If you poke a hole in a container full of gas, the gas will start leaking out. In this problem, you will make a rough estimate of the rate at which gas escapes through a hole. (This process is called effusion, at least when the hole is sufficiently small.)

1. Consider a small portion (area = $\mathit{A}$) of the inside wall of a container full of gas. Show that the number of molecules colliding with this surface in a time interval $\mathbf{\Delta }\mathit{t}$is role="math" localid="1651729685802" $\mathit{P}\mathit{A}\mathbf{\Delta }\mathit{t}\mathbf{/}\mathbf{\left(}\mathbf{2}\mathbf{m}\overline{{\mathbf{v}}_{\mathbf{x}}}\mathbf{\right)}$, where width="12" height="19" role="math">Pis the pressure, is the average molecular mass, and $\overline{{\mathbf{v}}_{\mathbf{x}}}$is the average $\mathit{x}$velocity of those molecules that collide with the wall.
   
2. It's not easy to calculate $\overline{{\mathbf{v}}_{\mathbf{x}}}$, but a good enough approximation is ${\left(\overline{v_x^2}\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}$, where the bar now represents an average overall molecule in the gas. Show that ${\left(\overline{v_x^2}\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\sqrt{\mathbf{k}\mathbf{T}\mathbf{/}\mathbf{m}}$.
   
3. If we now take away this small part of the wall of the container, the molecules that would have collided with it will instead escape through the hole. Assuming that nothing enters through the hole, show that the number $\mathit{N}$of molecules inside the container as a function of time is governed by the differential equation
   $\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\frac{\mathbf{A}}{\mathbf{2}\mathbf{V}}\sqrt{\frac{\mathbf{k}\mathbf{T}}{\mathbf{m}}\mathbf{N}}$
   Solve this equation (assuming constant temperature) to obtain a formula of the form $\mathit{N}\left(t\right)\mathbf{=}\mathit{N}\left(0\right){\mathit{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{r}}$, where $\mathit{r}$is the “characteristic time” for $\mathit{N}$(and $\mathit{P}$) to drop by a factor of $\mathit{e}$.
   
4. Calculate the characteristic time for gas to escape from a 1-liter container punctured by a $\mathbf{1}\mathbf{-}{\mathbf{mm}}^{\mathbf{2}}$? hole.
5. Your bicycle tire has a slow leak so that it goes flat within about an hour after being inflated. Roughly how big is the hole? (Use any reasonable estimate for the volume of the tire.)
6. In Jules Verne’s Around the Moon, the space travelers dispose of a dog's corpse by quickly opening a window, tossing it out, and closing the window. Do you think they can do this quickly enough to prevent a significant amount of air from escaping? Justify your answer with some rough estimates and calculations.

Area of a small portion of the wall of container which is full of gas is. A small hole is poked in a container so that gas will start leaking out.

The expression for force is

$F=\frac{\Delta p}{\Delta t}$ …… (1)

Here,$\Delta p$ is the change in momentum,$\Delta t$ is the change in time.

The expression for pressure is given by

$P=\Delta N\frac{F}{A}$ …… (2)

Here,$A$ is an area of cross-section.

Let us suppose that the average x component of the velocity of those molecules that are moving towards the hole is $\overline{v_x}$. It is assumed that the collision here is an elastic collision, so the change in velocity is

$\Delta v_x=2.{\overline{v}}_x$ …… (3)

Substitute $\frac{\Delta p}{\Delta t}$for $F$in equation (2)

$P=\Delta N\frac{\Delta p}{A\Delta t}$ …… (4)

But

$\Delta p=m\Delta v_x=2m{\overline{v}}_x$ …… (5)

Substitute $2m{\overline{v}}_x$for $\Delta p$in equation (4)

$P=\Delta N\frac{2m{\overline{v}}_x}{A\Delta t}$ …… (6)

Rearranging the above equation (6) for$\Delta N$

$\boxed{\Delta N=\frac{PA\Delta t}{2m{\overline{v}}_x}}$

Hence, the required number of molecules is $\Delta N=\frac{PA\Delta t}{2m{\overline{v}}_x}$.

The ideal gas equation can be expressed as

$PV=NkT$ …… (1)

Here,$P$ is the pressure of the gas,$V$ is the volume of gas,$N$ is the number of molecules of gas,$T$ its temperature,$k$ is the Boltzmann constant.

$P=\frac{Nm{v_x}^2}{V}$

$PV=Nm{v_x}^2$ …… (2)

Substitute$NkT$ for$PV$ from equation (1) in equation (2)

$\begin{array}{l}NkT=Nm{v_x}^2\\ {v_x}^2=\frac{kT}{m}\\ \boxed{\sqrt{{v_x}^2}=\sqrt{\frac{kT}{m}}}\end{array}$

Hence, the required expression is $\boxed{\sqrt{{v_x}^2}=\sqrt{\frac{kT}{m}}}$.

