Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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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. - Physics

Sum

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

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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 = 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)`

(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`

Concept: Kinetic Theory of Gases - Concept of Pressure
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 2 Kinetic Theory of Gases
Q 37 | Page 36
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