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31

The atmosphere actually loses gases to outer space. The average velocity $\bar v$ of gas molecules is determined by temperature $T$. However, not all the molecules travel with the same velocity. The probability of finding a molecule with a velocity near $v$ is described by the Maxwell distribution of speeds $$\begin{align} f{\left(v\right)}&=4\pi\sqrt{{\...


12

Yes. Any fluid with a temperature is above critical temperature and the pressure above the critical pressure is by defintion a supercritical fluid. Don't be mislead by all the claims that supercritical fluids are special and wonky with all sorts of amazing, bizarre properties. This is true of some supercritical fluids near the critical point, but the ...


11

Yes gravity pulls on gas molecules. That is why the atmosphere doesn't just float off into space. The gist is that the time between collisions is very short in the lower atmosphere, and the distances very short. The mean free path at atmospheric pressure is only about 70 nanometers. So the assumption is that gas particles travel in a straight line between ...


10

Your point 1 is mistaken (or incomplete) $$P=\frac{F}{A}=\frac1A\frac{\Delta p}{\Delta t}=\frac1A\times\frac{2nmv}{\Delta t}$$ This is assuming all particles hit perpendicularly (you can always modify this for all colissions by taking the average component of velocity in the perpendicular direction). So, pressure is proportional to number of impacts, mass, ...


9

If one has a probability density function $P(x)$, then the expectation value of a quantity $f(x)$ is given by $$\langle f \rangle = \int f(x)P(x)\,\mathrm{d}x$$ evaluated over the limits of the probability density function, i.e. if your PDF runs from $-\infty$ to $\infty$, then those are your limits of integration. In our case, the PDF is the Maxwell-...


8

I'm going to go against the grain here: the Maxwell distribution does describe the distribution of molecule speeds in any (3-D) matter, regardless of phase. Suppose we have a system of $N$ molecules with masses $m_i$, positions $\vec{r}_i$, and velocities $\vec{v}_i$ (with $i = 1, ..., N$). Assume that the total energy of this system is of the form $$ E(\...


8

A simple validation The result you quoted is the average translational kinetic energy for an ideal gas. First, let's sketch out a rough derivation for the average kinetic energy of a particles of an ideal gas using nothing more than high school physics. The gas molecules just undergo translations, and don't rotate/vibrate. Also, they don't interact with ...


8

Note: You can skip section I, and go straight to section II and/or the end of section III (specifically the conclusions subsection), if you are already familiar with the basic mathematical machinery/definitions I. Preliminaries Feel free to skip all of this if you are familiar with it First, we have an adiabatic expansion, i.e $\mathrm{q} = 0 $. Second, ...


7

I'm willing to put myself out there and say that no, a Maxwell-Boltzmann distribution will not be able to make any statement about the ability of liquids or solids to react. Meaning also that you could not study reaction rates with a Maxwell-Boltzmann distribution for the condensed phases. The reason why I am fairly confident this is true is because the ...


7

The average translational kinetic energy of a molecule is $3kT/2$ irrespective of whether the molecule is in the gas, liquid, or solid phase. In the liquid the motion giving rise to kinetic energy is restricted to a narrower range about the potential energy minimum than it is in the gas phase. The equipartition theorem is quite general. A derivation is ...


6

Pressure is the number of collisions with the container per unit area. Imagine a particle in a box of length $L$. Assuming for the moment that it only moves in the $x$-direction, whenever it collides with a wall of the container the wall will gain momentum from the particle: \begin{align} \Delta p &= p_{\mathrm{final}, x} - p_{\mathrm{initial}, x} \\ &...


6

Alright, here's my best attempt at it. Root Mean Squared $v_\mathrm{rms}$ The root-mean-square speed (r.m.s. speed) of the molecules. This quantity is defined as the square root of the mean value of the squares of the speeds,$v$, of the molecules. That is, for a sample consisting of $N$ molecules with speeds $v_{1}, v_2,...,v_N$, we square each speed, add ...


6

$$\ce{Kinetic ~Energy ~= ~1/2 (mass~of ~the~object~ [kg]) * (velocity~of~the~object~[m/sec])^2}$$


6

For a non-linear polyatomic molecule like methane, there are a total of $3N$ degrees of freedom, where $N = 4$ is the number of atoms in the molecule. This is broken down as follows: $$\begin{array}{c|c} \hline \text{Component} & \text{# Degrees of freedom} \\ \hline \text{Translational} & 3 \\ \text{Rotational} & 3 \\ \text{Vibrational} & ...


6

To my understanding, the speed distribution is wider for lighter molecules. The Wikipedia page for the Maxwell Boltzmann distribution has two images at the top that convey this very well. In the first plot, $a=\sqrt{k_bT/M}$ where $M$ is the molar mass and $k_b$ Boltzmann's constant. For a Maxwell-Boltzmann distribution, the variance is given by $$\sigma^2=...


