Problems | Kinetic theory of gases
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- Written by: Germán Fernández
- Category: Problems | Gas kinetic theory
- Hits: 149
For 1 mole of oxygen at 300K and 1 atm calculate:
a) The number of molecules whose speeds are between 500,000 and 500,001 m/s (since this speed interval is small, the distribution function practically does not vary, and we can consider it infinitesimal).
b) The number of molecules with $v_z$ between 150000 and 150001 m/s.
c) The number of molecules that simultaneously have $v_z$ and $v_x$ between 150,000 and 150,001 m/s.
Solution:
a) \begin{equation} \frac{dN_v}{N}=G(v)dv \;\rightarrow N_v=NG(v) \Delta v \end{equation} \begin{equation} G(v)=\ left(\frac{m}{2\pi kT}\right)^{3/2}e^{\frac{-mv^2}{2kT}}4\pi v^2 \end{equation}
Substituting, $v=500\;m/s$; $k=1.38x10^{-23}J/s$; $T=300K$; $m=\frac{0.032}{6x10^{23}}=5.33x10^{-26}\;kg$, we obtain: $G(v)=1.85x10^{-3}\;s/m$
Therefore, $N_v=6x10^{23}\cdot (1.85x10^{-3}\;s/m)\cdot (0.001\;m/s)=1.1x10^{18}\;molecules$
b) \begin{equation} \frac{dN_{vz}}{N}=G(v_z)dv_z \;\rightarrow N_{vz}=NG(v_z) \Delta v_z=7.75x10^{17}\;molecules \end{equation} \begin{equation} g(v_z)=\left(\frac{m}{2\pi kT}\right)^{1/2}e^{\frac{-mv_{z}^ {2}}{2kT}} \end{equation} It is solved analogously to part a)
c) \begin{equation} N_{v_xv_yv_z}=Ng(v_x)g(v_z) \Delta v_x \Delta v_z =9.22x10^{11}\;molecules \end{equation}
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- Written by: Germán Fernández
- Category: Problems | Gas kinetic theory
- Hits: 172
Calculate the probability density for the x component of the velocity of a sample of oxygen molecules at 300K in the interval $0<|v_x|<1000$m/s. Represent the resulting function.
Solution:
\begin{equation} g(v_x)=(1.04289x10^{-3}s/m)e^{(-6.4145x10^{-6}m^{-2}s^2)v_{x}^{ 2}} \end{equation}