A vertical cylindrical tank has its top end closed by a tightly fitted frictionless and massless piston of negligible weigh as shown in the figure. The pressure inside the tank is maintained at in equilibrium with atmospheric pressure. The piston is depressed by pouring mercury on it very slowly. Considering it as an isothermal process, find the depth upto which the piston descends before the mercury spills over the top of the tank is____m.

Answer

`01.14`

Explanation

Before pouring mercury,
`P_(1) = 1 "atm"` = 76 cm of Hg

Also, pressure inside the tank and outside will be equal

`therefore` Forces acting on the piston will be,
`P_(1)A = P_(0)A`
Where, A = area of cross-section of piston. After pouring mercury,

`P.= P_(0)` + pressure due to mercury
= 76 cm + x cm of Hg
= (76 + x) m of Hg
`P_(2)=` Pressure of the gas on the piston From figure (b),
`P_(2) = P. = (76 + x)` cm of Hg
As the gas compressed isothermally (temperature is kept constant),
PV = constant
`therefore P_(1)V_(1) = P_(2)V_(2)`
Here, `V = l xx A`
`therefore 76` of Hg `xx (190 xx A)`
`= [(76 + x)` of Hg] `xx [(190 -x)A]`
`therefore 76 xx 190 = (76 xx 190) + 190 x - 76 x -x^(2)`
`therefore 0 = -x^(2) + 114 x`
`therefore x^(2) = 114x`
`therefore x = 114 cm`
`therefore x = 1.14 cm`