Answer :

Let P be the pressure and n be the number of moles of gas inside the vessel at any given time.


As mentioned in question, pressure is decreasing continuously. So, suppose a small amount of gas ‘dn’ moles are pumped out and the decrease in pressure is ‘dP’.


So, pressure of remaining gas =P-dP


Number of moles of remaining gas =n-dn


Given


The volume of gas =Vo


Temperature of gas=T


We know that ideal gas equation


PV=nRT


Where V= volume of gas


R=gas constant =8.3JK-1mol-1


T=temperature


n=number of moles of gas


P=pressure of gas.


So, applying ideal gas equation for the remaining gas


(P-dP) Vo=(n-dn) RT


PVo-dPVo=nRT-dnRT …… (1)


Applying ideal gas equation, before gas was taken out


PVo = nRT ……. (2)


Using equation (2) in (1) we get


nRT - VodP = nRT - dnRT


VodP=dnRT …… (3)


According to question, pressure of gas being taken out is equal to inner pressure of gas always. So inner pressure is equal to P-dP


Let the volume of gas taken out dV.


Applying ideal gas equation to gas pumped out


(P-dP) dV=dnRT


PdV=dnRT …… (4)


Where we have ignored dPdV as it is very small and can be neglected.


From equation (3) and (4)


VodP=PdV



Given . Since volume is decreasing so rate should be negative.


Putting this value of dV in equation (5)



(a) Integrating equation (6) from Po to P and t=0 to t




Where we have used the formula



And





Taking exponential on both sides,



The pressure of the gas as a function of time is given as


(b) In second part the final pressure becomes half of the initial pressure



Putting this value of P in equation (7)





Taking natural logarithm on both side




The time taken before half the original gas is pumped out is


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