An insulated `0.2 m^(3)` tank contains helium at `1200 kPa` and `47^(@)C`. A value is now opened, allowing some helium to escaped. The valve is closed when one-half of the initial mass has escped. The temperature of the gas is `(root(3)(4) =16)`

Options

A. `100 K`
B. `200 K`
C. `73^(@)C`
D. `- 73^(@)C`

Answer

B, D

Explanation

Initial mass of air, `m=(P_(1)V)/(RT_(1))`
Final mass of air
`=(m)/(2)=(P_(2)V)/(RT_(2))`
`(P_(1)V)/(2RT_(1))=(P_(2)V)/(RT_(2))`
`((T_(2))/(T_(1)))=2((P_(2))/(P_(1)))`
As the tank is insulated, the process is adiabatic with `gamma=(5)/(3)`
`T_(2)=T_(1)((P_(2))/(P_(1)))^((gamma-1)/(gamma))implies ((T_(2))/(T_(1)))^((5)/(2))=((P_(2))/(P_(1)))`
`(T_(2))/(T_(1))=2((T_(2))/(T_(1)))^((5)/(2))impliesT_(2)=(T_(1))/(2)^((2)/(3))`
`=(320)/(2^((2)/(3)))=200K=-73^(@)C[` Given `root(3)(4)=1.6]`