A heat engine using a diatomic gas follows the cycle shown in ...

Question

A heat engine using a diatomic gas follows the cycle shown in FIGURE P21.55. Its temperature at point 1 is 20℃. Determine W_s, Q, and ∆E_th for each of the three processes in this cycle. Display your results in a table.

Accepted Answer

Hey, everyone in this column, a thermal engine uses helium as a working fluid at point A helium is initially at a temperature of 300 Kelvin. And the temperature at point C is 750. Kelvin. We have a PD PV diagram of the thermal engine cyclic transformation that's shown. And we're asked to calculate the heat Q during the transformation A B. If we look at this diagram, we're given the volume in 10 to the exponent negative 6 m cubed on the X axis and the pressure tend to the exponent five pascals on the Y axis. And we have the transformation from A and we have a volume of 20 pressure of 1.25. OK. We're gonna hold the volume steady, increase the pressure up to 3.75 to get to point B. OK. So that is the transformation we're interested in. We have four answer choices all in jewels, option A 5.6 option B 7.5 option C zero and option D 3.4. So let's start, we're gonna think about again this transformation from A to B. Now we mentioned when we looked at our diagram that the volume was constant. OK. If we have constant volume, then we know that this is an iso coric process when we have an iso process, recall that the work done is just equal to zero. And our change in internal energy delta U is equal to the heat Q. So in order to calculate our heat Q from point A to B, we know that it's equal to the change in internal energy. And we can write this as N multiplied by CV, multiplied by delta T. OK. So the number of moles and multiplied by the heat capacity, constant volume multiplied by the change in temperature. So there's a couple of things here that we need to calculate before we can move on with our calculation. OK. The first thing is that we need to know the number of moles. And next thing we need to find is our heat capacity CV. And then we're gonna need to find delta T and we know the temperature at point A but we don't know the temperature at point B. So we're gonna have to calculate that as well. So let's start with the number of moles. And, and we're gonna find N using the ideal Gospel, we have helium here. So we can treat that as an ideal gas. We can use our ideal gas law. And it's gonna tell us that the pressure multiplied by the volume is equal to the number of moles. And multiplied by the gas constant or multiplied by the temperature. Now, PV and T we've put subscripts of A through. So we're looking at point A, we know the pressure and the volume, we can get those from our graph. We were given the temperature in the question, we know the value of R that gas constant. So we can use this to calculate N and N is going to be equal to P A VA divided by RT. Let's go ahead and substitute in our values. We have uh it's going to be equal to 1.25 multiplied by 10 to the exponent five pascal. Again, taking this value from the graph multiplied by 20 multiplied by 10 to the exponent negative 6 m cubed. Again, taking that value from the graph, this is all divided by gas constant 8.314 five jewels. Her more Calvin multiplied by the temperature at point A which were given as 300 Kelvin. Now, the unit of Kelvin is gonna divide out in the denominator. We have past K meters cubed in the numerator which is equivalent to a jewel, right? So the unit of jewels will divide out. We're gonna be left with the unit of mole, which is exactly what we want. So N is going to be equal to 0.001 mole. All right. So we have the number of moles. And next thing we needed to find was see the heat capacity at constant volume. So let's look at that. Now, in this case, we're dealing with helium and we know that helium is a mono atomic gas. And since it is a mono atomic gas, that tells us that our heat capacity at constant volume CV is going to be three halves multiplied by the gas constant or all right. So we have CB, the last thing we needed to calculate was that change in temperature delta T. And what we really need to find in order to calculate that change in temperature is the temperature at point B. So let's find the temperature at point B. Now again, we're gonna use the ideal gas law. We have that PB VB is equal to N RT. So exact same as when we were finding it. But now we're looking at point B instead of point A and we wanna isolate for the temperature. So TB is going to be equal to PB VB divided by NR. Now, we can substitute all of the numbers that we have in here. We can also rearrange this in terms of the end value we calculated. And so if we do that, we get that our temperature TV is equal to PB VB divided by N which we know is P A VA divided by RT A multiplied by A. Now the RS are going to divide up, remember that we have constant volume. So the volume VA and BB are the same and they're gonna divide out as well. And so we get our temperature at point B is going to be PB divided by P A multiplied by T A. Now, doing it this way allows us to avoid a little bit of round off error, OK. We calculated the value and so there's a little bit of round off error that exists in that value if we write it the way we just did. Now, we're using values that we were given and we can avoid a little bit of extra round off error. All right. So we have our temperature at point B PB divided by P A. Well, this is gonna be 3.75 multiplied by 10 to the exponent five pass scales divided by 1.25 multiplied by 10 to the exponent five pass by both of those values coming from that diagram multiplied by 300 Kelvin. The temperature we were given. And so we get our temperature at point BC B is going to be 900 Kelvin. So we have that nice exact value for that temperature. OK. So we can put it all together. Now, remember we're looking for the heat during this process, you, we know that it's gonna be N CV delta T, we've calculated all of the pieces that we need. So we can get back to this equation and find that heat. So cute again and multiplied by CV multiplied by delta T. This is gonna be equal to N which is zero 0.001 mole multiplied by CV, three halves multiplied by 8.31 45 jewels for more Calvin multiplied by delta T. OK. So that's gonna be the final temperature minus the initial temperature. The final temperature we calculated was 900 Kelvin. And then we're subtracting that initial temperature of 300 Kelvin. So we have 0.001 multiplied by three halves, multiplied by 8.3145 jewels per M mole. Calvin multiplied by 900 Calvin minus 300 Kelvin. When we work this out, we get that our heat Q for this process from A to B is going to be equal to about 7.48 looking at units, the unit of mole will divide out, the unit of Calvin will divide out and we are just left with jewels, which is exactly what we want. So we have that the quantity of heat qab is about 7.48 joules. If we go back to our answer choices and we round this to two significant digits, we can see that the correct answer is going to be option BQAB is equal to 7.5 joules. Thanks everyone for watching. I hope this video helped see you in the next one.

A heat engine using a diatomic gas follows the cycle shown in FIGURE P21.55. Its temperature at point 1 is 20℃. Determine W_s, Q, and ∆E_th for each of the three processes in this cycle. Display your results in a table.

Key Concepts

Heat Engine

First Law of Thermodynamics

Diatomic Gas Properties

Watch next

A heat engine using a diatomic gas follows the cycle shown in FIGURE P21.55. Its temperature at point 1 is 20℃. Determine Ws, Q, and ∆Eth for each of the three processes in this cycle. Display your results in a table.

Key Concepts

Heat Engine

First Law of Thermodynamics

Diatomic Gas Properties

Watch next

A heat engine using a diatomic gas follows the cycle shown in FIGURE P21.55. Its temperature at point 1 is 20℃. Determine W_s, Q, and ∆E_th for each of the three processes in this cycle. Display your results in a table.