The working substance of a certain Carnot engine is 1.0 mol of...

Question

The working substance of a certain Carnot engine is 1.0 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine’s cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 6.2. The work output of the engine is 920 J in each cycle. Compute the temperatures of the two reservoirs between which this engine operates.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A 1.0 mole of helium gas power a Carno engine in the isothermal expansion phase, the gas volume doubles during the subsequent adiabatic expansion. The volume expands to 5.5 times its original size. Each cycle of the engine delivers a work output of 1000 jewels, determine the temperatures of the two reservoirs that the engine operates between. So that's our end goal. Ultimately, we're trying to figure out what the temperatures of these two reservoirs that the engine operates between. And those are the two values that we're ultimately trying to solve for. So what are these two temperature values that this engine operates between? So now that we're trying to or now that we know that we're trying to figure out what the temperature of the two reservoirs are that are in this engine that are, they're operating between. Let's read off our multiple choice answers to see what our final answers for th and TL will be noting that they're all gonna be in the same units of Kelvin. So A is 150 75. B is 150 82. C is 260 82 and D is 260 75. So first off, let us create a PV diagram to help us better visualize this problem. So looking at our PV diagram here, our Y access will be denoted by P, the pressure and our X axe will be denoted by V the volume. And we will call our highest point on our graph here to be A and A goes down to B and then B to C and then C to D where D is facing towards the origin point. Awesome. And the temperature of th will be from point A to point B and TL will be from point C to D. And let us quickly note that inside of our, it's sort, it's sort of shaped like a parallelogram, but it's like stretched out on the inside indicated in red. I'm going around here. So the inside of our graph will be W the work. Awesome. So this is the work I'll draw a little arrow. So this is the work. So the whatever is highlighted in red is the work and to make things even more simple will go around in blue. So it's gonna go um So it's 123 and four. So that's the step. So it's gonna go in this motion from A to BC to D. So 1234, so it's gonna be going in that flow, that system flow. So with that in mind, we can now discuss the carnal cycle stages. So our first stage, which is what we have for 123 and four. So those are our stages. So for the first stage, it's an isothermal expansion from A to B. So we need to recall that the gas expands isotherm at the high temperature th and the volume doubles from VA to VB. So let's to keep it all straight. Let's do our two. Let's write down all of our states and fill in our key pieces of information for each stage. So talking about stage one. So this is A to be and it's at the high temperature th and the volume doubles from VA to VB, which means that VP is equal to two multiplied by VA Awesome. So our second stage is the adiabatic expansion from B to C. And we need to recall that the gas will expand idiomatically from VB two VC and with the volume increasing by a factor of 5.5. So that will mean that VC is equal to 5.5 multiplied by VB Awesome. And our third stage is an isothermal compression from C to D. And the, we need to recall that the gas will be compressed isotherm at the low temperature TL and the volume will decrease from VC to VD. And finally, for our fourth stage, there will be an adiabatic compression from D to A. And we need to recall that the gas will be compressed adiabatic back to its initial state. So, so, all right, initial state. Fantastic. So now moving along here, we need to note that we are given that the number of moles, which we're gonna denote as N is equal to 1.0 moles. And we're also told that the work output which we're gonna call capital W is equal to 1000 jewels. And we're also should be able to recall that capital R which is the universal gas constant is equal to 8.314 and its units will be joules per mole multiplied by Calvin. Awesome. So we need to recall and note that the heat input during the isothermal expansion in order to solve for this or considering this, we need to recall and note that the heat input qh during the isothermal expansion at temperature, th will be as we should be able to recall that qh is equal to N multiplied by R multiplied by capital T where T is the temperature and specifically, th multiplied by the natural log of VB divided by VA. And we need to note, and let's make a little side note here after the side that sense VB is equal to two multiplied by va we could write that Qh is equal to N multiplied by R multiplied by th, multiplied by the natural log of two. Fantastic. So now we need to recall and use the equation to help us solve for the efficiency of a Carno engine, which we're gonna use E to denote the car, the efficiency and the efficiency E is equal to one minus TL divided by th which is also equal to W divided by Qh. So substituting in our value for Qh, we will determine that we can write, so we can write that and we're plugging in our known values. So we can write that. Qh, so let's do a little arrow here. So Qh is equal to and multiplied by R multiplied by th multiplied by the natural log of two. So plugging in this value, we will get that E is equal to one minus TL divided by T which we simplify rearrange this equation to isolate and solve for the work. We will find that the work is equal to drum roll. N multiplied by R multiplied by TH minus TL multiplied by the natural log of two. Awesome. Or we could simply write that th so th minus TL is all equal to W where, once again, W is our work divided by N multiplied by R multiplied by the natural log of two. Awesome. So now we consider the adiabatic process. So considering the adiabatic process, we need to recall that the pressure multiplied by the volume B so V beat to the power of gamma is equal to P multiplied by VC to the power of gamma, which we can expand this to write that N multiplied by R multiplied by th divided by VB multiplied by V bmmvb to the power of gamma is equal to N multiplied by R multiplied by TL divided by VC multiplied by VC to the power of gamma, which we can rearrange this equation to solve for th by itself. Or we can say that T is equal to T multiplied by VC divided by VB. So VC divided by VB all to the power of gamma minus one. And we need to quickly recall and let's write this as a note off to the side. Let us note that gamma for a mono atomic gas will mean that gamma is equal to five thirds. So we can write that T is equal to. So th is equal to TL multiplied by. And this is plugging in her known values 5.5 to the power of when we simplify. So basically we took gamma minus one and when we take gamma minus one, we'll get 5.5 to the power of two thirds. Awesome. So that will mean when we solve for TH and TL, we will get that TH minus TL is equal to plugging in our known variables. 1000 jewels divided by 1.0 no multiplied by eight point 314. And don't forget that our units will be jewels per move multiplied by Calvin multiplied by. And don't forget that we're multiplying by the natural log of two. Awesome. So this will mean that this is all equal to TL multiplied by five 0.5 to the power of two thirds minus TL. Which will mean that when we plug this into a calculator, we need rearrange this equation to isolate and just solve for TL by itself, we will find that TL is approximately equal to, let me plug it into our calculator 82 Calvin. Awesome. So now solving for th we need to take and note that when we initially type in for TL into our calculator, we will get 82.01 Calvin, but we just rounded to the nearest phone number. But let's use this expanded number to get a more accurate answer. So we need to take 82 0.01 and we need to and don't forget that the units are Calvin and we need to multiply this value by 5.5 and we need to take it to the power of two thirds. And when we plug into our calculator and we round to the nearest full number, we will get 260 Kelvin or specifically that it's approximately equal to 260 Kelvin. So using our approximate symbol and that's it, that's our final answer. So our two final answers for TL is 82 Calvin and for our th value it's equal to 260 Calvin. So looking at our multiple choice answers, the correct answer has to be multiple choice answer letter CTH equals 260. Kelvin and TL is equal to 82. Kelvin. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Carnot Engine

Ideal Gas Law

Work Done by a Gas

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