equation of state for ideal gas
micro view
$pV=NkT$,where $k=\frac{R}{N_A}$ is Boltzmann Constant.
In addition, we usually denote the mass of a molecule by $m$.
MACRO VIEW
$pV=nRT$,$R=kN_A$,$M=mN_A$
equation of state for real gases
- the Virial expansion
$pV=nRT(1+B_1\frac{n}{V}+B_2(\frac{n}{V})^2+...)$, where $B_1,B_2,...$ are called $Virial\ coefficients$. - the van der Walls equation
$(p+\frac{an^2}{V^2})(V-nb)=nRT$
pressure
$p=\frac{1}{3}\rho\bar{v^2}$($\mathbb{NOTICE}$:mean after square)
- deviration
$collision:v_x\rightarrow-v_x\Rightarrow F_{ix}\Delta t=-2m_iv_{ix},\Delta t=\frac{2x}{v_{1x}}$
$F_i=-F_{ix}=\frac{m_iv_{ix}^2}{x},p=\frac{\Sigma_iF}{A}=\frac{\Sigma_i m_iv_{ix}^2}{xA}=\frac{m_{total}\bar{v_x^2}}{V}=\rho\bar{v_x^2}$
$suppose:\bar{v_x^2}=\bar{v_y^2}=\bar{v_z^2},therefore:p=\frac{1}{3}\rho\bar{v^2}$
kinetic energy $\bar{\epsilon _t}=\frac{1}{2}m\bar{v^2}$
Also $p=\frac{2}{3}\frac{N}{V}\bar{\epsilon _t}$
root mean square speed $v_{rms}$
$v_{rms}=\sqrt{\bar{v^2}}=\sqrt{\frac{3p}{\rho}}$
mean free path $\lambda$
$\lambda = \frac{kT}{\sqrt{2}\pi d^2p}$, where $d$ be the diameter of the gas molecule
what if such a $\lambda > L$, where $L$ is the length of volume?
If so, we let $\lambda = L$
- deviration
speed distribution law
$f(v),N(v)=Nf(v)$, where $N$ is the number of gas molecule, $v\in[0,+\infty)$
- explanation
The percentage of gas molecule whose $v \in [v_a,v_b]$ is $\int_{v_a}^{v_b}f(v)dv$
The number of ... is $\int_{v_a}^{v_b}N(v)dv=N\int_{v_a}^{v_b}f(v)dv$
most probable speed $v_p$
$v_p:f(v_p)=max\{f(v):v\in[0,+\infty)\}$
average speed $v_{av}$
$v_{av}=\int_0^{+\infty}vf(v)dv$
root mean square speed $v_{rms}$
$v_{rms}=\sqrt{\int_0^{+\infty}v^2f(v)dv}$
Maxswell speed distribution law
$$ f(v)=4\pi {(\frac{m}{2\pi kT})}^{\frac{3}{2}}v^2e^{-\frac{mv^2}{2kT}}\\ v_p=\sqrt{\frac{2kT}{m}}=\sqrt{\frac{2RT}{M}}\\ v_{av}=\sqrt{\frac{8kT}{\pi m}}=\sqrt{\frac{8RT}{\pi M}}\\ v_{rms}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3RT}{M}}\\ $$
energy distribution law
$f(E)=f(v)\frac{dv}{dE},E=\frac{1}{2}mv^2,v=\sqrt{\frac{2E}{m}},\frac{dv}{dE}=\sqrt\frac{1}{2mE}$
- Maxswell-Boltmann energy distribution law $f(E)=\frac{2}{\sqrt\pi}\frac{1}{{(kT)}^{\frac{3}{2}}}E^{\frac{1}{2}}e^{-\frac{E}{kT}}$
- gravity field omitted, which is in the note & PPT
intermolecular force
potential energy $U(x) = \frac{\lambda}{x^s}-\frac{\mu}{x^t}$
diatomic molecule
$U(x)=\frac{a}{x^{12}}-\frac{b}{x^6}$