p(h):pressure n(h):number of particles in unit volume F(h):force applied to the each particle U(h):potential energy of the particle From equation of state, we obtain:
p(h)V=NRT→p(h)=n(h)kbT
Considering equilibrium of forces in the gas between h and h+dh, we obtain:
Sp(h+dh)+n(h)F(h)Sdh=p(h)S→p(h+dh)+n(h)F(h)dh=p(h)→p(h+dh)−p(h)dh=−n(h)F(h)→dpdh=−n(h)F(h)
By substituting (1) to (2),we can derive differential equation (3).
kbTdndh=−n(h)F(h)
ODE (3) can be solved by dividing both sides by n.
1ndndh=−F(h)kbT
Integrating both sides by h results in following.
∫1ndn=−1kbT∫F(h)dh
\begin{equation} \leftrightarrow \int \frac{1}{n}dn=-\frac{1}{k_b...
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