Two kilograms of nitrogen (N2) at 25°C is contained in a 0.62 m3 rigid tank. This tank is connected by a valve to a 0.16 m3 rigid tank containing 0.8 kg of oxygen (O2) at 127°C. The valve is opened, and the gases are allowed to mix, achieving an equilibrium state at 87°C. initial pressures of N2 is 5.7293 bar and O2 is 5.2 bar.
the final pressure is 6.44 bar.
the magnitude of the heat transfer for the process is 162.8 kJ, and the direction of energy flow is going in.
What is the entropy change for the mixing process, in kJ/K?

Explanation:

For entropy change the formula is

ΔS = ΔQ / T

ΔQ = Δ H

ΔS = Δ H / T

Given

Δ H = + 162.8 kJ

We can take equilibrium temperature as average temperature of the whole process

So, T = 273 + 87 = 360 K

ΔS = Δ H / T

=  162.8 kJ  / 360

= +  0.508 kJ / K .

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!