24

$\ce{H2}$ cannot be liquified at room temperature, whatever the pressure. Generally speaking, all gases can only be liquified when the temperature is under its critical value.


22

Hydrogen critical temperature is $\pu{32.938 K, resp. -240.21 ^{\circ}C}$. Above this temperature, it cannot be liquified. So to answer your question, you can get as high pressure as you can produce and the container can withstand, as there is no condensation reducing the pressure. WARNING: An accidental explosive container rupture can easily cause severe ...


9

There is one importing principle to consider. Air does not store vapor. Space does. Less space means less vapor capacity, no matter how much of air is there. (*) If air is being compressed, the partial vapor pressure increases. It it reaches the saturated vapor pressure at given temperature, water vapor starts to condensate. When air is then expanded, its ...


8

I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa. You cannot generally ...


6

As others have said, hydrogen cannot be liquified above its critical temperature, which my source (Wolfram Alpha chemical database*) says is $\pu{32.97 K} = \pu{-240.18 ^\circ C}$ However, with sufficient pressure, the molecules can be squeezed together until they have a liquid-like density**, and are thus no longer considered a gas, but rather a ...


4

A supercritical fluid is just a state of matter. like solid, liquid or gas. Hexane is no more or less toxic if it has become supercritical and then brought back to standard temperature and pressure, as would water, if it changed from liquid to ice or to supercritical "steam" and back to water. Further, supercritical $\ce{CO2}$ is used for dry-...


4

Old question, but still seems to attract some interest. In general, the total pressure of the system is not taken into considerations for corrections. The pressure at which you do the corrections is the partial pressure / concentration of the molecule. I recommend calculating the thermal correction for $G$ at the temperature of interest and the default ...


3

Using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I found the saturation point of nitrous oxide at equilibrium for a temperature of $T=-45.000\ \mathrm{^\circ C}$ at a pressure of $p=783110\ \mathrm{Pa}=7.8311\ \mathrm{bar}$; for $T=-78.000\ \mathrm{^\circ C}$, it is $p=...


3

The lower right part of graph indicates that real gases exert less pressure than ideal gas for a fixed volume. This happens because the ideal gas molecules are not attracted by surrounding molecules (therefore if an ideal gas molecule move at velocity $V_1$, It will strike the wall with same velocity, provided it doesn't collide with another molecule) while ...


3

Summary At sufficiently low volumes, the excluded volume effect causes the van der Waals (VDW) pressure of all gases to exceed the ideal gas (IG) pressure. At higher volumes, the VDW pressure may or may not cross below the IG pressure. This depends on the temperature, and the relative size of the VDW a and b terms. If the temperature is sufficiently low, ...


3

According to their website, the manufacturers say "The Ultra and Max models of Instant Pot come with the ability to adjust pressure settings to account for altitude." So, yes, the timings will be affected by altitude, even in a pressure cooker. This is probably because the pressure inside the cooker is controlled by some method that uses the ...


3

Suppose initially you had 1 mole of $\ce{N2}$ and 3 moles of $\ce{H2}$, and that $n$ moles of $\ce{N2}$ had reacted. Then you would have, at equilibrium, $1-n$ moles of $\ce{N2}$, $3(1-n)$ moles of $\ce{H2}$, and $2n$ moles of $\ce{NH_3}$. So the total moles at equilibrium would be $4-2n$, and the mole fractions (equal to the partial pressures in atm) ...


2

You have differential equations. Differential equations are ALL about the initial conditions, and A PROCESS. So, what’s the initial condition? You have a piston with a given volume of gas. What’s happening then? I apply an additional force with an object. (The atmospheric pressure already was there) And what if you want to know the work made by the ...


2

As usual, I suggest to start with a RICE table to get an overview, denoting initial partial pressures with $p_0$ and the change in partial pressure with $x$: $$ \begin{array}{lccccc} &\text{R} &\ce{&CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)}\\ &\text{I} && p_0(\ce{CH3OH}) && p_0(\ce{NOCl}) &...


2

Said by other words, you do not believe that $3 + 2 = 3 + 2$ There are 2 $\ce{Hg}$ columns, sharing the same pressure at the bottom. Both have the same height, so the pressure difference is the same. Yet you have doubts the pressures at their tops are the same. Pressure at A = Pressure at B +pgh It should be $p_\mathrm{A} = p_\mathrm{B} + \rho gh$ where $\...


2

What I understood: Let's say $\ce{H2O}$ has a (saturated) vapour pressure of $x$ atm at a certain temperature. Now Relative humidity (RH) is the ratio of the partial pressure of water vapour to the equilibrium vapour pressure of water at a given temperature. So what this means is that if RH = 100% then the maximum amount of water vapours is present in the ...


