# an adiabatic chamber has friction less piston(mass m).mass of remaining chamber including gas is 4m.there are n moles of ideal gas at atmospheric tempereature and pressure.piston is at rest in equilibrium and confined to move along the length of cylinder .a particle of mass m strikes elastically  with piston. change in temperature of the gas when compression of the gas is maximum isa)2mv^2 /9nRb) 4mv^2 /15nRc)7mv^2 /15nRd) mv^2 /nR

Anish Karn
24 Points
6 years ago
Since it is adiabatic processSo, dQ=0 i.e. dU=-dWNow, since collision is elasticmv=5mv°=>v°=v/5Change in kinetic energy=net work donedW=dK=(1/2)5m*(v°)^2-(1/2)mv^2On solving,dW=-2/5mv^2dU=nCvdT=-dWn*fR/2*dT=-dK3nRdT/2=2/5mv^2=>dT=4mv^2/15nRb) is correct
Gaurav Singh
29 Points
6 years ago
I just have one confusion.... Why will the chamber acquire the same velocity as the piston after the collision???....
Gaurav Singh
29 Points
6 years ago
If it does acquire the same velocity as the piston, then there will be no compression!!....so why have we taken "v" as the velocity for both chamber and piston????
Gaurav Singh
29 Points
6 years ago
I mean why have we taken **"v/5" as the velocity for both chamber and piston after collision?????...please explain
Gaurav Singh
29 Points
6 years ago
And since the collision is with the piston , which is free to move, why have we considered the entire mass for momentum conservation????
Krishnendu Bera
19 Points
5 years ago
The particle strikes the piston with velocity v and since the collision is elastic, it comes to rest and piston(mass m) moves with velocity v. But due to pressure of the gas, the whole system(mass 5m) moves with a certain velocity. Conserving linear momentum we find the velocity is v/5. Change in kinetic energy is -2/5mv^2.....then proceed as above.