The system shown in the figure is in equilibrium. The piston is massless, frictionless and insulated. All walls of the chamber are also insulated. When heat is generated inside the lower chamber the piston slowly moves upwards by `2` m and the liquid comes out through an orifice so that it can rise to a maximum height of `5` m above the orifece level. The lower chamber contains `2` moles of an ideal monoatomic gas at `500k` .
Area of orifice a `=0.05 m ^(2)`
Area of the piston `A=1 m^(2)`
Density of the liquid, `rho=10^(3) kg//m^(3)` , `g=10 m//s^(2)`
Atmospheric pressure, `P_(atm)=10^(5) N//m^(2)`

The final pressure of the gas is

Options

A. `1.8 atm`
B. `0.18 atm`
C. `1.5 atm`
D. none of these

Answer

A

Explanation

Assume datum at the position of the orifice. Initial energy of the liquid

`V=(a)/(A)sqrt(2gh)=(0.05)/(1)sqrt(2xx10xx5)=0.5m//s`
`E_(i)=-(1)/(2)rho gAH_(1)^(2)=(1)/(2)(10^(3))(10)(5^(2))=-1.25xx10^(5)J`
and final energy `E_(r)=-(1)/(2)rhoA(H_(1)-h)^(2)+rhogAhh_(0)`
`=-(1)/(2)(10^(3))(10)(1)(5-1)^(2)+(10^(3))(10)(1)(5)`
`-0.45xx10^(5)+10^(5)`
Work done by the expanding gas is
`W=E_(r)-E_(i)=1.25xx10^(5)-0.45xx10^(5)+10^(5)`
`=1.8xx10^(5)J`
Initial pressure of the gas , `p_(1)=P_("atm")+rhogH_(1)`
`=10^(5)+(10^(3))(10)(5)=1.5xx10^(5)Pa`
Final , pressure of the gas k,
`p_(2)=P_("atm")+rho(H_(1)-h+h_(0))`
`=10^(5)+(10^(3))(10)(5-2+5)=1.8xx10^(5)Pa`
Initial volume, `V_(1)=AH_(2)=(1)(2)=2m^(3)`
Final volume, `V_(2)=A(H_(2)+h)=(1)(2+2)=6m^(3)`
Since`(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2)) , or T_(2)=(p_(2)V_(20))/(p_(1)V_(1)) T_(1)=((1.8)(4))/((1.5)(2))`
`T_(1)=((1.8)(4))/((1.5)(2))(500)`
Increases in internal energy, `DeltaU=nC_(v)DeltaT`
here, `n=2,C_(v)=(3R)/(2),DeltaT=1200-500=700K`
`:. Delta u(2)(3//2)(8,3)(700)=17.43kJ`
`Q=DeltaU+W=17.43+180=197.43kJ`
Time taken `=(2m)/(0.5m//s)=4s`
Average power `=(197.43kJ)/(4s)=49.36KW`