91影视

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

$F=\frac{螖p}{螖t}$ 鈥︹�� (1)

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

The expression for pressure is given by

$P=螖N\frac{F}{A}$ 鈥︹�� (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 $\stackrel{炉}{v_x}$. It is assumed that the collision here is an elastic collision, so the change in velocity is

$螖v_x=2.{\stackrel{炉}{v}}_x$ 鈥︹�� (3)

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

$P=螖N\frac{螖p}{A螖t}$ 鈥︹�� (4)

But

$螖p=m螖v_x=2m{\stackrel{炉}{v}}_x$ 鈥︹�� (5)

Substitute $2m{\stackrel{炉}{v}}_x$for $螖p$in equation (4)

$P=螖N\frac{2m{\stackrel{炉}{v}}_x}{A螖t}$ 鈥︹�� (6)

Rearranging the above equation (6) for$螖N$

$\boxed{螖N=\frac{PA螖t}{2m{\stackrel{炉}{v}}_x}}$

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

The ideal gas equation can be expressed as

$PV=NkT$ 鈥︹�� (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.

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

$PV=Nm{v_x}^2$ 鈥︹�� (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 $螖N$in a container over a time interval $螖t$is

$\frac{螖N}{螖t}=鈭�\frac{PA}{2m\stackrel{炉}{v_x}}$ 鈥︹�� (1)

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

The average velocity of molecules is expressed as

$\sqrt{{v_x}^2}=\sqrt{\frac{kT}{m}}$ 鈥︹�� (2)

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

The ideal gas equation can be expressed as

$PV=NkT$ 鈥︹�� (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)

$\frac{螖N}{螖t}=鈭�\frac{PA}{2m}\sqrt{\frac{m}{kT}}$ 鈥︹�� (4)

Equation (3) can be simplified as

$P=\frac{NkT}{V}$ 鈥︹�� (5)

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

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

$\boxed{\frac{螖N}{螖t}=鈭�\frac{A}{2V}\sqrt{\frac{kT}{m}}N}$ 鈥︹�� (6)

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

Differentiate equation (6) on both sides of it,

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

$\ln \left(\frac{N\left(t\right)}{N\left(0\right)}\right)=鈭�\frac{A}{2V}\sqrt{\frac{kT}{m}}t$ 鈥︹�� (7)

Take exponential on both the sides of equation (7)

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

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

Mass of molecules can be obtained as

$m_{molecules}=\frac{\text{尘补蝉蝉鈥塷蹿鈥塷苍别鈥刴辞濒别鈥塷蹿鈥塧颈谤}}{\text{苍耻尘产别谤鈥塷蹿鈥尘辞濒别肠耻别濒蝉鈥塱苍鈥塷苍别鈥尘辞濒别鈥�}{N}_A}$ 鈥︹�� (1)

Characteristic time at room temperature$T=293\text{鈥块}$ is

$蟿=\frac{2V}{A}\sqrt{\frac{m}{kT}}$ 鈥︹�� (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$28.969脳{10}^3$ for the mass of one mole of air and$6.02脳{10}^{23}$ for$N_A$ in equation (1)

role="math" localid="1651732809570" $\begin{array}{l}{m}_{molecules}=\frac{28.969脳{10}^3}{6.02脳{10}^{23}}\\ =4.81脳{10}^{鈭�26}\text{鈥塳驳}\end{array}$

Characteristic time at room temperature $T=293\text{鈥块}$is calculated by substituting $1脳{10}^{鈭�3}$kg for volume, $1脳{10}^{鈭�6}\text{鈥尘}$for the area, $4.81脳{10}^{鈭�26}\text{鈥塳驳}$for the average mass of molecules in equation (2)

$\begin{array}{l}蟿=\frac{2脳1脳{10}^{鈭�3}}{1脳{10}^{鈭�6}}\sqrt{\frac{4.81脳{10}^{鈭�26}}{1.38脳{10}^{鈭�23}脳293}}\\ \boxed{蟿=6.9\text{鈥塻}}\end{array}$

Hence, the characteristic time is $\boxed{蟿=6.9\text{鈥塻}}$.

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

Characteristic time at room temperature$T=293\text{鈥块}$ is

$蟿=\frac{2V}{A}\sqrt{\frac{m}{kT}}$ 鈥︹�� (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

$d=2\sqrt{\frac{A}{蟺}}$ 鈥︹�� (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. $蟿=3600\text{鈥塻}$.

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

$A=\frac{2V}{蟿}\sqrt{\frac{m}{kT}}$ 鈥︹�� (3)

Substitute$1.4脳{10}^{鈭�3}{\text{鈥尘}}^{\text{3}}$ for volume, 3600 s for $t$,$4.81脳{10}^{鈭�26}\text{鈥塳驳}$ for $m$,$1.38脳{10}^{23}{\text{鈥塉碍}}^{\text{-1}}$ for $k$, and$293\text{鈥块}$ for$T$ in equation (3)

$\begin{array}{l}A=\frac{2脳1.4脳{10}^{鈭�3}}{3600}\sqrt{\frac{4.81脳{10}^{鈭�26}}{1.38脳{10}^{鈭�23}脳293}}\\ A=2.68脳{10}^{鈭�9}{\text{鈥尘}}^{\text{2}}\end{array}$

The hole diameter is calculated by substituting$2.68脳{10}^{鈭�9}{\text{鈥尘}}^{\text{2}}$ for$A$ in equation (2)

$\begin{array}{l}d=2脳\sqrt{\frac{2.68脳{10}^{鈭�9}}{蟺}}\\ \boxed{d=0.058\text{鈥尘尘}}\end{array}$

Hence, the required size of the hole is $\boxed{d=0.058\text{鈥尘尘}}$.

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

Characteristic time at room temperature $T=293\text{鈥块}$is

$蟿=\frac{2V}{A}\sqrt{\frac{m}{kT}}$ 鈥︹�� (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

$A=蟺r^2$ 鈥︹�� (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 $V=150{\text{鈥尘}}^{\text{3}}$. They could open the window, toss out the dog and close the window again within 1 s.

The area of a porthole is

$A=蟺脳{0.2}^2=0.125{\text{鈥尘}}^{\text{2}}$.

Characteristic time is obtained by substituting $150{\text{鈥尘}}^{\text{3}}$for $V$, $0.125{\text{鈥尘}}^{\text{2}}$for $A$, $4.81脳{10}^{鈭�26}\text{鈥塳驳}$for $m$, $1.38脳{10}^{23}{\text{鈥塉碍}}^{\text{-1}}$for$k$and$293\text{鈥块}$for$T$ in equation (1)

$\begin{array}{l}蟿=\frac{2脳150}{0.125}\sqrt{\frac{4.81脳{10}^{鈭�26}}{1.38脳{10}^{鈭�23}脳293}}\\ 蟿=8.234\text{鈥塻}\end{array}$

The ratio of the number of molecules is

$\frac{N\left(t\right)}{N\left(0\right)}=e^{鈭�\frac{1}{蟿}}$ 鈥︹�� (3)

Substitute 8.234 s for $蟿$in equation (3)

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

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