How is Chemical Potential Affected by Altitude in an Ideal Gas?

benjibutton
Messages
7
Reaction score
0

Homework Statement



Consider a monatomic ideal gas that lives at a height z above sea level, so each molecule has potential energy mgz in addition to its kinetic energy.(a) Show that the chemical potential is the same as if the gas were at sea level, plus an additional term mgz:μ(z) = -kT ln [V (2πmkT)3/2] + mgz

[N ( h2 ) ](You can derive this result from either the definition μ = -T(∂S/∂N)U,V or the formula μ = (∂U/∂N)S,V).(b) Suppose you have two chunks of helium gas, one at sea level and one at height z, each having the same temperature and volume. Assuming that they are in diffusive equilibrium, show that the number of molecules in the higher chunk isN(z) = N(0)e-mgz/kT

Homework Equations


So, what am I attempting to show? That kinetic potential is constant? What will I differentiate with respect to?

The Attempt at a Solution

 
Physics news on Phys.org
μ(z) = -kT*ln[V/N * (2πmkT/h2)3/2 ] +mgz
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top