Two cylinder A and B having piston conneted by massless rod (as shown in figure). The cross-sectional area two cylinders are same & equal to 'S'.The cylinder A contains m gm of an ideal gas at Pressure P & temperature `T_(0)` The cylinder B contain identical gas at same temperature `T_(0)` but has different mass. The piston is held at the state in the position so that volume of gas in cylinder A & cylinder B are same and is equal to `V_(0)`. The walls & piston of cylinder A are thermally insulated, whereas cylinder B is maintained at temperature `T_(0)` reservoir. The whole system is in vacuum. Now the piston is slowly released and it moves towards left & machanical equilibrium is reached at the state when the volume of gas in cylinder A becomes V_(0)/2 Then (here gamma for gas =1.5)

The mass of gas in cylinder B

Options

A. `2sqrt(2)` m
B. `3sqrt(2)` m
C. `sqrt(2)` m
D. none

Answer

B

Explanation

`P_(0)V_(0)=m/M RT` ina cylinder....(1)
`P_(B)V_(0)=m_(B)/MRT` in Bcylinder.....(2) adiabatic
`P_(0)V_(0)^(gamma)=P_(f)((V_(0))/2)^(gamma)`
`P_(f)=2sqrt(2)P_(0)=P_(B)`
Divide((1))//(2) `(P_(0)V_(0))/(P_(B)V_(B))=m/m_(B) V_(B)=3/2V_(0)`
`(P_(0)V_(0))/(2sqrt(2)P_(0)(3/2V_(0)))=m/m_(B)`
`m_(B)=3sqrt(2)m`
`DeltaU=-W=((P_(f)V_(f)-P_(i)V_(i))/(gamma-1))=(2P_(0)V_(0)/2xx2sqrt(2)-P_(0)V_(0))/((3//2)-1 )=2(sqrt(2)-1)P_(0)V_(0)`
`F_(compressive)=P_(f)Area=2sqrt(2)Ps`
`W_(A)=((P_(i)V_(i)-P_(f)V_(f))/(gamma-1))=(P_(0)V_(0)-2sqrt(2)P_(0)(V_(0)//2)) /((3//2)-1)=2P_(0)V_(0)(1-sqrt(2))`
`|W_(A)|=0.828P_(0)V_(0)=P_(0)V_(0)(2(1-11.414)=P_(0)V_(0)(-0.828)=-0.828P_(0)V_(0)`
`T=C=nRT_(0)l n(V_(f)/V_(i))=P_(f)V_(f)l n ((3V_(0))/(2V_(0)))=2sqrt(2)P_(0)((3V_(0))/2)l n (3/2)=3sqrt(2)P_(0)V_(0)l n(3/2)`
`=[3sqrt(2)l n(1.5)P_(0)V_(0)]=(4.242xx0.40)P_(0)V_(0)`
`W_(B)=1.719P_(0)V_(0)`
`|W_(B)|=1.719P_(0)V_(0)`
`|W_(A)|lt|W_(B)|`