2

Assumptions: Reaction of carbon dioxide with water is neglected. Vapour pressure of water is negligible. Volume of solution does not change on dissolution of carbon dioxide. Moles of carbon dioxide dissolved in water is very less as compared to moles of water and thus, $X_{\ce{CO2{(aq)}}}≈\frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}}$ Initially no $\ce{CO2}$ is ...


2

So you need $$K_p = \frac{p_{\ce{H2S}}^3}{p_{\ce{H2}}^3}$$ where $p_{\ce{H2S}}$ and $p_{\ce{H2}}$ are partial pressures of $\ce{H2S}$ and $\ce{H2}$ respectively. Partial pressure is nothing but the mole fraction of the substance times the total pressure. You already have mole fractions, so that part is done. Now notice that in the expression of $K_p$ , you ...


2

Diffusion is a fundamental molecular phenomenon and it does not have a direction. CASE-1 : I cut open the top of cylinder, now the gas inside experiencees a uniform 1 atm pressure and exerts the same pressure too. As there is no pressure difference so Carbon dioxide should not diffuse out! Why not? Recall if you open a bottle of perfume in a room, ...


2

Fundamentally, the problem is in the problem statement. The proper form for the Van der Waals equation is $[P+(a/V_m^2)](V_m\color{blue}{-b})=RT$ which would give a cubic equation when we try to isolate the molar volume. It is this cubic equation that usually has one or three roots that can be correlated with the possible phases, as described in the other ...


2

As the approximation, the saturated vapour pressure does not depend on the total pressure, only on the substance and temperature. (*) If there is nothing but the vapor in the gaseous phase under the piston then the amount of vapor and its pressure accomodate to external pressure. (**) If the external pressure is kept greater than saturated vapor pressure ...


2

Two carbon dioxide scrubbers come to mind: lithium hydroxide and soda-lime. (https://en.wikipedia.org/wiki/Lithium_hydroxide)"Lithium hydroxide is used in breathing gas purification systems for spacecraft, submarines, and rebreathers to remove carbon dioxide from exhaled gas by producing lithium carbonate and water: 2 LiOH•H2O + CO2 → Li2CO3 + 3 H2O or ...


1

You're right. The piston can gain kinetic energy and overshoot its equilibrium position, but then, the force of the gas on the piston will drop until the piston comes back the other way. The piston will undergo an oscillation back and forth, but this oscillation will be damped as a result of viscous dissipation stresses in the gas. Eventually, the ...


1

The easiest way to prove the equality in pressure is by taking a cross-section of the narrow tube and finding the forces acting on it. By extending this, we can prove that the pressures are equal. Take a cross-section area of the narrow tube. We assume this is at steady state and there is no average motion of particles in the entire system. Now, since there ...


1

Why should they ( supposing ideal gas behaviour ) ? If a volume, containing $\pu{4 mol}$ $\ce{N2}$ and $\pu{1 mol}$ $\ce{O2}$ has pressure $\pu{5 atm}$, partial pressures are: $\ce{N2} : \pu{4 atm}$ $\ce{O2} : \pu{1 atm}$ If you add $\pu{5 mol}$ $\ce{Ar}$, the total pressure raises to $\pu{10 atm}$, with partial pressures : $\ce{Ar} : \pu{5 atm}$ $\ce{N2} : ...


1

What happens during a reaction is important insofar as reversibility is concerned (a reversible path being a special way of performing a reaction in which equilibrium is sustained throughout), but not when computing $\Delta H$. The condition $\Delta H = q$ is derived from the basic law of conservation of energy and the definition of enthalpy. Because H is a ...


1

At the interface between the gas and its surroundings, by Newton's 3rd law, the pressure exerted by the gas on its surroundings is equal to the pressure exerted by the surroundings on the gas. So we can use either. For a reversible expansion, the gas pressure is given by the ideal gas law. But the ideal gas law only applies at thermodynamic equilibrium ...


1

Your terminology makes seeing the correct picture harder. Part of the problem is that the pressure at A is could be at any point in the column of mercury and that is not constant. The pressure in the column of mercury varies with height (depending on the weight of mercury above that point). When you fill the tube with mercury and invert it there is no ...


1

Jon Custer helpfully noted (in a comment) two common production processes that result in compression of the final product. I did some research to discover the typical pressures involved. Tempered Glass Glass may be tempered by cooling the outer layer quickly, while letting the inner layer cool slowly. As the inner layer cools, it shrinks; but because the ...


1

The total work ( and its value for the constant force case ) for pressure $$p=p_\mathrm{force} + p_\mathrm{atm}$$ $$W_\mathrm{tot} = - \int_{V1}^{V2}{p \cdot \mathrm{d}V} $$ is shared between the source of the explicit mechanical force acting on piston and atmosphere. Atmosphere would do work $$W_\mathrm{atm} = - \int_{V1}^{V2}{p_\mathrm{atm} \cdot \...


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