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## Hot answers tagged gas-laws

71

The other answers here, describing oxygen toxicity are telling what can go wrong if you have too much oxygen, but they are not describing two important concepts that should appear with their descriptions. Also, there is a basic safety issue with handling pressure tanks of high oxygen fraction. An important property of breathed oxygen is its partial pressure....

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It does. You would find the average percentage of the atmosphere that is argon is very slightly higher at the floor of valleys. However, bear in mind first of all it wouldn't be anywhere near a complete stratification -- a layer of pure argon, then another of pure N2, and so on. A mixture of nearly ideal gases doesn't do that, at least at equilibrium, ...

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The common saying is a hold over from when STP was defined to be $\pu{273.15 K}$ and $\pu{1 atm}$. However, IUPAC changed the definition in 1982 so that $\pu{1 atm}$ became $\pu{1 bar}$. I think the main issue is a lot of educators didn't get the memo and went right along either teaching STP as $\pu{1 atm}$ or continuing with the line they were taught ("$\pu{... 31 The ideal gas law is a very good approximation of how gases behave most of the time There is no logical flaw in the laws. Most gases most of the time behave in a way that is close to the ideal gas equation. And, as long as you recognise the times they don't, the equation is good description of the way they behave. The ideal gas equations assume that the ... 27 Our body is used to the environment around us. Once you change part of the environment, you have to be ready for the consequences. Inhaling pure oxygen is the cause for what is known as oxygen toxicity. Oxygen toxicity is a condition resulting from the harmful effects of breathing molecular oxygen$\ce{(O2)}$at increased partial pressures. High ... 24 Essentially, because the carbon dioxide sublimates from solid (dry ice) to gas at a very low temperature (roughly −78 °C at 1 atm), it causes water vapour in the air to condense, causing a visible fog. Thus what you are seeing is not carbon dioxide, but rather water. When we exhale and it is reasonably warm, the carbon dioxide expelled is roughly body ... 18 The differences in acceleration due to gravity is not the main factor in comparing how accurate the approximation is for each planet. The main factor is the mass of gas each planet's atmosphere contains. Mercury has almost no atmosphere. The total mass of all gas in Mercury's atmosphere is only 10000 kg! The pressure is less than$10^{-14}$bar. The ... 18 A big point of confusion is that it is still taught (at least in the mid-2000's) that STP is defined with respect to$\pu{273 K}$and$\pu{1 atm}$of pressure, or$\pu{1.01325 bar}$of pressure, even though IUPAC changed their definition to be with respect to$\pu{1 bar}$of pressure. By using the ideal gas law on the old STP definition, you get that the ... 17 You must consider this: The question whether a physical system follows a particular law is not a "yes or no" question. There is always an error when you compare what you measure with what the law predicts. The error can be at the 17th digit, but it's still there. Let me quote a very insightful passage by H. Jeffreys about this: It is not true that ... 16 While most everything the previous answer states is correct, I would point out that taking four times the volume of a single particle has nothing to do with experiment and arises mathematically. In deriving the VDW equation, the particles are still assumed to be hard spheres, but this assumption is corrected for with the parameter$a$. The hard sphere ... 16 I didn't know that balloons expanded during the fly because of thermodynamics, and I didn't know how high they can fly, but a rapid search tells that a partially unfilled regular balloon can fly until an altitude of around$\pu{25 km}$. Now,$\pu{25 km}$means that it reaches the first part of the stratosphere, with temperatures of$\pu{-60 ^\circ C}$, that ... 15 Preliminaries Consider$U = U(V,T, p)$. However, assuming that it is possible to write an equation of state of the form$p = f(V,T)$, I don't have to explicitly address the$p$dependence of$U, and I can write the following differential: $$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial V} \right)_T}_{\pi_T} \mathrm{d}V + \underbrace{\... 15 That's because of two reasons. One is entropy, the ultimate force of chaos and disorder. Sure, gases would like to be arranged according to their density, but even above that, they would like to be mixed, because mixing creates a great deal of entropy. If you prevent the mixing, then they would behave just as you expected. Indeed, a balloon filled with \ce{... 14 If the balloon is closed, then yes, both volume and pressure will increase when the gas inside is heated. Let's look at two simpler cases first. If the gas were completely free to expand against ambient pressure (say, inside of a container sealed with a freely moving piston, with no friction), then the heated gas would expand until it created as much force ... 