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

Question: Can $\ce{SiO2}$ melt at $\pu{20 ^\circ C}$? According to experimental and calculated data values, my answer is no. See the phase diagram of pure silica based on the experimental and calculated data given in Ref.1: Reference 1 states that: An internally consistent data set on the thermodynamic properties of the silica polymorphs stable up to $\pu{...


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 ...


5

Calderon & Mohazzabi give an excellent summation of the various theories proposed through the years to explain why ice is so slippery in their 2018 paper "Premelting, Pressure Melting, and Regelation of ice revisited" in the Journal of Applied Mathematics and Physics. They offer both theoretical and experimental evidence that neither pressure ...


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 ...


4

The reaction of interest is: $$\ce{XY(s) <=> X(g) + Y(g)} \tag1$$ Thus, $K_p = P_\ce{X}\cdot P_\ce{Y} = 4.1$ since $P_\ce{XY} = 1$ in given condition because it is a solid. If we assume $\ce{X}$ and $\ce{X}$ are real gases, the given conditions are such that $P_\ce{X} = \frac{n_\ce{X} RT}{V} = n_\ce{X}$ and $P_\ce{Y} = \frac{n_\ce{Y} RT}{V} = n_\ce{Y}$,...


4

The problem is that, when the flask is empty, it weighs $85.431\ \mathrm g$. But this weight is the weight of the empty flask plus the air included. So the first thing to do is to calculate the weight of the air, and the corresponding number of moles of air in the flask before adding $\ce{CO2}$. The volume of the air is the same as the volume of water when ...


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-...


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

This is an approach to Question 2. The reaction equilibrium condition for the system is: $$\frac{V}{RT}\frac{n_C}{n_A n_B}=K_P\tag{1}$$Let $n_{A0}$, $n_{B0}$, and $n_{C0}$ be the number of moles at equilibrium for each species when the volume is equal to $V_0$. Then we have $$\frac{V_0}{RT}\frac{n_{C0}}{n_{A0} n_{B0}}=K_P\tag{2}$$. For a small change in ...


3

There's a very nice—and fairly accessible—discussion of your third question in Nature Magazine's News and Views Section[1]. In this article, Bonn discusses the results of a fairly recent work that appeared in Physical Review X[2]. Bonn summarizes the key points as follows: The idea that a thin film of meltwater wets the surface of ice has been accepted ...


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

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

Yes, I agree with your teacher's comment that upon increasing the total pressure of a system, there are select reactions "more dependent on collisions", and such reactions have been a matter of study. For example, per this 2008 work reported in 'The Journal of Physical Chemistry', The Temperature and Pressure Dependence of the Reactions H + O2 (+M)...


2

The pressure underwater increases by roughly 1 bar every 10 meters, that is, ambient pressure is doubled every 10 meters compared to standard atmospheric pressure. When you breathe air underwater using a SCUBA tank, your lungs expand against the surrounding ambient pressure in order to increase their volume. As the lungs expand they fill with gas which props ...


2

The temperature and the volume of the inner ear are constant. When your ears pop during descent, air from the cabin goes into the ear, increasing the pressure. The law is the following: $$n / P = const$$ You can derive this from the ideal gas law. It has no special name.


2

There is no contradiction. Each law is valid under a certain conditions. This is because each law of these assumes the constancy of one of the three following variables (the pressure , the volume , the absolute temperature) and then studies the relation between the other two. Isn’t this a contradiction? This is how scientist design experiments. Suppose a ...


2

The root of your confusion, as I understand, is that you are not much familiar with other kinds of work done. Although, $p\,\mathrm{d}V$ is everyone's favourite but it's not the only one in the league. Always remember, the $p\,\mathrm{d}V$ is work done during an expansion/compression. You can change the pressure or concentration to get electric work.


2

Since you are given the volume, temperature and initial amount of each gas, you can compute initial partial pressures as $n_iRT/V$ and from these the product $Q_p=p_X\cdot p_Y$ and compare this to $K_p$. If $Q_p>K_p$ then solid will form: a. $\pu{5.0 mol}$ of $\ce{X}$, $\pu{0.5 mol}$ of $\ce{Y}$ $Q_p=2.50 \rightarrow$ no solid is formed b. $\pu{2.0 mol}$ ...


2

Assuming $\ce{A, B,}$ and $\ce{C}$ are gases and behave like ideal gases we can conclude following by applying $P_xV=n_xRT$ where $x$ is either $\ce{A, B,}$ or $\ce{C}$, and $V$ and $T$ are constants: $$[\ce{A}] = \frac{n_A}{V}=\frac{P_A}{RT}; \quad [\ce{B}] = \frac{n_B}{V}=\frac{P_B}{RT}; \quad \text{and } \quad [\ce{C}] = \frac{n_C}{V}=\frac{P_C}{RT} $$ ...


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

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

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

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

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, ...


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