6

What your book states is not generally true. Two counter-examples: Ammonia ($\ce{NH3}$) will liquefy at room temperature and a pressure of approximately $\pu{10bar}$. (CRC Handbook of Chemistry and Physics 44th ed; information as cited on Wikipedia’s data page). Butane ($\ce{C4H10}$) will liquefy at room temperature and a pressure of just over $\pu{2bar}$. (...


6

Molecules have internal energy due to intermolecular interactions, as well as translational kinetic energy, rotational energy, vibrational energy, electronic energy (and if you care to include them, nuclear energy and the mass-energy of the protons, neutrons and electrons themselves). By saying "ideal", energy due to intermolecular interactions is ...


6

Following up on user andselisk's answer ... The problem is weird for several different reasons. First the solution depends on assuming that the velocity of the gas molecules follow the Maxwell-Boltzmann distribution which is weird for only 5 molecules. The problem also doesn't state that the 5 molecules are a sample from a much larger number of molecules. ...


5

The second approach looks more correct to me. The first approach is using raw percentages without converting to a ratio between isotopes, and the ratio rates is going to be directly related to the ratio of isotopes, not raw percentages. From the comments, ed: Well, $1.0043$ is the ratio between rates, so each stage will cause the ratio between isotopes to ...


5

Rather than answer the question numerically I have outlined the four different cases, reversible / irreversible and isothermal / adiabatic. In adiabatic changes no energy is transferred to the system, that is the heat absorbed or released to the surroundings is zero. A vacuum (Dewar) flask realises a good approximation to an adiabatic container. Any work ...


5

It helps to rewrite $$\text{rate} \propto \frac{PA}{\sqrt{TM}}$$ by assuming the ideal gas law holds, as follows: $$\text{rate} \propto \frac{RA}{V_m}\sqrt{\frac{T}{M}}$$ where $V_m$ is the molar volume or inverse of molar particle density. Written this way it is clear that if the particles occupy the same volume, increasing their temperature increases ...


4

The activation energy varies a lot for various reactions. As long as T > 0 K, there is still disordered kinetic energy in the molecules. But the reason gasoline doesn't spontaneously combust, is that the temperature isn't high enough to overcome the activation energy - a spark is needed. So you are onto something, just remember that the activation energy ...


4

I'd go about such a question looking at the easiest thing first: ideal gases. So, what do you know about the relations between $C_{p}$ and $C_{v}$ (both being the molar heat capacities at constant pressure and constant volume, respectively) for ideal gases? For one thing, it is well known that \begin{equation} C_{p} - C_{v} = R \end{equation} with $R$ is ...


4

$$E_k = \frac{1}{2}mu^2$$ where, $m$ is molar mass (unit - kg/mol), and $u$ is root mean square speed. According to wikipedia, The molar mass of atoms of an element is given by the atomic mass of the element multiplied by the molar mass constant. See this page (from wikipedia) for complete explanation.


4

The marked location corresponds to a level of kinetic energy in the reactants sufficient to result in a successful collision (energy wise, it says nothing about orientation). The energy required for a successful collision is the gap in potential energy between the reactants and transition state. A heterogeneous system is a system where the reactants are in ...


4

Actually some do. There is a problem of helium escaping the earth's atmosphere. https://en.wikipedia.org/wiki/Atmospheric_escape There are many other sources about this very problematic occurrence.


4

To add to previous answers, all molecules and atoms are affected by gravity and so the density of the atmosphere is greater at the surface of the earth compared to higher up, which is why climbing on Everest most climbers take extra oxygen (although, remarkably, it has been done without this aid). At room temperature gas molecules have an average thermal ...


4

There is, indeed, a function that plots the graph. There are two common types of graph, one with kinetic energy as the $x$-axis, and one with speed. For the most part, they are qualitatively the same, although there is a slight difference in curvature near the origin. The speed function is much more commonly used, because one can more easily extract useful ...


4

I would like to first point out that the pressure you've determined is too low, most likely because you used only molar mass of protons $\ce{H+}$ instead of average mass for star matter in order to fulfill the condition of plasma quasineutrality: $$\bar{M} = \frac{M(\ce{H+}) + M(\ce{e-})}{2} = \frac{\pu{(1 + 0) g mol-1}}{2} = \pu{0.5 g mol-1} = \pu{5e-4 kg ...


4

The volume stays the same, because the containers stay the same. If the volume of one container would increase, that would mean it would expand. If in a stable container, the gas cannot take more volume. It cannot take less, because a gas always fills the whole container. Temperature can change, pressure can change (for instance by forcing more gas in the ...


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