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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 ...


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 ...


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 ...


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} ...


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

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. ...


6

Yes, the vibrational modes are present at any temperature, including absolute zero where the lowest vibrational energy levels only are populated; the zero-point levels. The molecule is linear ( OCO ) so it has $3N-5=4$ vibrational modes for $N=3$ atoms. The symmetric stretch corresponds to both CO bonds stretching in phase, and asymmetric stretch to one CO ...


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 ...


5

I think the author of this problem forgot to add units to $K_c$ since it's not a dimensionless entity: $$K_c = \frac{[\ce{NO}]^2}{[\ce{N2O}][\ce{O2}]^{0.5}}$$ and should be $K_c = \pu{1.7e-13 mol^{0.5} L^{-0.5}}$. In general $$[K_c] = \mathrm{dim}(c)^{Δn}$$ where square brackets denote the dimensions of the quantity $K_c$, $c$ is concentration and $Δn$ ...


5

There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less ...


5

Here is a cookbook recipe for determining the change in entropy for a system that has suffered an irreversible process: THE RECIPE Apply the First Law of Thermodynamics to the irreversible process to determine the final thermodynamic equilibrium state of the system Totally forget about the actual irreversible process (entirely), and focus instead ...


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

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 ...


5

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, ...


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 ...


4

How do we know that one mole of an ideal gas occupies (exactly) 22.4 litres? It doesn't. The current recommended value for the molar volume of an ideal gas at a temperature of $T=273.15\ \mathrm K$ and a pressure of $101.325\ \mathrm{kPa}$ is $V_\mathrm m=22.413962(13)\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$ (source). Note the given uncertainty; i.e. it is ...


4

For a reversible process in a closed system (no mass entering or leaving), the general equation for the expansion work done by a real gas on the surroundings is the same for an ideal gas, namely $$W = \int p\,\mathrm{d}V$$ However, for a real gas, we use the equation of state for that gas $p=p(n,V,T)$ rather than $p=\frac{nRT}{V}$, the equation of state for ...


4

Regarding the universal gas constant $R$, you may find it expressed in multiple units -- and consequentially, sometimes numerically quite different -- in textbooks about Physical Chemistry, or even right on top right of the corresponding entry of wikipedia. (source) Regarding your second point, of course you may use cubic metres as unit of volume, or non-...


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

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 ...


3

Well, I can see all the relations you require knowledge of in the question itself! Calculate the effective molecular mass from $$M = \frac{dRT}{P}$$ $R$ is known, $d$ given. $P$ and $T$ are available from the fact that it is at STP. $M$ comes out to be ${20.176 u}$. Effective molar mass is easily calculated below: $$M_\mathrm{eq} = M_1x_1 + M_2x_2$$ ...


3

As matt_black commented, there is no need to use ideal gas law as you are given the molar volume already. Volume of the air $V_\mathrm{air}$ can be found from the volume fraction $\varphi(\ce{O2})$: $$V_\mathrm{air} = \frac{V(\ce{O2})}{\varphi(\ce{O2})}$$ Volume of the oxygen can be found from the molar volume $V_\mathrm{m}(\ce{O2})$ and the amount of ...


3

Collision is predominantly a bimolecular act. The probability of three gas molecules simultaneously colliding is incredibly low, therefore it quite accurate to say that the excluded molar volume $b$ is overcounted by a factor of 2.


3

I haven't dived into your math, but I think there are 3 things that are confusing: intermediate calculations, even though you weren't asked to do any; missing units and inconsistent notations (e.g. base 10 is not always shown); ambiguous phrase "atoms occupy a volume of $2\times 10^{-4}$." I always find it cleaner and more productive to derive the ...


3

Yes, for many practical purposes, you may use the ideal gas law also as an approximation for water vapour. However, you need two independent quantities to describe the state of the gas, for example temperature and pressure. The given temperature of $T=298\ \mathrm K$ alone is not enough. The ideal gas law does not give you a value for the second quantity. ...


3

Let's derive the appropriate realtionship... $$PV = nRT\tag{1}$$ so $$\dfrac{PV}{T} = nR\tag{2}$$ but $nR$ is constant for the given sample of gas, thus $$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\tag{3}$$ For the sample the volume is 137 ml, the dry gas has a pressure of 753-22=731 mm Hg, and the temperature is 22 °C = 295 °K. Now here is a bit of ...


3

One way to proceed is as follows. Start from two things you know from the problem statement: You have an ideal gas The process is isothermal Those two conditions mean the energy of the system does not change ($\Delta U=0$) during the process. It follows, from the first law, that $$ q_\mathrm{sys} = \Delta U -w_\mathrm{sys} = -w_\mathrm{sys} $$ You are ...


3

In my textbook the opposite meaning is given to $a$ (an attractive parameter) and $b$ (a size parameter), but this may be a matter of differences in labelling, not in meaning. I will for the sake of consistency stick to your convention. For $\ce{He}$ the van der Waals parameters have the following values: $b = \pu{3.4598 J mol-1 M-1}$ $a = \pu{0.023733 M-1}...


3

The "solution" to the quadratic equation makes no sense to me. Assuming the Van der Waal equation with b=0, I agree with the solution to the equation: $$\mathrm{PV}_m^2 - 24\mathrm{V}_m + 2 = 0\tag{1}$$ I'll point out that I'm following the notation of the gievn solution, but $\mathrm{V}_m$ seems odd to me. I'd think that $\mathrm{V}_m$ would be ...


3

That expression is incorrect. For one mole of an ideal gas, $$ V_\mathrm{m} = \frac{RT}{p}$$ where $p$ is the pressure of the gas. $pV$ work is defined as $$ W_{pV} = -\int_{V_i}^{V_f} p_\mathrm{ext}\mathrm{d}V_\mathrm{m}$$ where $p_\mathrm{ext}$ is the applied pressure against which work must be done. By this sign convention work done by the system ...


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