Question

If the gas inside the vessel had molecular weight 9 times the gas in the previous example and area of orifice was doubled and temperature maintained at 4T, time required for pressure to fall to 1/e times of its initial value would be (t = answer of previous option)

• A. 1.33 t
• B. 4.24 t
• C. 0.75 t
• D. 1.125 t

Answer 1

The correct option is C 0.75 t

M=9m

A=2A${}_0$

T=4t

$t_1={\left[\frac{2\pi \left(9m\right)}{k\left(4t\right)}\right]}^{\frac{1}{2}}\frac{V}{2A_0}$

$=\frac{1}{2}{\left(\frac{9}{4}\right)}^{\frac{1}{2}}\left[{\left(\frac{2\pi m}{kT}\right)}^{\frac{1}{2}}\frac{V}{A_0}\right]$

$=\frac{1}{2}\times \frac{3}{2}\times t$

$=0.75t$