A vessel of volume V0 contains an ideal gas at pressure p0 and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out.
Use R = 8.3 J K-1 mol-1
Let P be the pressure and n be the number of moles of gas inside the vessel at any given time t .
Suppose a small amount of gas of dn moles is pumped out and the decrease in pressure is dP .
Applying equation of state to the gas inside the vessel , we get
`(P - dP) V_0 = (n - dn) RT`
⇒ `PV_0 - dPV_0 = nRT - dnRT`
But `PV_0 = nRT`
⇒ `V_0dP = dnRT` ..............(1)
The pressure of the gas taken out is equal to the inner pressure .
Applying equation of state , we get
`(P - dP) dV = dnRT`
⇒ `PdV = dnRT` ..........(2)
From eq. (1) and eq. (2) , we get
`V_0dP = PdV`
⇒ `(dP)/P = (dV)/V_0`
`(dV)/dt = r`
⇒ `dV = rdt`
⇒ `dV = -rdt` ........(3) [Since pressures decreases , rate is negative]
`(dP)/P = (-rdt)/V_0` [From eq. (3)]
Integrating the equation P = P0 to P = P and time t = 0 to t = t , we get
`∫_(P0)^P = ∫_0^t`
⇒ `In P - In P_0 = - (rt)/V_0`
⇒ `In (P/P_0) = - (rt) /V_0`
⇒ `P = P_0e^((-rt)/V_0)`
`P = P_0/2`
`P_0/2 = P_0e^((-rt)/V_0)`
⇒ `e^((rt)/V_0) = 2`
⇒`(rt)/V_0 = In 2`
⇒ `t = (V_0 In 2)/r`