The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partition. both sides contain one mole of a monatomic ideal gas (γ=5/3) with the initial temperature being 525 k on the left and 275 k on the right. the partition is then allowed to move slowly to the right until the pressures on each side of the partition are the same. find the final temperatures on the left and right side.

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

T1 = 525 K

T2 = 275 K

We know that

n and v remain same at both side. so we have

..............1

let final pressure is P and temp  

..................2

similarly

.............3

divide 2 equation by 3rd equation

thus, temperature on left side is 1.48 times the temperature on right

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i just did this and got a 100.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.