Change in the number of molecules $\Delta N$in a container over a time interval $\Delta t$is

$\frac{\Delta N}{\Delta t}=-\frac{PA}{2m\overline{v_x}}$ …… (1)

Here,$\Delta N$ is a number of changes in molecules,$\Delta t$ is the time interval,$P$ is pressure and$A$ is an area of cross-section,$m$ is the mass, and$\overline{v_x}$ is the average velocity of molecules.

The average velocity of molecules is expressed as

$\sqrt{{v_x}^2}=\sqrt{\frac{kT}{m}}$ …… (2)

Here,$k$ is Boltzmann constant,$m$ is mass of molecules of gas,$T$ is temperature.

The ideal gas equation can be expressed as

$PV=NkT$ …… (3)

Here,$P$ is the pressure of the gas,$V$ is the volume of gas,$N$ is the number of molecules of gas,$T$ is temperature,$k$ is the Boltzmann constant.

Since no molecules will pass through into the container so the change in the number of molecules in the container over the time interval is given by equation (1), here minus sign represent$N$ decreases with time.

Substitute$\sqrt{\frac{kT}{m}}$ for$v_x$ in equation (2)

$\frac{\Delta N}{\Delta t}=-\frac{PA}{2m}\sqrt{\frac{m}{kT}}$ …… (4)

Equation (3) can be simplified as

$P=\frac{NkT}{V}$ …… (5)

Substitute$\frac{NkT}{V}$ for$P$ in equation (4)

$\begin{array}{c}\frac{\Delta N}{\Delta t}=-\frac{NkTA}{2mV}\sqrt{\frac{m}{kT}}\\ =-\frac{NA}{2V}\sqrt{\frac{kT}{m}}\end{array}$

$\boxed{\frac{\Delta N}{\Delta t}=-\frac{A}{2V}\sqrt{\frac{kT}{m}}N}$ …… (6)

Now to solve the equation (6) over the interval$t=\left[0,t\right]$

Differentiate equation (6) on both sides of it,

$\begin{array}{l}\underset{0}{\overset{t}{\int }}\frac{\Delta N}{\Delta t}=-\frac{A}{2V}\sqrt{\frac{kT}{m}}\underset{0}{\overset{t}{\int }}\Delta t\\ {\left[\ln \left(N\left(t\right)\right)\right]}_0^t=-\frac{A}{2V}\sqrt{\frac{kT}{m}}t\end{array}$

$\ln \left(\frac{N\left(t\right)}{N\left(0\right)}\right)=-\frac{A}{2V}\sqrt{\frac{kT}{m}}t$ …… (7)

Take exponential on both the sides of equation (7)

$\begin{array}{l}\exp \left(\ln \frac{N\left(t\right)}{N\left(0\right)}\right)=\exp \left(-\frac{A}{2V}\sqrt{\frac{kT}{m}}t\right)\\ \frac{N\left(t\right)}{N\left(0\right)}=e^{-\frac{A}{2V}\sqrt{\frac{kT}{m}}t}\\ \boxed{N\left(t\right)=N\left(0\right)e^{-\frac{A}{2V}\sqrt{\frac{kT}{m}}t}}\end{array}$

Hence, the required differential equation which is followed by the number of molecules inside the container is$\boxed{\frac{\Delta N}{\Delta t}=-\frac{A}{2V}\sqrt{\frac{kT}{m}}N}$ and its solution is $\boxed{N\left(t\right)=N\left(0\right)e^{-\frac{A}{2V}\sqrt{\frac{kT}{m}}t}}$.

Mass of molecules can be obtained as

$m_{molecules}=\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}one\hspace{0.33em}mole\hspace{0.17em}of\hspace{0.17em}air}}{\text{number\hspace{0.17em}of\hspace{0.17em}molecuels\hspace{0.17em}in\hspace{0.17em}one\hspace{0.17em}mole\hspace{0.17em}}{N}_A}$ …… (1)

Characteristic time at room temperature$T=293\text{\hspace{0.17em}K}$ is

$\tau =\frac{2V}{A}\sqrt{\frac{m}{kT}}$ …… (2)

Here, $V$is the velocity of molecules, $A$is an area of cross-section, $m$is mass of molecules, $k$is Boltzmann constant, and $T$is temperature.