14 The heat capacities are defined as$$C_p = \left(\frac{\partial H}{\partial T}\right)_{\!p} \qquad \qquad C_V = \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{1}$$and since H = U + pV, we have$$\begin{align} C_p - C_V &= \left(\frac{\partial H}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{2} \\ &= \... 14 As a certified SCUBA diver, I learned that breathing pressurized pure oxygen leads to oxygen toxicity, which can be fatal. However, I'm not anywhere near an expert on the mechanism of oxygen toxicity, but I believe it has to do with resulting in a lot more reactive oxygen species which can cause oxidative stress and lipid peroxidation. I'm not really ... 14E=\frac 12mv^2 \implies 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$, and as there are 3 independent ... 13 Does this mean that both 1 mole of$\ce O$would occupy$22.4~\mathrm L$(or if this doesn't usually occur in nature, say 1 mole of$\ce{He}$or another monoatomic gas) 1 mole of$\ce{O2}$would occupy$22.4~\mathrm L$Yes, it means exactly that. And you're right, a stable gas of$\ce O$atoms is a pretty exotic thing, so$\ce{He}$is a much better ... 13 This is merely a shard of a fact which does not make much sense in and by itself. After all, in systems with gas/liquid equilibrium there is nothing really special about$\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. On the contrary, this is pretty typical. See all those points where the blue lines (isotherms) are horizontal? They make up a ... 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 ... 12 You may recall the ideal gas law: $$PV = nRT.$$ Here,$P$is pressure,$V$is volume,$n$is the amount of gas present (in moles),$R$is the ideal gas constant, and$T$is temperature. In an enclosed system, with no gas flowing in or out,$n$is constant (as is also, obviously,$R$). We can rearrange the equation above to pull all the constant terms to ... 11 Carbon Dioxide (CO2) readily dissolve in water and form Carbonic Acid (i.e H2CO3 (aq) ) This is the formation of bonds. Then Carbonic Acid (i.e H2CO3 (aq) ) dissociate in water as follows. So water gets H+ ions, so that cause water acidic. The following shows dissociation of Carbonic Acid (i.e H2CO3 (aq) ) more clearly. Carbon Monoxide (CO) do not ... 10 You're actually on the right track. Looking at the percent composition, you've correctly identified that the ratio of$\ce{C}$to$\ce{F}$atoms is 1:1, however, you cannot assume that the formula is just$\ce{CF}$(which isn't a known compound), it could be any compound with that ratio,$\ce{C2F2}$,$\ce{C3F3}$,$\ce{C4F4}$, etc. The way to narrow it down ... 9 There is a liquid state for carbon dioxide. Borrowing the$\ce{CO2}$phase diagram from Wikipedia, we can see that$\ce{CO2}$will condense at a few atmospheres, dependent on temperature. At still higher pressures, the liquid will solidify. Below the triple point temperature, the gas will transition directly to solid. 9 General estimates have placed a can of Coca-Cola to have 2.2 grams of$\ce{CO2} $in a single can. As a can is around 12 fluid ounces, or 355 ml, the amount of$\ce{CO2}\$ in a can is: $$\text{2.2 g} \ \ce{CO2}* \frac{\text{1 mol} \ \ce{CO2}}{\text{44 g} \ \ce{CO2} } = 0.05 \ \text{mol}$$ $$\text{355 mL} * \frac{\text{1 L}}{\text{1000 mL}} = 0.355 \ \text{L} ... 9 If one rearranges the ideal gas law equation, you can obtain the following (assuming n and T are non-zero):$$\frac{PV}{nT} = R$$R is a constant, and there are in fact infinitely many possible sets of values (P, V, n, T) that satisfy the equation. Let (P_1, V_1, n_1, T_1) denote one such set, and let (P_2, V_2, n_2, T_2) denote a second one. ... 9 TL;DR: Spray cans don't actually get colder when shaken. However, shaking a can does increase heat conduction from your hand to the can, making it feel colder. Humans don't actually sense external temperature directly; our thermoreceptors are located under the skin, and thus effectively measure the rate at which body heat is lost through the skin. This is ... 9 Avogadro's law, which can be written as V \propto n, where V is the volume of the gas and n is the amount of substance of the gas (measured in moles), can be thought of as just another manifestation of the ideal gas law rewritten as follows,$$ V = (RT/p) n \, . $$Consequently, strictly speaking, Avogadro's law is applicable only for a hypothetical ... 9 Note: You can skip section I, and go straight to section II and/or the end of section III (specifically the conclusions subsection), if you are already familiar with the basic mathematical machinery/definitions I. Preliminaries Feel free to skip all of this if you are familiar with it First, we have an adiabatic expansion, i.e \mathrm{q} = 0 . Second, ... 9 This question requires a simplistic notion of real gas behavior. The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction b is negative in the equation and the pressure correction, a is positive. The formula is$$(P + a/V_\mathrm{m}^2)(V_\mathrm{m} ...

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