A vessel of volume V_{0} contains an ideal gas at pressure p_{0} 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}

#### Solution

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]

Now ,

`(dP)/P = (-rdt)/V_0` [From eq. (3)]

(a)

Integrating the equation P = P_{0} 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)`

(b)

`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`