The heat engine shown in FIGURE P21.62 uses 2.0 mol of a monatomic gas as the working substance.

c. What is the engine's thermal efficiency?

Question

The heat engine shown in FIGURE P21.62 uses 2.0 mol of a monatomic gas as the working substance.

c. What is the engine's thermal efficiency?

Accepted Answer

Hey, everyone. So this problem is dealing with heat engines. Let's see what it's asking us. We have a thermal engine that undergoes cyclic process that consists of three steps and utilizes 1.5 moles of neon gas as its operating fluid. The engine's PV diagram is given below. And we're asked to determine the engine's efficiency. Our multiple choice answers are a 2.1%. B 3.3% C 7.6% or D 8.9%. So we're asked to determine the engine's efficiency. We can recall that the equation for efficiency is given by beta is equal to our network divided by our exhausted heat. We can recall that when we're using a PV diagram, our total work is the area under the curve and our network is the area enclosed. So first we can solve for our network. So that will be the area enclosed in this right handed or right angled triangle. And so we can recall the equation for the area of a triangle is one half multiplied by the base multiplied by the height. And when we look at this diagram, our base is going from 200 to 510 to the negative 5 m cued. So that is a distance or a length of the base of 300 times 10 to the negative 5 m cubed. And then our height is going from 1.25 to 3. past gap times 10 to the five past gals. That's on our Y axis. So 1.25 to 3.75 is 2.5. So it'll be 2.5 times 10 to the five past gaps. And when we plug that into our calculator, we get 375 jules. So that is our network. Next, we have to find our Q sub age or our exhausted heat. We can recall the equation for heat is equal to our total work plus our change in thermal energy and our total work is equal to the area under the curve and are like that better. Delta E is equal to and our number of moles multiplied by CV, our specific heat when we have a constant volume multiplied by our delta T or our change in temperature. And for an ideal gas, which we're working with our CV is equal to three halves multiplied by R or gas constant. And so we're going to find Q for each of these three steps from A to BB to C and then C back to A and we'll use that to find our exhaust heat. So for Q from A to B, we are dealing with an iso coric process, which means we have a constant volume. So we can recall that with a constant volume, our work is zero. So we have zero for work plus our number of moles. So that's 1.5 from the problem multiplied by three halves, multiplied by our gas constant, which we can recall is 8. Jews per Kelvin mole then multiplied by our delta T. So from the diagram, we can see that the temperature at A is Kelvin and we're going to be the temperature of B is Kelvin. So our delta T is 900 minus or 600. Kelvin. We plug that into our calculator and we get 11,224 jewels. And so next, we'll go from B to C. And so the area under the um line from B to C is going to be the area of the triangle which we've already sold for, that's W net plus the area of this rectangle that goes from the bottom of the triangle down to Rx axis. And so we can recall that the area of a rectangle is simply the base multiplied by the height. So that base we've already determined is times 10 to the negative 5 m cubed multiplied by the height. And so this height is going from 0 to 1.25. So that's 1.25. And the unit on that X er, excuse me, on that Y axis is 10 to the five pascals plus the area of the triangle which we've already determined is 375 jewels and then plus our delta E term. And so that will be 1.5 moles multiplied by three halves, multiplied by 8.314 jules per Kelvin mole multiplied by our delta T, our delta T from B to C. So B we have a temperature of 900 Kelvin. And at C, we have a temperature of 750 Kelvin. So our delta T is a negative 150 K. We plug all of that into our calculator and we get a heat from B to C of negative 2056 jewels. And lastly, we'll go from C to A and now the area under the line from C to A is simply the area of that rectangle that we found before. So that will be 300 times 10 to the negative 5 m cubed multiplied by 1.25 times 10 to the five pascals plus our delta E term. So that's 1.5 moles multiplied by three halves, multiplied by 8.314 joules per Kelvin mole and multiplied by our delta T. Now again, our delta or our temperature at point C 750 Kelvin, our temperature at point A, it's 300 Kelvin. And so our delta T is a negative Kelvin. And that gives us a, a heat from C to a of negative 8043 jewels. So how do we use each of these individual heats to find our exhausted heat? We can recall that our exhausted heat must be greater than zero. So it's heat that is exhausted into the environment, which means it has to be positive value. And so the only one of these three individual heats that's positive is this Q from A to B. And so our Q of age is equal to 11,000 224 jewels. Now, if we had multiple positive heats, we would sum those to find our exhausted heat. And so now we have everything we need to find our efficiency denoted by a our network, 375 jewels divided by our of age 11,224 jewels. And that equals 3.3%. And so that is the final answer for this problem. And when we look at our multiple choice answers that aligns with answer choice B. So that's all we have for this one. We'll see you in the next video.

The heat engine shown in FIGURE P21.62 uses 2.0 mol of a monatomic gas as the working substance. c. What is the engine's thermal efficiency?

Verified Solution

Key Concepts

Thermal Efficiency

First Law of Thermodynamics

Monatomic Gas Behavior