A gas undergoes two processes. In the first, the volume remains constant at 0.200 m^3 and the pressure increases from 2.00 * 10^5 Pa to 5.00 * 10^5 Pa. The second process is a compression to a volume of 0.120 m^3 at a constant pressure of 5.00 * 10^5 Pa. (b) Find the total work done by the gas during both processes.

Video transcript

everyone in this problem. We have an ideal gas in a container with a piston that's taken through a two step process. Okay, so in the initial step, volume is kept constant at 20.8 m cubed and the pressure is raised from 0.98 times 10 to the five pascal's to 4.2 times 10 to 5 pascal's. Okay. And in the second step we have the gas compressed, The pressure is kept constant at 4.2 times 10 to the five Pascal's To a final volume of .45 m cubed. Now we're asked what is the total work done during this two step process? Okay, so let's write out what we have. Okay, We have two steps. So we have our initial state. We'll call it a. Okay, where we start? We undergo some process to end up in step B or stage B. Okay. And then we undergo a second process and we'll end up at our final state C. So we have three stages and our processes go. Two processes. One to take you from A to B. No one to take you from B to C. Now, we're told that at a our initial volume. Okay. is 0.83 m cubed. Sorry, 0.8 m cubed. not 0.83 0.8 m Cube. Okay. And we're told that this is kept constant through the first step. Okay, so the volume at B is also going to be 0.8 m cute. Okay. What about the pressure? We're told the pressure is raised from 0.98 times 10 to the five pascal's. Well that means that the pressure at A. is equal to 0.98 times 10 to the five Pascal's And it's raised to 4.2 times to the five Pascal's. So the pressure at B is 4. Times 10 to the five Pascal's. And then at sea. Well what happens to the volume? What happens to the pressure? Well the pressure work is kept constant at 4.2 Times 10 to the five Pascal's. Okay. And the final volume is .45 m cubed. So now we have all of the volumes and pressures we've been given written down and we're asked to find the total work. Okay so the total work that we want to find is going to be equal to the sum of the work of each process. Okay so we're gonna get the work from A. To B. Plus the work from B to C. Okay and this total work is what we're looking for. Now let's start with the work from A. To B. Okay so the process from A. To B. We know that it has constant volume. And when we have a constant volume process, recall that this means that there is no work done. Okay this is an ice A cork process. We have no work done which means that the work from A to B is just equal to zero. Alright so we found the first component of our total work by looking at the process from A to B. Now let's do the second process from B to C. Okay so from B to C instead of having constant volume we have constant pressure. The pressure at B and C is the same. So we have constant pressure and what do we know about a process with constant pressure? Well this tells us that the work and in this case it's the work from B to C is equal to the pressure times of change in volume V two minus V. One. Alright well the pressure is 4.2 times 10 to the five pascal's The final volume 0.4 five m cute. And the initial volume 0.8 m scooped. We work this out, we get negative 1.47 Times 10 to the five jewels. So now we found our work from B to C by using the fact that this is a constant pressure process. We found that it's negative. Okay so let's go ahead and calculate our total work. Now remember from the top we wrote that work total is equal to the work baby Plus the work from B to C which is just equal to zero plus negative 1.47 times 10 to the five jewels. And so we get that the work total Is equal to negative 1.47 times 10 to the five jewels. Okay so we broke that work up by looking at those two separate processes, and what we found is that we get answer B. The work is equal to negative 1.47 times 10 to the five jules. Thanks everyone for watching. I hope this video helped see you in the next one.