Q. 375.0( 1 Vote )
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
The volume of gas =Vo
Temperature of gas=T
We know that ideal gas equation
Where V= volume of gas
R=gas constant =8.3JK-1mol-1
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
PdV=dnRT …… (4)
Where we have ignored dPdV as it is very small and can be neglected.
From equation (3) and (4)
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
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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