13

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 ...


13

$E=\frac 12mv^2 => v=\sqrt{\frac{2E}{m}} $ is valid for translational kinetic energy and the speed of the centre of mass. Vibrational or rotational energy does not count. An object may vibrate or rotate even if it's centre of mass has zero speed. As each available degree of freedom has the mean energy $E=\frac 12kT$, and as there are 3 independent ...


9

The authors attempt to explain that in many sources the term "ideal gas" is used in place of "perfect gas" to indicate a gas following the ideal gas law and which has the property that the molecules do not interact. They equate the absence of an intermolecular interaction with the statement that the "interactions are zero". In other words, there is no ...


7

Why thermodynamical equations are just for gases? They are not. The equation $\Delta U = q + P\Delta V$ applies to any phase (gas, liquid, solid...) when only pV work is done. In the particular form of the equation you present, the pressure is in addition constant during the work. Gases are (1) an easy way to introduce thermodynamics concepts because ...


6

Yes. But with several complications you don't normally use the gas laws to deal with The ideal gas law is useful in laboratory situations where you can control the various components in the law (volume, pressure, amount of gas, temperature). And it usually describes the equilibrium reached when those factors are held constant. In the situation you describe, ...


6

I give you a case. Suppose you opened an airtight jar containing an ideal gas. Suppose by some mysterious power, you are able to observe the molecules of the gas. What do you think you'll see on opening the jar? Obviously you'll see the molecules dissipating in space(the correct word would be diffusing). Initially they suppose on opening occupied 100ml, then ...


6

In the simplest model, a gas is called ideal when its particles are point-like (no volume) and have no interactions. Real gases behave like ideal gases at low pressure (where the particle volume is neglible compared to the total volume) and high temperature (where condensed phases, i.e. interatomic or intermolecular interactions are disfavored). The size-...


5

The typical gas sensor marketed to hobbyists and enthusiasts (except for $\ce{CO2}$) are of the metal oxide semiconductor type, not to be confused with MOS as in MOSFET. How do these work? In these detectors, a semiconductor material made of the oxide of a metal, usually tin. I'll quote from this question: ...sintered composite based on the ...


5

Poor carbon dioxide has been unnecessarily defamed as a greenhouse effect. This is not only gas in the atmosphere which is an infrared absorber. Guess what, water vapor is a major culprit as well along with methane (recall cows digestive system). Nature has a very delicate balance, carbon dioxide is plant's food as well. I recall some people doing PhDs on ...


5

For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium: $$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$ Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$: $$p_1 = x_1p\tag{1}$$ To find ...


5

According to the equipartition theorem, energy is shared equally among all accessible degress of freedom. A monatomic ideal gas has three translational degrees of freedom, so each translational degree of freedom has an energy of $\frac12 k T$, for a total translational energy of $\frac32 k T$. A diatomic ideal gas has three translational degrees of ...


4

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

In case you're actually trying to design a serious scheme for extracting and storing carbon dioxide, you might want to start by reading the Wikipedia articles on carbon capture and storage and carbon sequestration. For the impatient, the summary is that it's simply not practical to store that much $\ce{CO2}$ in containers of any kind. Instead, you either: ...


4

There is no work done on moving a piston against zero external pressure (not if it is ideal, ie lacks mass and there is no friction in its movement). You might consider the definition of $pV$ work $$w = -\int_{V_\mathrm{ini}}^{V_\mathrm{fin}}p_\mathrm{ext} \mathrm{d}V$$ Obviously if the external pressure $p_\mathrm{ext}=0$ then $w=0$. Conceptually, the ...


4

The heat capacity comprises contributions from translational motion (3 terms for each of $x, y, z$) plus (whole body) rotational motion ( 3 axis directions). Each motion accounts for $(1/2)R$ by equipartition theorem making the $(3/2)R + (3/2)R$ above. Then there is a contribution from the fact that the molecule is vibrating. Non-linear molecules have $3N-...


4

A straightforward way to evaluate ideality is to compute the compressibility Z: $$Z=\frac{PV_m}{RT}$$ Z equals 1 for an ideal gas, so deviations from this condition serve as a measure of non-ideality. If you examine a plot of compressibility for a real gas you will in general notice the existence of two regimes: at low pressure the compressibility is ...


4

Given you know and understand Charles' and Gay-Lussac's laws, it's not about chemistry, rather, simple ratios: $$ \begin{cases} T \propto V\\ T \propto p \end{cases} \implies p \propto \frac 1 V $$ which, as Zenix commented, is a math form of Boyle's law.


