Q. 375.0( 1 Vote )

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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 =V_{o}

Temperature of gas=T

We know that ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^{-1}mol^{-1}

T=temperature

n=number of moles of gas

P=pressure of gas.

So, applying ideal gas equation for the remaining gas

(P-dP) V_{o}=(n-dn) RT

PV_{o}-dPV_{o}=nRT-dnRT …… (1)

Applying ideal gas equation, before gas was taken out

PV_{o} = nRT ……. (2)

Using equation (2) in (1) we get

nRT - V_{o}dP = nRT - dnRT

V_{o}dP=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)

V_{o}dP=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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