91影视

The average distancebetween particles is proportional cube root of me volume$\mathit{V}$per atom.

$\mathit{d}\mathbf{鈭�}{\mathbf{\left(}\frac{\mathbf{V}}{\mathbf{N}}\mathbf{\right)}}^{\mathbf{1}\mathbf{/}\mathbf{3}}\mathbf{鈥�}\mathbf{鈥�}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here$\mathit{N}$is the number of particles.

A variation of the ideal gas law states that the product of the pressure$\mathit{P}$and volume$\mathit{V}$is equal to the product of the number of particles$\mathit{N}$Boltzmann's constant${\mathit{k}}_{\mathbf{B}}$and temperature$\mathit{T}\mathbf{.}\mathit{P}\mathit{V}\mathbf{=}\mathit{N}{\mathit{k}}_{\mathbf{B}}\mathit{T}\mathbf{鈥�}\mathbf{鈥�}\mathbf{.}$(2)

The de Broglie's wavelength$\mathbf{\left(}\mathit{位}\mathbf{\right)}$of an object is inversely proportional to the root of the mass$\mathit{m}$.

Boltzmann's constant${\mathit{k}}_{\mathbf{B}}$and temperature$\mathit{T}$of the particle.

$\mathit{位}\mathbf{=}\frac{\mathbf{h}}{\sqrt{\mathbf{3}\mathbf{m}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}}$

Here,$n$is Planck's constant.

The condition to see it an object can be treated classically is,

${\mathbf{\left(}\frac{\mathbf{位}}{\mathbf{d}}\mathbf{\right)}}^{\mathbf{3}}\mathbf{<}\mathbf{<}\mathbf{1}$

(a)

Given data:

Convert of the molecular mass of nitrogen from$u$to$\text{kg}$.

$\begin{array}{c}m=28u\\ =\left(28u\right)\left(\frac{1.66脳{10}^{鈭�27}\text{kg}}{1u}\right)\\ =4.65脳{10}^{鈭�26}\text{kg}\end{array}$

Rearrange equation (2) for$\frac{V}{N}$.

$\frac{V}{N}=\frac{k_{B}T}{P}$

Substitute$\frac{k_{B}T}{P}$for$\frac{V}{N}$in equation (1).

$d鈭�{\left(\frac{k_{B}T}{P}\right)}^{1/3}鈥�$

Substitute$1.38脳{10}^{鈭�23}\text{J}/K$for$k_B,300K$for$1.01脳{10}^5\text{N}/{\text{m}}^2$andfor

$\begin{array}{c}P.\frac{V}{N}=\frac{(1.38脳{10}^{鈭�23}J/K)\left(300K\right)}{1.01脳{10}^5\text{N}/{\text{m}}^2}\\ =4.09脳{10}^{鈭�26}\text{N}/{\text{m}}^2\end{array}$

Since the nitrogen is$80\%$of the atmosphere, its partial pressure will be$80\%$that of atmospheric pressure.

$\begin{array}{c}{P}_N=80\%\left(P\right)\\ =\left(0.80\right)(1.01脳{10}^5\text{n}/{\text{m}}^2)\\ =0.808脳{10}^5\text{N}/{\text{m}}^2\end{array}$

Substitute$1.38脳{10}^{鈭�23}\text{J}/\text{K}$for$k_B,300\text{K}$for$T$and$0.808脳{10}^5\text{N}/{\text{m}}^2$for partial pressure due to nitrogen$\left(P\right)$in equation (3).

$\begin{array}{c}d={\left(\frac{(1.38脳{10}^{鈭�23}J/K)\left(300K\right)}{0.808脳{10}^{5}N/m^2}\right)}^{\frac{1}{3}}\\ =3.71脳{10}^{鈭�9}\text{m}\end{array}$

(b)

Substitute$6.63脳{10}^{鈭�34}\text{j}.\text{s}$ for$h,4.65脳{10}^{鈭�26}\text{kg}$for$m,1.38脳{10}^{鈭�23}\text{J}$

I $K$for role="math" localid="1658385704993" $k_B$and 300 k for$T$

$\begin{array}{c}位=\frac{(6.63脳{10}^{鈭�34}j.s)}{\sqrt{3(4.65脳{10}^{鈭�26}\text{kg})(1.38脳{10}^{鈭�23}\text{J}/K)\left(300K\right)}}\\ =2.76脳{10}^{鈭�11}\text{m}\end{array}$

$d$(c)

The condition to see it an object can be treated classically is,

${\left(\frac{位}{d}\right)}^3<<1$

Here,$d$is the separation that it has with other objects. If the statement is true, then the object can be treated as a classical gas. Otherwise, it's a quantum gas.

Calculation:

Substitute$2.76脳{10}^{鈭�11}\text{m}$for$位$and$3.71脳{10}^{鈭�9}\text{m}$forin left hand side of in equality.

$\begin{array}{c}{\left(\frac{位}{d}\right)}^3<<1{\left(\frac{位}{d}\right)}^3\\ ={\left(\frac{2.76脳{10}^{鈭�11}\text{m}}{3.71脳{10}^{鈭�9}\text{m}}\right)}^3\\ =4.12脳{10}^{鈭�7}\end{array}$