- #1

Heisenberg2001

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- Homework Statement
- Consider a monoatomic gas, whose momentum relation is ##\vec{p}=m\vec{v}##, where ##\vec{v}## is the three-dimensional velocity of the gas particles.

1. Show that the classical partition function can be written as $$z=\frac{4\pi m^3}{h^3}\int dv v^2e^{\beta \frac{mv^2}{2}}.$$

2. Let ##f(v)=Av^2e^{\beta \frac{mv^2}{2}}.## Find ##A## such that ##\int_{0}^{\infty}dv f(v)=1.##

- Relevant Equations
- ##\vec{p}=m\vec{v}\, , \,z=\frac{4\pi m^3}{h^3}\int dv v^2e^{\beta \frac{mv^2}{2}}##

##f(v)=Av^2e^{\beta \frac{mv^2}{2}}\, , \,\int_{0}^{\infty}dv f(v)=1##

1.

##\vec{p}=m\vec{v}##

##H=\frac{\vec{p}^2}{2m}+V=\frac{1}{2}m\vec{v}^2##

##z=\frac{1}{(2\pi \hbar)^3}\int d^3\vec{q}d^3\vec{p}e^{-\beta H(\vec{p},\vec{q})}##

##z=\frac{Vm^3}{(2\pi \hbar)^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}##

##z=\frac{Vm^3}{(2\pi \frac{h}{2\pi})^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}##

##z=\frac{Vm^3}{h^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}##

I am lost at this point of the solution. Can someone assist me with this?

2.

##\int_{0}^{\infty}dv f(v)=1##

##\int_{0}^{\infty}Av^2e^{\beta \frac{mv^2}{2}}dv=1##

##A\int_{0}^{\infty}v^2e^{\beta \frac{mv^2}{2}}dv=1##

##A\frac{1}{4\left(\frac{\beta m}{2} \right)}\sqrt{\frac{\pi}{\frac{\beta m}{2}}}=1##

##A=\sqrt{\frac{4\beta ^3m^3}{2\pi}}##

##A=4\pi \sqrt {\left(\frac{\beta m}{2\pi} \right)^3}##

##A=4\pi \left( \frac{m}{2\pi k_BT} \right)^\frac{3}{2}##

##\vec{p}=m\vec{v}##

##H=\frac{\vec{p}^2}{2m}+V=\frac{1}{2}m\vec{v}^2##

##z=\frac{1}{(2\pi \hbar)^3}\int d^3\vec{q}d^3\vec{p}e^{-\beta H(\vec{p},\vec{q})}##

##z=\frac{Vm^3}{(2\pi \hbar)^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}##

##z=\frac{Vm^3}{(2\pi \frac{h}{2\pi})^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}##

##z=\frac{Vm^3}{h^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}##

I am lost at this point of the solution. Can someone assist me with this?

2.

##\int_{0}^{\infty}dv f(v)=1##

##\int_{0}^{\infty}Av^2e^{\beta \frac{mv^2}{2}}dv=1##

##A\int_{0}^{\infty}v^2e^{\beta \frac{mv^2}{2}}dv=1##

##A\frac{1}{4\left(\frac{\beta m}{2} \right)}\sqrt{\frac{\pi}{\frac{\beta m}{2}}}=1##

##A=\sqrt{\frac{4\beta ^3m^3}{2\pi}}##

##A=4\pi \sqrt {\left(\frac{\beta m}{2\pi} \right)^3}##

##A=4\pi \left( \frac{m}{2\pi k_BT} \right)^\frac{3}{2}##

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