Question

A conducting and closed container of capacity 100 liter contains an ideal gas at a high pressure. Now using a pump, the gas is taken out at a constant rate of 5 liter/sec. Find the time taken in which the pressure will decrease to \[\dfrac{{{P}_{initial}}}{100}\]? (Assume isothermal condition).
A. 45 sec
B. 92 sec
C. 118 sec
D. 146 sec

Answer 1

Hint: We are given the volume of a conducting closed container which contains an ideal gas at high pressure. It is said that a certain volume of gas is taken out of the container per second using a pump. We can solve this question using the ideal gas equation.

Formula used:
$PV=nRT$

Complete step by step answer:
In the question we are given a conducting and closed container with a capacity of 100 liter which contains an ideal gas at high pressure.
Therefore we have volume, $V=100l$
It is said that the gas is taken out of the container using a pup at a rate of 5 liter/sec.
Therefore we can write, $\dfrac{dV}{dt}=5$
$\Rightarrow dV=5dt$
Therefore we will get the small number of moles of the gas ejected out as per second as,
$\dfrac{dn}{dt}=\dfrac{n}{V}\times 5$
$\Rightarrow dn=\dfrac{n}{V}\times 5dt$
We have the ideal gas equation as,
$PV=nRT$, were ‘R’ is the ideal gas constant, ‘n’ is the number of moles of the gas, ‘T’ is the temperature of the gas, ‘V’ is the volume of the gas and ‘P’ is the pressure of the gas.
Therefore for a small pressure $\left( dP \right)$ and small number of moles $\left( dn \right)$, we can write the ideal gas equation as,
$\Rightarrow \left( dP \right)V=\left( dn \right)RT$
Since $dn=\dfrac{n}{V}\times 5dt$, we get the above equation as,
$\Rightarrow \left( dP \right)V=-\left( \dfrac{n}{V}\times 5dt \right)RT$
By solving this we can write,
$\Rightarrow \left( dP \right)V=-5dt\times \dfrac{nRT}{V}$
From the ideal gas equation we know that pressure can be written as, $P=\dfrac{nRT}{V}$. Therefore the above equation will become
$\Rightarrow \left( dP \right)V=-5dt\times P$
We can simply this and we will get,
$\Rightarrow \dfrac{dP}{P}=-\dfrac{5}{V}dt$
Since, $V=100$ we will get
$\Rightarrow \dfrac{dP}{P}=-\dfrac{5}{100}dt$
$\Rightarrow \dfrac{dP}{P}=-\dfrac{1}{20}dt$
In the question we are asked to find the time taken for the pressure to decrease to $\dfrac{{{P}_{initial}}}{100}$. Therefore let us integrate the above equation from initial pressure to final pressure. Thus we can write,
$\Rightarrow \int\limits_{{{P}_{initial}}}^{{{P}_{final}}}{\dfrac{dP}{P}=\int\limits_{0}^{t}{-\dfrac{1}{20}dt}}$
$\Rightarrow \int\limits_{{{P}_{initial}}}^{{{P}_{final}}}{\dfrac{dP}{P}=-\dfrac{1}{20}\int\limits_{0}^{t}{dt}}$
By integrating this we will get,
$\Rightarrow \dfrac{{{P}_{final}}}{{{P}_{initial}}}={{e}^{-{}^{t}/{}_{20}}}$
$\Rightarrow {{P}_{final}}={{P}_{initial}}^{{{e}^{-{}^{t}/{}_{20}}}}$
Since, ${{P}_{final}}=\dfrac{{{P}_{initial}}}{100}$, we can write the above equation as,
$\Rightarrow \dfrac{{{P}_{initial}}}{100}={{P}_{initial}}^{{{e}^{-{}^{t}/{}_{20}}}}$
From this we will get the time,
$\Rightarrow t=20\text{In}\left( 100 \right)$
$\Rightarrow t=92.103\sec $
$\therefore t\approx 92\sec $
Therefore we get the time taken as 92 sec.

So, the correct answer is “Option B”.

Note:
Let us consider a gas of volume ‘V’ at a temperature ‘T’ exerts a pressure ‘P’. Let the number of moles of the gas be ‘n’.
Then from Boyle’s law we know that at constant number of moles and temperature, the volume of the gas will be inversely proportional to the pressure exerted, i.e.
$V\propto \dfrac{1}{P}$
And also from Charles’ law, we know that for constant number of moles and constant pressure the volume of the gas will be proportional to temperature, i.e.
$V\propto T$
And from Avogadro’s law we know that for constant pressure and temperature of a gas the volume will be directly related to the number of moles, i.e.
$V\propto n$
By combining all these equations we will get,
$\Rightarrow V\propto \dfrac{nT}{P}$
This can be written as,
$\Rightarrow PV\propto nT$. The proportionality in this equation is replaced with a constant known as the ideal gas constant R.
Therefore we get the ideal gas equation as,
$\Rightarrow PV=nRT$