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

b. Make a table that shows ∆Eₜₕ, Wₛ, and Q for each of the three processes.

Question

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

b. Make a table that shows ∆Eₜₕ, Wₛ, and Q for each of the three processes.

Accepted Answer

Hey, everyone. So this problem is dealing with PV diagrams and the second law of thermodynamics. Let's see what it's asking us. A heat engine undergoes a cyclic process consisting of three steps A to BB to C and C to A. The engine utilizes one mole of helium as its working fluid. We're asked to calculate the net change in internal energy, the net work done by the engine and the net heat exchange by the engine during one cycle. We have our PD diagram here showing our process from A to BB to C and then C back to a, our multiple choice answers are a zero jewels, 30 jewels and 30 jewels, B, 60 jewels, 120 jewels and negative 60 jewels, C 60 jewels, 60 jewels and zero jewels or D zero jewels, 30 jewels and 60 jewels. So the key to solving this problem is recalling that Q or heat exchanged is equal to work plus our delta eth or our change in thermal energy. We can recall then in turn that our change in internal energy is equal to and our number of moles multiplied by CV multiplied by delta T and work is the area enclosed in our PD diagram. We need to solve for our delta T. First, we have initial T for T A as 300 degrees Kelvin as shown in our diagram, but we don't have our TB or TC. And once we find those, then we can solve for our internal or net change in internal energy. So four an ideal gas, we can recall that PV equals N RT where P is our pressure, V is our volume N is our number of moles, R is our gas constant and T is our temperature. So for point A and point B, we can rewrite this as P A VA divided by N RT A is equal to PB VB divided by and our TV. Because for the same ideal gas under two different conditions, this equation remains constant. And so we have T A N and R are constant, we also have based off of our figure or diagram, we have VA and VB. So we can in turn solve for TB isolating that on the left hand side of the equation gives us TB is equal to PB multiplied by BB multiplied by T A all divided by PAV multiplied by VA. And so when we plug in our known values for that equation, we get TB is equal to 2.5 times 10 to the five pascals for PB again, taken from our graph multiplied by VB which is 0.45 times 10 to the negative three meters cubed multiplied by T A which is 300 degrees. Kelvin all divided by our pressure at point A which is 1.5 times 10 to the fifth pascals multiplied by R VA which is 0.2 times 10 to the negative 3 m cubed. We plug that into our calculator and we get a temperature at point B of 1125 degrees. Kelvin, we're going to take the same approach to solve for our TC. So TC equals PC VC multiplied by T A divided by P A VA. And so similarly, we can essentially copy our P A VA and T A terms. And the only thing that, that changes is PC and VC from the PV diagram, our pressure at point C is 1.5 times 10 to the fifth pascals. And our volume at point C is 0.8 times 10 to the negative three meters cubed. So we solve for our temperature at point C and we get 1200 degrees Calvin. Now, we have everything we need to solve for our delta th I'm sorry, delta eth or, or change in thermal energy. Again, we can recall that that equation is N CV delta T. So N is our number of moles from the problem we're told it was one mole are specific heat capacity of helium. You can recall the CV is equal to three halves multiplied by R and then our delta T and so we can use this base equation to find our change in internal energy at each of the points from A to BB to C and then C back to A. So delta E from A to B is going to be that one mole three halves multiplied by our gas constant, which is we can recall 8.31 Joel, sorry, just 8.31 Joel per degree Calvin multiplied by mole. And then our delta T from A to B is going to be 1125 degrees Kelvin minus 300 degrees Kelvin. And so we plug that into our calculator and we get 10,000 284 jewels. Now, the only thing that changes from A to B to B to C is our delta T. So using the temperatures, we had just calculated our delta T is going to be 1200 Kelvin minus 1125 Kelvin. And that makes our internal energy from B to C equal to 935 jewels. And then again, similarly from C to A, the only thing that changes is our delta T. So from C to A, we have 300 Kelvin as our temperature at point A or TF and then 1200 Calvin as our temperature at point C or T I. And we plug that in and we get a change of negative 11,219 shorts. And so our final or net change in internal energy is going to be are changed from A to B plus our change from B to C plus our change from C to A. And when we sum these three together, we get zero jewels. So our total heat we can recall is our work plus our net change in internal energy. So we've found our net change in internal energy is zero and we can calculate our network which is simply the area enclosed. So we can recall that the area of a triangle is one half multiplied by the base, multiplied by the height. And so our base you can see from the diagram is from 0.8 to 0.2. So 0.6 times 10 to the negative 3 m cubed. And then our height is from 1.5 to 2.5 or just one times 10 to the five pass gals. And we can calculate then that our work is equal to 30 tools. And so back to our initial equation where net heat exchanged is equal to 30 jewels plus zero Jews, which equals of course 30 jewels. And so this these three steps, this is the answer to our problem. And we can see then that, that aligns with answer choice. A. So answer choice A is the correct one for this problem. 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. b. Make a table that shows ∆Eₜₕ, Wₛ, and Q for each of the three processes.

Verified Solution

Key Concepts

First Law of Thermodynamics

Monatomic Gas Properties

Thermodynamic Processes