A heat engine with 0.20 mol of a monatomic ideal gas initially fills a 2000 cm³ cylinder at 600 K. The gas goes through the following closed cycle:

Isothermal expansion to 4000 cm³.
Isochoric cooling to 300 K.
Isothermal compression to 2000 cm³.
Isochoric heating to 600 K.

How much work does this engine do per cycle and what is its thermal efficiency?

Question

A heat engine with 0.20 mol of a monatomic ideal gas initially fills a 2000 cm³ cylinder at 600 K. The gas goes through the following closed cycle:

Isothermal expansion to 4000 cm³.
Isochoric cooling to 300 K.
Isothermal compression to 2000 cm³.
Isochoric heating to 600 K.

How much work does this engine do per cycle and what is its thermal efficiency?

Accepted Answer

Hey, everyone. So this problem is dealing with thermal engines in the second law of thermodynamics. We're going to utilize a PV diagram and I will be drawing that as we work through the problem text. So we have a lab experiment where we have 0.3 moles of helium used as the working fluid. The helium undergoes isothermal expansion from an initial volume of 2.5 times 10 to the negative three cubic meters to 7.5. So we'll mark that our volume is Rx axis. And so we'll say show that this is in units of 10 to the negative three cubic meters. And isothermal expansion, we know that we are increasing in volume. If we're increasing in volume and keeping our temperature constant, then we are decreasing in pressure. So from this first step, we'll call it from A to B. That's what that first um curve looks like on our PB diagram. All right. And that's happening at a temperature of 400 degrees Kelvin. Next, we have an ISOC co cooling to 320 degrees Kelvin. And so when we don't change our volume, but continue to um reduce our temperature our pressure again will drop. And then we have isothermal compression back to the volume of 2.5 times 10 to the negative 3 m cubed. And finally iso choric heating. So again, iso coric no change in volume, which means we have an increase in pressure were asked to calculate the work done in one cycle and then the thermal efficiency of that engine. And so the total work done by this engine is going to be the work done at each of these steps on our diagram. So the work from A to B plus the work from B to C plus the work from C to D plus the work from D to A for an ideal gasp which helium is for an isothermal transition, we have, our work is equal to N RT multiplied by the Ln of V two divided by V one from B to C RV two and V one are the same. So isoc coric processes will have V two is equal to V one, that's the definition of iso coric. And so the Ln of one is equal to zero and therefore the work is equal to zero. And so for our two isoc Coric processes, so from B to C and from D to A, those both go to zero. And then for our isothermal processes, we can use this equation and actually plug in um all of the values to find our work. So work from A to B is equal to N. That's our number of moles that was given as 0.3 in the problem multiplied by our, that's our gas constant. We can recall 8.314 oops, sorry, 8.314 jules per Kelvin multiplied by mole multiplied by tr temperature of 400 Kelvin and then multiplied by the Ln of RV 27.5 times 10 to the negative 3 m cubed divided by V 12.5 times 10 to the negative 3 m cubed. We plug that in to our equation or into our calculator. And we get a work from A to B of 1096 jewels. And so similarly, the work from C to D, it's going to have the same molar. The number of moles 0.3 gas constant is a constant. Our temperature from C to D is 320 Kelvin. And then RV, one and V two are flipped because from C to D, we're going from 7.5 to 2.5. So B two is 2.5 times 10 to the negative three and B one is 7.5 times 10 to the negative three. And that's in units of meters cubed, you plug that in and we get negative 877 Jews. And so our total work for the cycle is equal to 1096 jewels minus 877 jewels, which is equal to 219 jewels. So that is the answer to the first part of our problem. For the second part of our problem, we're asked to find our efficiency of the engine. So we can recall that our efficiency denoted by A A is equal to our network divided by RQ sub H. So we found our work, but now we need to find our Q of H or exhaust heat. And so we can recall that Q is equal to our work plus NC sub V delta T. So for our isothermal processes where we have no delta T Q is equal to work in for our iso choric processes where we have no network, Q is equal to NC sub V delta T. So for each of these steps, similarly to how we did, how we solve for work, we'll have Q from A to BQ, from B to CQ, from C to D and Q from D back to A. So for our isothermal processes, we can simply plug those values in. So Q A to B is 1096 joules and then Q from C to D was negative 877 joules. So those were already calculated. And then for our isoc choric processes, we will solve for that exhausted heat. So N is still 0.3 moles or C sub B, we can recall for mono atomic Adams is three halves multiplied by R. So that's three halves times 8.314 jewels per Calvin mole. And then our delta T from B to C is negative 80 degrees Kelvin, we plug that in to our calculators and we get negative 299 jewels. And then our exhausted heat from D to A is the same 0.3 moles multiplied by three halves multiplied by 8.314 jewels per Kelvin mole. And then the delta T from D to A is a positive 80 degrees Kelvin. And that comes out to positive 299 Joel. So the last trick here is we are not summing all of these, the heat exhausted is just the positive values. So we ignore the negative values. And so our Q sub H is 1096 or 1096 jewels plus 299 jewels or 1395 jewels. Put that back into our efficiency equation and we have our network of 219 jewels divided by our exhausted heat of 1395 jewels. And we get in efficiency of 16%. And so when we look at our multiple choice answers for this problem, we can see that, that aligns with answer choice A 219 joules with a 16% efficiency. That's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Ideal Gas Law

Thermodynamic Processes

Thermal Efficiency