If this trough can hold all of the mercury, then I think that during the first phase, mercury is pushed up at a constant and slow rate, therefore the amount of work done by the gas is equal to the weight of the mercury times height displacement?
Sorry, I was interrupted. I agree with supposing that the mercury be expelled at a low rate, it help simplify the problem. Initially, I divide the process into two phase:
The first one being heating up the gas to the hottest temperature: a certain amount of mercury has been expelled. Can I...
Firstly I figure out that there is a point where the temperature of the gas is the hottest and beyond that no more amount of heat is necessary, descibed as following inequation:
Thank you all for all of your help <3, I've managed to express the respective height of each piston, substituting all that air with water. This is just a random physics problem I picked up online so it can be quite "broken" sometimes.
I can't help thinking that since both pistons have the same mass m, their weight should cancel each other when equilibrium is achieved, keeping both pistons at the initial height h.
Thank you for asking! I figured out that the air pressure inside the pipe must be constant such that its volume remains constant in order for Pascal's law to work.
I have tried using Pascal's law for the pipe but it turned out that the law only works for incompressible fluids, whereas this pipe contains air, which is compressible. I thought then this problem might be the same as one with a cylinder containing air sealed with a piston, but there might be...
I've been able to found this wonderful forum with lots of insightful and helpful people who are always willing to help and share their knowledge for everyone else. I'm currently an undergraduate student who has a little hobby of solving high school physics problem. I'm not that good at physics...