Tag Info

2

I think there may be some confusion as to terminology: compressibility is defined as $$-\frac 1V \frac{\partial V}{\partial p}$$ where something like temperature or entropy is held constant, whereas the compressibility factor is defined as a certain ratio. (ref: https://en.wikipedia.org/wiki/Compressibility) In the ideal gas, particles do not interact ...

1

The compressibility factor was originally derived from empirical testing of gases to correct for the observed non-ideal behavior at more extreme pressures and temperatures. Although it cannot be derived from first principles in the kinetic theory of gases, the experimentally derived factors can be reconciled using Van Der Waal's equation that deals with the ...

0

You said it yourself the molecules have direction rather than randomly moving about. Picture a wide spot in a river, slow water, maybe eddy currents, random water flow. River narrows, water is directed through the slot. I agree with you, speed before pressure differential. What do you think about this, molecules entering Venturi creating vacuum which sucks ...

0

The question may be about the covariance or otherwise of temperature, in which case have a look here. As well, have a look at the paper "Temperature in special relativity" by J. Lindhard, Physica Volume 38, Issue 4, 5 June 1968, Pages 635-640.

3

In the kinetic theory of gases, you only really define the temperature for molecules that are in constant, random, and rapid motion. So if you have a container with a gas at temperature $T$ you don't change the internal energy of the gas by uniformly moving the container. Uniformly moving the container gives all the molecules a non-zero average motion, but ...

1

I think you can avoid all these troubles if you define the temperature as proportional to the variance of velocity, i.e. $$E[(v-\overline{v})\cdot(v-\overline{v})]=E(v\cdot v)-\overline{v}\cdot\overline{v}$$ Here $E$ means expected value, $v$ ranges over the velocities of the individual particles, and $\overline{v}=E(v)$. Clearly this is frame-...

1

All kinds of weird things happen if you try to define temperature in a moving object. The paradox to me (not a generalized accepted answer) resolves by realizing that temperature should only be defined as measured when the object is stationary. Not only is not a scalar but it is not even well defined for areference frame in relative motion. Is temperature ...

10

Temperature is related to kinetic energy in the rest frame of the fluid/gas. In non-relatvistic kinetic theory the distribution function is $$f(p) \sim \exp\left(-\frac{(\vec{p}-m\vec{u})^2}{2mT}\right)$$ where $\vec{u}$ is the local fluid velocity. The velocity can be found by demanding that the mean momentum in the local rest frame is zero. Then $\vec{u}... 7 [June 19,2016: thoroughly revised, giving a more detailed, comparative presentation and better references] General case. In relativistic thermodynamics, inverse temperature$\beta^\mu$is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the ... 1 Let me take a stab at this (I am not claiming this is correct... but it may guide your thinking). The "proper" definition of the Ohnesorge number is $$\rm{Oh} = \frac{\mu}{\sqrt{\rho\sigma L}}$$ If we multiply top and bottom by$v$- a "characteristic velocity" and by$L^2$- a "characteristic area" and rearrange a bit, we get$$\rm{Oh} = \frac{\mu L^2 \... 2 I think it is wrong to define the temperature by the average energy of the molecule in all frames of reference. The reason for that is clear: take all of your particles and send them at$100 m/s$to the north. This won't make the gas hotter, just like the fan does not cool/heat the air (another great mystery!). The organized movement does not participate in ... 0 The gas as a whole is moving in the$x$direction everywhere, but at different heights it is moving different speeds, so we have$u_x(y)$. Even though the gas as a whole has velocity in the$x$direction only, individual molecules have motion in the$y$direction (and$z$-direction, but that doesn't matter here). So let's suppose that at$u_x(y_0) = 2 m/s$... 1 In order to input in your terms 500rpm you are required to input the actual shaft rpm Plus 500 you seek to add. You must charge your input to the systems energy level. So to answer your question yes your input of 500rpm would slow down the shaft rpm. 0 A gas can be approximated as an ideal gas. You then assume that the particles don't feel each other and that they are infinitely small. The potential is zero. The particles can only have kinetic energy. If you would make a simulation of such a system the particles can literally move through each other. If you make this ideal gas approximation, it is ... 0 You can model the gas a a collection of hard spheres of some radius$r$, and do the correction relative to the limit as$r\to\infty$perturbatively. What you find is that for a fixed pressure and temperature amd number of molecules the first order correction to the law replaces$V$with$V-\frac{4n\pi r^3}{3}\$. This tells you that the volume to use is the ...

Top 50 recent answers are included