Work & Equilibrium of Insulated Box: Why W=0?

  • Thread starter Thread starter princejan7
  • Start date Start date
  • Tags Tags
    Work
AI Thread Summary
In the insulated box scenario, the work done (W) is considered zero because the system is isolated, meaning there is no heat transfer or external work acting on it. The piston is frictionless and thermally conducting, allowing for equilibrium without energy exchange with the surroundings. Since both sides of the piston are at the same temperature (400K) after equilibrium is reached, there is no net work done between the two air chambers. The absence of external forces and heat transfer confirms that W=0 for the entire system. Therefore, the conclusion that W=0 is valid in this context.
princejan7
Messages
93
Reaction score
0

Homework Statement



"An insulated box is initially divided by a frictionless, thermally conducting piston. One side of the piston is 1.5(m^3) of air at 400k, 4bar. The other side is 1.5(m^3) of air at 400K,2 bar
The piston is released and equilibrium is attained, with the piston experiencing no change of state"


the solution assumes W=0. why is this?

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
What solution? You didn't post any solution.
 
dauto said:
What solution? You didn't post any solution.

here
 

Attachments

  • Untitled.png
    Untitled.png
    7 KB · Views: 418
That's the work for the entire system, not the work between the two chambers. Since the entire system is in a isolated box, there is no external work or heat.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top