The gasoline engine in your car can be modeled as the Otto cycle shown in FIGURE CP21.73. A fuel-air mixture is sprayed into the cylinder at point 1, where the piston is at its farthest distance from the spark plug. This mixture is compressed as the piston moves toward the spark plug during the adiabatic compression stroke. The spark plug fires at point 2, releasing heat energy that had been stored in the gasoline. The fuel burns so quickly that the piston doesn't have time to move, so the heating is an isochoric process. The hot, high-pressure gas then pushes the piston outward during the power stroke. Finally, an exhaust value opens to allow the gas temperature and pressure to drop back to their initial values before starting the cycle over again. a. Analyze the Otto cycle and show that the work done per cycle is

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A typical gas power generator that is used in a portable power applications operates on the auto cycle model. In this type of engine, the combustion chamber is usually a single cylinder with a piston that moves up and down to complete each of the four strokes. During the compress compression stroke A to B, the piston moves up, compressing the air in the cylinder. This process is adiabatic as there is no exchange of heat between the system and its surroundings, the temperature and pressure of the air increase and the volume decreases as the piston moves upward, the spark plug ignites B to C, the compressed air causing it to rapidly combust and generate heat. The volume of the cylinder remains constant. During the power stroke C to D, the pressure increases and pushes the piston downward. Finally, during the exhaust stroke D to A, the piston moves upward again, expelling the exhaust gasses from the cylinder through the exhaust valve. This process is iso coric at the end of the cycle. The thermodynamic state at point A returns to its initial state. The auto cycle in the gas powered generator is a thermodynamic process that converts the chemical energy stored in the fuel into work. Calculate the work produced in terms of the temperature at points ABC and D, the number of moles N the heat capacities ratio gamma and the gas constant R OK. So that's our end goal is to calculate the work produced in terms of the temperature at points ABC and D. The number of moles N, the heat capacities ratio gamma and the gas constant R capital R. OK. So we're given some multiple choice answers. Let's read them off to see what our final answer might be. A is N multiplied by R divided by one minus gamma. All multiplied by TB plus TD plus T A plus TCB is N multiplied by R divided by one minus gamma, all multiplied by TB plus TD plus T A minus TC C is N multiplied by R divided by one minus gamma, all multiplied by TB plus TD minus T A plus TC and D is N multiplied by R divided by one minus gamma, all multiplied by TB plus TD minus T A minus TC. OK. So first off, let us assume that the gas inside the cylinder is an ideal gas. Thus, we can recall and apply the ideal gas law equation. Which states that the pressure multiplied by the volume is equal to the number of moles multiplied by the universal gas constant multiplied by the temperature. So we now we need to note that we can substitute P subscript capital, A multi and V subscript capital A with N RT A. So P capital A, let's write that down multiplied by V capital A. So P subscript capital, A multiplied by V subscript capital A is equal to N multiplied by R multiplied by T subscript capital A. OK. When we perform our calculations, so before we dive into our calculations, let us quickly define all of our variables. So note that we need to use T A TB TC and TD to denote the temperatures that points ABC and D respectively. And that gamma is equal to the heat capacity ratio and that R is our universal gas constant. OK. So now we need to calculate the work done during each process. OK. So the work done during the adiabatic compression from A to B is, and let's call it W subscript A B is equal to PB multiplied by VB minus P A multiplied by VA all divided by one minus gamma, which is equal to the number of moles multiplied by the universal gas constant multiplied by T B minus T A all divided by one minus gamma. OK. So the work done during the Iso Cork combustion from B to C is zero since the volume is constant. So let's make a note that delta V is equal to zero. So the work done from B to C equals the pressure multiplied by delta V. The change in volume is equal to zero. So the work done during the adiabatic expansion from C to D can be written as W CD is equal to PD multiplied by VD minus PC multiplied by VC, all divided by one minus gamma, which is equal to the number of moles multiplied by the universal gas constant multiplied by T D minus TC, all divided by one minus gamma. OK. So the work done during the Iso Cork combustion from D to A is zero since the volume is constant. So the work done from D to A. So W subscript D A is equal to the pressure multiplied by the change in volume which is equal to zero. Thus, the net produced in the system is, and let's call it W subscript net. So W net is equal to WAB plus W BC plus W CD plus WD A Awesome. So this is the net, the net work done and that's adding together all the values we just found. So let's plug in all of our knowns. So for the network, so W net is equal to number of moles multiplied by the universal gas constant multiplied by TB minus T A all divided by one minus gamma plus N multiplied which is, and the number of moles multiplied by the universal gas constant multiplied by TD minus TC, divided by one minus gamma. So note that we found earlier that W BC and WD A equals zero. So the only values we need to plug in are the values for WAB and W CD. OK. So that's where we got these two pieces of information that we added together. OK. So when we add these two together and we simplify, you will get that the net work is equal to N R divided by one minus gamma, all multiplied by T B plus TD minus T A minus TC. So the expression for the work produced is bracket nr divided by one minus gamma, bracket multiplied by bracket TB plus TD minus T A minus TC ray. We did it. So that is our final answer. So let's go back to look at our multiple choice answers to see which letter is our final answer. So that means the correct answer has to be the letter D N multiplied by R divided by one minus gamma, all multiplied by TB plus TD minus T A minus TC. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.