonna be equal to the mass M divided by the molar mass capital M. In this case, we have 0.2 g and we have neon. And so the molar mass, you can look up in a table in your textbook for neon is going to be 20.18 g per mole. And so we get zero 0.00 991 four. OK. So we have that value, let's keep going. Now the pressures are gonna cause a little bit of an issue. If we take a look at our diagram, we don't have any values for the pressures. OK. We're given PC on the graph and then the pressure at point A and B we're told is 0.4 PC. So what we're gonna do is we're gonna start by finding PC. OK. So let's find PC. We're gonna need that in our calculations as well. So finding PC, we can use the ideal gas law recall that PV is equal to N RT, OK. Because we have neon, we can treat it as an ideal gas and we're using subscript C to indicate at point C here. Now, the pressure at point C, well, that's just PC, that value we're looking for the volume is 2 L. We wanna convert to our standard unit. So we take that two, we're gonna multiply it by 10 to the exponent negative three to convert to meters cubed. This is gonna be equal to the moles, 0.00991 moles multiplied by R the ideal gas constant 38.3145 joules per mole Calvin multiplied by the temperature at point C which we're told is 1720. Kelvin on our diagram. Now to solve for PC, we're gonna divide by two multiplied by 10 to the exponent negative 3 m cubed on both sides. And when we put that into our calculator, we get that the pressure at point C is gonna be about 70,861 0.1577 has scales. All right. So we have that pressure PC that allows us to read these values off of our graph. Now, let's get started. We're gonna take a look at each process one by one starting with the process from A to B. OK. So from A to B, what do we have? Well, the volume is changing but the pressure is staying constant, right? If we have constant pressure, then our call that we can call this a isobaric process. OK. This is an isobaric process. And when we have an iso process, the heat is given by the following. OK. We're gonna have QE from A to B equal to the number of mole N multiplied by CP. So the heat capacity at constant pressure multiplied by delta T the change in temperature now, and we know delta T, we can calculate, OK, we know the temperature at A and the temperature at B we actually know as well. OK. We're told that the path from B to C is isothermal. That means that the temperature stays the same, the temperature is constant. So that means the temperature at point B must be the same as the temperature at point C. And so the temperature at B is going to be 1720 Kelvin. OK. So the only thing in this equation, we don't know yet is that CP value, let's go ahead and find that now and we can do that because we know what gas we're dealing. OK. We're dealing with neon. Neon is a mono atomic gas. And so CV, the heat capacity at constant volume is gonna be equal to three halves multiplied by the ideal gas constant. And again, because we're dealing with a mono atomic gas now, because we have an ideal gas or we can consider it ideal CP, the value we're looking for, it's just gonna be that CV value plus R three halves, R plus R gives us five halves R. All right. So we have now everything we need to calculate qab. Let's go back to our equation and seven, our values. So N 0.00991 more multiplied by CP five paths multiplied by R which is 8.3145 jewels for more go then multiplied by delta T 1720 Calvin minus 688 Kelvin. When we work all of this out, we get that Q from A to B as equal to 212.5835 jewels. All right, we're gonna put a red box around this because at the end, we're gonna need to add all of these values up. So we don't wanna lose track of it. So we're done with one process now and now we're gonna keep going. We're gonna go to the next one and the next one is gonna be this isothermal process from B to C. OK. So from B to C, we're told this is an isothermal process. So we have constant temperature. And when we have an isothermal process, the quantity of heat Q so Q from B to C is just gonna be equal to the work which we can write as N multiplied by R multiplied by T multiplied by lawn of the volume at the second point at the end point. So VC divided by the initial volume, which is VB. In this case, we know all of these values N RT and both of these volumes are given on our diagram or we've already calculated them. And so we can substitute those in Q BC is going to be equal to 0.00991 no multiplied by 8.3145 jewels per mole. Calvin multiplied by 1720. Calvin multiplied by one of 2 L divided by 5 L. Now, typically in these calculations, we convert our leaders to our standard unit of meters cubed. But in this case, because we're just dividing them by each other, the units are gonna cancel it. OK? So doing that conversion is just an extra step. We don't need to do the units are gonna cancel out and we're gonna get this ratio 2/5 inside of our lawn, which is everything we need. Now, if we work this out, Q BC is going to be equal to negative 129.86 joules. OK. So that's for that second process. So we've got two down and one to go. And the last one is gonna be this process from C to A. So let's go back to our diagram and look at our process from C to A and we can see that the volume is remaining constant and the pressure is changing. So we're gonna work this out from C to A and we have a constant volume and when we have constant volume, the quantity of heat Q so Q from C to A just going to be equal to N DB delta T. So it's almost the exact same as the quantity of heat from A to B when we dealt with the constant pressure situation. But now, instead of using the heat capacity at constant pressure, we use it at constant volume. Substituting in our values here, 0.00991 more multiplied by three halves, multiplied by 8.3145 joules per mole multiplied by 688 Kelvin minus 1720 cup. OK? And be very careful when you're doing your change in temperature that you've got it going the right way we take the final temperature and subtract the initial temperature. So in this case, delta T is going to be negative and that's gonna give us a negative Q value. So QC A is going to be equal to when we work all of this out negative 127 0.55 jus. All right. So we're done with three processes. Now we get back to calculating cube net. Remember that Q net is just going to be all three of them added together. So let's go ahead and do that. Q net is going to be equal to 212.5835 jewels minus 129.86 jewels minus 127.55 jewels. When we work all of this out, we get that, that net heat that we're interested in finding is gonna be negative 44 8265 jewels approximately. And that is the final answer we were looking for. If we round to the nearest jewel, what we'll find is that our answer corresponds with option a negative 45 jewels. Thanks everyone for watching. I hope this video helped you in the next one.

FIGURE P19.62 shows a thermodynamic process followed by 120 mg of helium. c. How much heat energy is transferred to or from the gas during each of the three segments?

Verified Solution

Key Concepts

Thermodynamic Processes

Heat Transfer

First Law of Thermodynamics