A vessel of volume V contains ideal gas having mass density `rho` at temperature T and pressure P. After a portion of the gas is let out, the pressure in the vessel is decreased by `Delta P.` The mass of the released gas is

Options

A. `rho V Delta P//P`
B. `(DeltaP )/(P)`
C. `(rho)/(P)`
D. `(rhoV)^(2) Delta P//P`

Answer

A

Explanation

Given,
Volume of vessel, `V _(1) =V`
Mass density of ideal gas `= rho`
Temperature of gas, `T_(1) ne T`
Pressure of gas, `P _(1) =P`
Pressure in the vessel is decreased by `Delta P.`
When gas is let out, `Delta V` of volume decreases. Thus,
`V_(2) = V_(1) - Delta V = V- Delta V`
`P _(2) = P _(1) - Delta P = P - Delta P`
`T _(2) = T_(1) =T`
From ideal gas equation,
`(P _(1) V_(1))/(T_(1))= (P_(2) V_(2))/(T_(2))`
`(PV)/(T) = ((P -Delta P ) (V - Delta V))/(T)`
`PV = PV - P Delta V - Delta PV + Delta V Delta P`
`Delta V ( Delta -P) = Delta PV`
`Delta V = (Delta PV)/(Detla P-P)`
`therefore` Mass of the released gas `= rho Delta V`
`= rho ( Delta PV)/(Delta P -P)`
` ~~ (rho Delta PV)/(P)`