Question

The rate of change of pressure (p) of a gas at constant temperature and constant external pressure due to effusion of gas from a vessel of constant volume is related to rate of change of number of molecules present by $\frac{dp}{dt}$=$\frac{kT}{V}$ $\frac{dN}{dt}$ where k=Boltzmann constant,T=temperature,V=volume of vessel and N =No of molecules and $\frac{dN}{dt}$=$\frac{-p{A}_0}{(2\pi mkT{)}^{1/2}}$, where $A_0$= area of office and m =mass of molecule.

The time required for pressure inside vessel to reduce to 1/e of its initial value in (In e=1)

• A. ${(\frac{2\pi m}{kT})}^{1/2}$ $\frac{V}{A_0}$
• B. ${\left(\frac{kT}{2\pi m}\right)}^{1/2}$ $\frac{V}{A_0}$
• C. ${\left(\frac{2\pi mkT}{A^0}\right)}^{1/2}$
• D. $\frac{2\pi m}{kT}$ $\frac{V}{A_0}$

Answer 1

The correct option is A ${(\frac{2\pi m}{kT})}^{1/2}$ $\frac{V}{A_0}$

Given equation :

$\frac{dP}{dt}=\frac{kT}{V}\frac{dN}{dt}$

Rewriting the above equation we get,

$\frac{-V}{kT}\frac{dP}{dt}=\frac{dN}{dt}$

$\frac{-V(2\pi mkT{)}^{\frac{1}{2}}}{kTp{A}_0}dp=dt$

$\frac{-V}{p{A}_0}(\frac{2\pi m}{kT}{)}^{\frac{1}{2}}dp=dt$

$\frac{V}{p{A}_0}(\frac{2\pi m}{kT}{)}^{\frac{1}{2}}{\int }^{\frac{1}{e}}\frac{-1}{p}dp={\int }^{t}dt$

$\frac{V}{p{A}_0}(\frac{2\pi m}{kT}{)}^{\frac{1}{2}}+[lnp{]}_1^{\frac{1}{e}}=t$

t=$(\frac{2\pi m}{kT}{)}^{\frac{1}{2}}\frac{V}{p{A}_0}$