The average mass of air molecules is obtained by substituting$28.969\times {10}^3$ for the mass of one mole of air and$6.02\times {10}^{23}$ for$N_A$ in equation (1)

role="math" localid="1651732809570" $\begin{array}{l}{m}_{molecules}=\frac{28.969\times {10}^3}{6.02\times {10}^{23}}\\ =4.81\times {10}^{-26}\text{\hspace{0.17em}kg}\end{array}$

Characteristic time at room temperature $T=293\text{\hspace{0.17em}K}$is calculated by substituting $1\times {10}^{-3}$kg for volume, $1\times {10}^{-6}\text{\hspace{0.17em}m}$for the area, $4.81\times {10}^{-26}\text{\hspace{0.17em}kg}$for the average mass of molecules in equation (2)

$\begin{array}{l}\tau =\frac{2\times 1\times {10}^{-3}}{1\times {10}^{-6}}\sqrt{\frac{4.81\times {10}^{-26}}{1.38\times {10}^{-23}\times 293}}\\ \boxed{\tau =6.9\text{\hspace{0.17em}s}}\end{array}$

Hence, the characteristic time is $\boxed{\tau =6.9\text{\hspace{0.17em}s}}$.

The bicycle tire has a slow leak so it goes flat within about an hour after being inflated.

Characteristic time at room temperature$T=293\text{\hspace{0.17em}K}$ is

$\tau =\frac{2V}{A}\sqrt{\frac{m}{kT}}$ …… (1)

Here,$V$ is the velocity of molecules,$A$ is an area of cross-section,$m$ is mass of molecules,$k$ is Boltzmann constant, and$T$ is temperature.

Diameter of the hole is

$d=2\sqrt{\frac{A}{\pi }}$ …… (2)

Here,$A$ is an area of the cross-section of the hole.

Assume that a typical bicycle tire has a volume of 1.4 liters and the tire goes flat after 1 hour i.e. $\tau =3600\text{\hspace{0.17em}s}$.

Therefore, the hole size is obtained by simplifying equation (1)

$A=\frac{2V}{\tau }\sqrt{\frac{m}{kT}}$ …… (3)

Substitute$1.4\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}}$ for volume, 3600 s for $t$,$4.81\times {10}^{-26}\text{\hspace{0.17em}kg}$ for $m$,$1.38\times {10}^{23}{\text{\hspace{0.17em}JK}}^{\text{-1}}$ for $k$, and$293\text{\hspace{0.17em}K}$ for$T$ in equation (3)

$\begin{array}{l}A=\frac{2\times 1.4\times {10}^{-3}}{3600}\sqrt{\frac{4.81\times {10}^{-26}}{1.38\times {10}^{-23}\times 293}}\\ A=2.68\times {10}^{-9}{\text{\hspace{0.17em}m}}^{\text{2}}\end{array}$

The hole diameter is calculated by substituting$2.68\times {10}^{-9}{\text{\hspace{0.17em}m}}^{\text{2}}$ for$A$ in equation (2)

$\begin{array}{l}d=2\times \sqrt{\frac{2.68\times {10}^{-9}}{\pi }}\\ \boxed{d=0.058\text{\hspace{0.17em}mm}}\end{array}$

Hence, the required size of the hole is $\boxed{d=0.058\text{\hspace{0.17em}mm}}$.

The space travelers dispose of the dog’s corpse very quickly by opening a window, tossing it out, and closing the window on the spot.

Characteristic time at room temperature $T=293\text{\hspace{0.17em}K}$is

$\tau =\frac{2V}{A}\sqrt{\frac{m}{kT}}$ …… (1)

Here,$V$ is the velocity of molecules,$A$ is an area of cross-section,$m$ is mass of molecules,$k$ is Boltzmann constant, and$T$ is temperature.

Area of the window through which the dogs corpse is

$A=\pi r^2$ …… (2)

Here, is the radius of the window.

Assume that it is a small dog, so the porthole has a diameter of 40 cm and the spaceship has a volume of $V=150{\text{\hspace{0.17em}m}}^{\text{3}}$. They could open the window, toss out the dog and close the window again within 1 s.

The area of a porthole is

$A=\pi \times {0.2}^2=0.125{\text{\hspace{0.17em}m}}^{\text{2}}$.

Characteristic time is obtained by substituting $150{\text{\hspace{0.17em}m}}^{\text{3}}$for $V$, $0.125{\text{\hspace{0.17em}m}}^{\text{2}}$for $A$, $4.81\times {10}^{-26}\text{\hspace{0.17em}kg}$for $m$, $1.38\times {10}^{23}{\text{\hspace{0.17em}JK}}^{\text{-1}}$for$k$and$293\text{\hspace{0.17em}K}$for$T$ in equation (1)

$\begin{array}{l}\tau =\frac{2\times 150}{0.125}\sqrt{\frac{4.81\times {10}^{-26}}{1.38\times {10}^{-23}\times 293}}\\ \tau =8.234\text{\hspace{0.17em}s}\end{array}$

The ratio of the number of molecules is

$\frac{N\left(t\right)}{N\left(0\right)}=e^{-\frac{1}{\tau }}$ …… (3)

Substitute 8.234 s for $\tau$in equation (3)

$\frac{N\left(t\right)}{N\left(0\right)}=e^{-\frac{1}{8.234}}=0.885$

Hence, they will lose 0.115$\left(1-0.885\right)$ of their air. They will probably survive.