Thermal Physics> an adiabatic chamber has friction less pi...

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 is

a)2mv^2 /9nR

b) 4mv^2 /15nR

c)7mv^2 /15nR

d) mv^2 /nR

Navjyot Kalra , 12 Years ago

Grade 10

6 Answers

Anish Karn

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

Last Activity: 9 Years ago

Gaurav Singh

I just have one confusion.... Why will the chamber acquire the same velocity as the piston after the collision???....

Last Activity: 9 Years ago

Gaurav Singh

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????

Last Activity: 9 Years ago

Gaurav Singh

I mean why have we taken **"v/5" as the velocity for both chamber and piston after collision?????...please explain

Last Activity: 9 Years ago

Gaurav Singh

And since the collision is with the piston , which is free to move, why have we considered the entire mass for momentum conservation????

Last Activity: 9 Years ago

Krishnendu Bera

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.

Last Activity: 8 Years ago