4

You can liquify ethane at standard pressure (1 bar) simply by lowering the temperature (cooling), as indicated by its phase diagram (see for instance here). On the other hand, if you look at the phase diagram of $\ce{CO2}$ you find that starting at standard pressure (1 bar) and room temperature, it cannot be cooled into the liquid state. Rather it will ...


3

Yes, they are related. The first comes directly from the conservation of number of moles of the solute in a dilution, $$n_1 = n_2 $$ Since $n = MV$, $$M_1 V_1 = M_2 V_2$$ The second is related to the conservation of the total number of moles in an isothermal compression or expansion, $$n_1 = n_2$$ Using the ideal gas law $n = pV/RT$, $$\frac{P_1 V_1}{...


3

Usage of ppm/ppb etc units is discouraged, as they are ambiguous without the explicit context. As generally, 1 ppm (w/w) <> 1 ppm (V/V) <> 1 ppm (n/n) <> 1 ppm (w/V). It is recommended to use explicit units, like e.g mg/L. Expecting ideal gas behaviour $pV = nRT$ is not reasonable. Ideal gases are not in equilibrium with their liquid phase. ...


3

A combination of gravity and the suction you apply with your lips ("suck") help empty the cup, assuming as you do that the coffee in the cup leaves a vacuum as it exits. This is similar to the "problem" of emptying a bottle of liquid when you invert it (the mouth facing downward). The challenge is most evident with honey or ketchup, viscous liquids or those ...


3

I'm a bit confused by your question -- your diagram shows the path of refrigerant through the cooling system, but you seem to be discussing water's state changes. Liquid refrigerant under pressure enters the metering device. As it leaves the device at a lower pressure, it boils, producing some vapor, but not evaporating completely. The rest of the liquid ...


3

I would approach this a little differently. I would integrate the pressure between $V_D$ and $V_B$, such that the average pressure is 17.76 bars: $$\frac{\int_{V_D}^{V_B}{PdV}}{(V_B-V_D)}=17.76$$or, $$17.76(V_B-V_D)=RT\ln{\frac{(V_B-b)}{(V_D-b)}}-a\left[\frac{(V_B-V_D)}{(V_B-b)(V_D-b)}\right]$$ This would provide an equation for expressing a in terms of b. ...


3

Well, an ideal gas has no attraction force between its molecules and does not have a boiling point, as you cannot make it liquid. Also, the sum of molecule own volumes is supposed negligible wrt the gas volume. Water does not form anything close to an ideal gas because of its hydrogen bonds. You have to heat water to 100 °C to make it boiling to overcome ...


3

No answer is correct, but "A" is closest to the correct answer. Heavier molecules diffuse slower than lighter ones because the same amount of thermal energy has to move more mass. However, this relation (diffusion rate is inversely proportional to the square root of molecular weight) actually leads to 44 seconds not 64. The professor who told you "D" ...


2

The acidity of HCl is so great compared to H2CO3 that you have no equilibrium between Na2CO3 and HCl; the equilibrium is only between CO2 and water and the vapor space. H2CO3 is one of the forms of CO2 in water, but it is unnecessary to delve into this. After your reaction, you will have a solution of NaCl in which some of your CO2 is dissolved; the rest of ...


2

Both, since atmospheric pressure does not exist without gravity. Atmospheric pressure is essentially the weight of air above something, which happens because of gravity.


2

Reason for rewrite: refer to edit summary. From Wikipedia on Disulfur ($\ce{S2}$) This violet gas is commonly generated by heating sulfur above 720 °C Since the sulfur is well above $720^\circ\pu{C}$, it would most likely be in the diatomic state, more like an analogous structure to dioxygen ($\ce{O2}$). We can also judge this from the molar ratios ...


2

Interaction between atom or molecule is neglected True. In an ideal gas the gas particles are unaware of each other's existence. The transformation to liquid or solid is observed under appropriate condition False. As mentioned in the answer to (1), the particles don't sense each other. Since there are no attractive interactions there is ...


2

As in how much you could compress $\ce{CO2}$ would be the point when it gets converted to dry ice. Surely a solid would take even lesser space than liquid. As for liquid, you need to maintain a pressure of approx 5 atm at 31.1 °C to liquify $\ce{CO2}.$ If you want, you can have a look at the article World can ‘safely’ store billions of tonnes of CO2 ...


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