For those who have not studied Electric Feilds yet and are a tad bit confused about what Mr.Doner means by the independence of path, I think (correct me if I am wrong) it may be helpful to think of when we have a direct transformation of Gravitational Potential Energy to Kinetic energy. Regardless of the path, an object takes to drop from a height (assuming all paths lead to the same height), the KE of the object will be the same at the end of each path.
Good point. In fact, my students have not learned about electric potential when they study thermomdynamics. Eventually I will get around to making a change.
Hi Mr Donor. I was an IB Physics student from the graduating class of 2019. Got a 7 in Physics despite getting a 5 on my IA XD. Anyways, a huge fan of yours and thank you for your excellent videos. Definitely helped me a lot. Our teacher chose Option D Astrophysics probably because it was easier and he liked it but out of my own interest, I'm watching your videos on the other Options as well. At 18:11 , where the change in U = 0. Doesn't U, the total internal energy include both the Kinetic and the Potential energy? Whenever there is a state change, isn't there a change in the potential energy? Thus to make the total energy change zero, shouldn't the temperature change? Thanks.
In an ideal gas, we assume no interactions between the molecules and hence the potential energy is zero, and does not change. Therefore, all changes must be changes in KE or temperature.
Hey Mr. Doner, I find your videos really helpful, however my school is doing option C and it would be really helpul for me and my classmates (as we are all mostly studying from your videos :) ) if you do a video on option C. Thank you so much!
At 19:55, if it is under isothermal situation, then how can adding the heat increase the molecular speed in the system and therefore having higher pressure than the adiabatic one? (Since constant temperature means constant kinetic energy and therefore constant molecular speed)
For isothermal, the pressure drops since the speed stays the same and the volume increases (more time between molecules striking walls). In the adiabatic situation, no heat is added so the pressure decreases more because both the molecules both have more distance to travel to meet the walls, but also because the temperature decreased so the molecules are no moving as quickly.
At 15:40 If the condition is isothermal, would not the energy absorbed by the gas = 0 ? If energy is absorbed by the gas, wouldn't that mean an increase in temperature.
+Charlie Ivarsson For an ideal gas, isothermal implies that the internal energy of the gas does not change. (internal energy would equal the sum of the KEs of all molecules.) The gas would have to be on a heat reservoir at a given temperature. Heat would would enter the chamber to restore the KE of molecules due to loses caused by the molecules striking a piston that is moving outwards as the gas expands.
Saying that each time you have a freely moving piston you get an isobaric process is a bit oversimplified isn't it? I mean look at the PV diagram of an engine, that doesn't look like constant pressure
+nutellabananenmist It is more helpful if you let me know what time in the video you are referring to. The discussion was of a gas in a closed piston subject to single processes, not cyclic processes involving changes to the number of molecules and turbulence. In general, I have found that my students are less confused if I begin simple, and introduce exceptions as more concepts are introduced.
I think what Mr. Doner is trying to say is that the freely moving piston helps keep the pressure constant. If changes are made to the gas, the piston can automatically adjust (since it's freely moving) to the changes on the system. Thus, pressure should stay constant, making it an isobaric process.
You are a God at teaching physics... I wish you'd have been my teacher from the beginning
For those who have not studied Electric Feilds yet and are a tad bit confused about what Mr.Doner means by the independence of path, I think (correct me if I am wrong) it may be helpful to think of when we have a direct transformation of Gravitational Potential Energy to Kinetic energy. Regardless of the path, an object takes to drop from a height (assuming all paths lead to the same height), the KE of the object will be the same at the end of each path.
Good point. In fact, my students have not learned about electric potential when they study thermomdynamics. Eventually I will get around to making a change.
Hi Mr Donor. I was an IB Physics student from the graduating class of 2019. Got a 7 in Physics despite getting a 5 on my IA XD. Anyways, a huge fan of yours and thank you for your excellent videos. Definitely helped me a lot. Our teacher chose Option D Astrophysics probably because it was easier and he liked it but out of my own interest, I'm watching your videos on the other Options as well.
At 18:11 , where the change in U = 0. Doesn't U, the total internal energy include both the Kinetic and the Potential energy? Whenever there is a state change, isn't there a change in the potential energy? Thus to make the total energy change zero, shouldn't the temperature change? Thanks.
In an ideal gas, we assume no interactions between the molecules and hence the potential energy is zero, and does not change. Therefore, all changes must be changes in KE or temperature.
Hey Mr. Doner, I find your videos really helpful, however my school is doing option C and it would be really helpul for me and my classmates (as we are all mostly studying from your videos :) ) if you do a video on option C. Thank you so much!
There is a chance that I do this in the future but it won't be in time for your class.
@@donerphysics Okay! Thanks anyways :D
At 19:55, if it is under isothermal situation, then how can adding the heat increase the molecular speed in the system and therefore having higher pressure than the adiabatic one? (Since constant temperature means constant kinetic energy and therefore constant molecular speed)
For isothermal, the pressure drops since the speed stays the same and the volume increases (more time between molecules striking walls). In the adiabatic situation, no heat is added so the pressure decreases more because both the molecules both have more distance to travel to meet the walls, but also because the temperature decreased so the molecules are no moving as quickly.
@@donerphysics This makes sense, thank you very much sir
At 15:40
If the condition is isothermal, would not the energy absorbed by the gas = 0 ? If energy is absorbed by the gas, wouldn't that mean an increase in temperature.
+Charlie Ivarsson
For an ideal gas, isothermal implies that the internal energy of the gas does not change. (internal energy would equal the sum of the KEs of all molecules.) The gas would have to be on a heat reservoir at a given temperature. Heat would would enter the chamber to restore the KE of molecules due to loses caused by the molecules striking a piston that is moving outwards as the gas expands.
Saying that each time you have a freely moving piston you get an isobaric process is a bit oversimplified isn't it? I mean look at the PV diagram of an engine, that doesn't look like constant pressure
+nutellabananenmist
It is more helpful if you let me know what time in the video you are referring to. The discussion was of a gas in a closed piston subject to single processes, not cyclic processes involving changes to the number of molecules and turbulence. In general, I have found that my students are less confused if I begin simple, and introduce exceptions as more concepts are introduced.
+Chris Doner (C. Doner's IB Physics) I was referring to the part at 8:35.
I think what Mr. Doner is trying to say is that the freely moving piston helps keep the pressure constant. If changes are made to the gas, the piston can automatically adjust (since it's freely moving) to the changes on the system. Thus, pressure should stay constant, making it an isobaric process.
You can hear his kid talking at ~ 7:30 :)