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It may be helpful to look at a related value $k_\mathrm{B}$, the Boltzmann constant, which is widely used in thermodynamics. These two are related by $R = k_\mathrm{B}N_\mathrm{A}$, allowing the ideal gas law to also be written: $$PV = Nk_\mathrm{B}T$$ where $N$ is the number of particles, as opposed to the number of moles. The units are $\pu{J\cdot K^{-1}}$....

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It's not a coincidence at all! If you do an online search for "derivation of osmotic pressure", you'll see how $R$ enters into the derivation. Indeed, that's one of the beauties of the van 't Hoff equation for osmotic pressure – it reveals that (under the modest simplifying assumptions of the van 't Hoff equation derivation) the osmotic pressure created ...

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Charles' law says that at constant pressure the volume and temperature of an ideal gas are related as $$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$ If $V_2=V_1+dV$ and $T_2=T_1+dT$ then $$\frac{V_1}{T_1}=\frac{V_1+dV}{T_1+dT}=\frac{V_1}{T_1}\left(\frac{1+dV/V_1}{1+dT/T_1}\right)$$ which can be rearranged into $$\frac{dV}{dT}=\frac{V_1}{T_1}$$ But Charles' law says ...

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The answer to your question is yes and no. You are correct in supposing that as an ideal gas expands the entropy will increase as it has more space to occupy and so the number of ways the molecules can be placed in the total space has increased. (Try to not think of entropy in terms of randomness but in the number of ways molecules can be positioned in space,...

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Here's your confusion: You need to consider two different things: The momentum transfer per particle per collision. There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that ...

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$R$ is a proportionality constant between the units we use to measure temperature (kelvins) and the units we use to measure energy (joules) and number of particles (moles). If you measure temperature in energy units and use the number of particles instead of the number of moles, the constant $R$ (and $k_\mathrm{B}$) disappears from your formulas. $k_\mathrm{... 7 The whole point of the graph is that$pV/(RT)$is no longer equal to$n$for a real gas.$n$isn't changing (as you point out, this is physically impossible), but the ratio$pV/(RT)$is changing to some value that is not the same as$n$. Once you're freed from the ideal gas equation, there's no need for$n$to "follow" the changes in pressure / ... 6 In the simplest model, a gas is called ideal when its particles are point-like (no volume) and have no interactions. Real gases behave like ideal gases at low pressure (where the particle volume is neglible compared to the total volume) and high temperature (where condensed phases, i.e. interatomic or intermolecular interactions are disfavored). The size-... 6 Let's start from the definition of enthalpy change (not going to formal definition because it's very non-intuitive) here's how I like to define it: Enthalpy change is the amount of heat absorbed ($\mathrm dH\gt0$) or given ($\mathrm dH\lt0) by the system at constant pressure of system. $$\mathrm dH=Q_p$$ $$\begin{array}{l} \therefore \mathrm d U=Q_p+W \... 6 Assume the work is against constant pressure and that the final pressure is that which is worked against. Then$$\begin{align} w&=\Delta U = C_V \Delta T= C_V (T_f-T_i) \\ &= -p_\textrm{ext} \Delta V =-p_f(V_f-V_i) \\ &= -\frac{nRT_f}{V_f}(8V_i-V_i)\\ &= -\frac{7RT_fV_i}{8V_i}\\ &= -\frac{7RT_f}{8} \end{align}$$Solving for T_f and ... 5 In ideal gases no intermolecular forces, therefore no potential energy. Thus, internal energy is equal to total kinetic energy (KE) of the system. Consider N monoatomic particles in a cubical box of side \ell (assumption: ideal gasses consist of monoatomic point particles). The amount of ideal gas in the box is \frac{N}{N_A} = n \ \pu{mol} where N_A ... 5 The 2/3 factor is from the math of the kinetic theory of gases. Think of a cube with six faces (i.e. three orthogonal directions), the particles have equal probability of hitting a face in the x, y or z direction. The pressure is exerted equally among the three orthogonal directions, hence a factor of 1/3. Wikipedia also shows the steps in deriving the ... 4 Compressing a gas by applying an external force means to work against the force the gas exerts on the walls of its container, i.e. its pressure. The definition of compressibility$$\beta = -\frac{1}{V}\frac{\partial V}{\partial p} \tag{1}$$for an ideal gas with the well-known equation of state$$p V = n R T \tag{2}$$gives$$\beta = -\frac{1}{V}\frac{\... 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 His name was Amedeo Avogadro, not Amidio nor Amadeo nor Avegadro nor Avagadro. The full name was Lorenzo Romano Amedeo Carlo Avogadro, Count of Quaregna and Cerreto. Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. More exactly, it also includes single atoms, e.g. for a case of ... 3 In general for a perfect or ideal gas, $$C_p=C_V + R'$$ (using your notation) where the heat capacities are molar quantities. It follows that for a perfect gas mixture(C_p)_\text{mix}=(C_V)_\text{mix} + R'$. 3 When you decrease the temperature of an ideal gas held at constant volume, what you are doing is transferring energy as heat from the gas into the surroundings. You do this by keeping the surroundings at a targeted final temperature and placing the gas in thermal contact with the surroundings, allowing heat to dissipate. When you reach the final state, you ... 3 The expression you've written can be obtained from the equipartition theorem $$U = n_\mathrm{d.f.} \times \frac{k_\mathrm B T}2$$ stating that the energy is evenly distributed over all available degrees of freedom$n_\mathrm{d.f.}$, where each degree of freedom contributes an amount of energy equal to$k_\mathrm B T/2$. For a monoatomic gas of$N$particles, ... 3 Because it gives simpler-to-derive laws which are often very good approximations Clearly real gases do not always follow the ideal gas laws. They mostly liquefy under some conditions, for example, and, under those conditions they are clearly not ideal. But in practice gas laws are used for things far away from those non-ideal regions. When we are applying ... 3 OP has given a good effort to solve the problem using correct path. Only loose point was not considering the diatomic nature of the gases as Safdar Faisal pointed out in a comment. Suppose the amount of$\ce{O2}$in the gas mixture is$n_1 \ \pu{mol}$and the amount of$\ce{N2}$in the gas mixture is$n_2 \ \pu{mol}$in volume of$\pu{1.0 L}$container. Thus,... 3 If$V=Nb$, then container volume is 4 times total volume of molecules, according to the geometry used in the reasoning of van der Waals real gas model. Molecules of a gas ( or rather a fluid ) at such a state are not able to freely move anymore, forming for each other not passable obstacles. Therefore the statistical pressure model breaks down. Signs of ... 2 This is an example of using partial pressures for a mixture of ideal gases. We are making a big assumption - that water is an ideal gas - but once we make that assumption, the way to proceed is to treat each gas in the mixture as if it were by itself in the container. As you pointed out, it isn't very obvious from the way the question is worded that we ... 2 It doesn't matter if there is a container. I'm picturing bubbling the nitrogen through the water, for example the water is in a flask, and there is a hose going down into the water, maybe a frit at the end of the hose. The important assumption is that nitrogen reaches equillibrium with the water, which is a big assumption. But adding 1.2 grams of water ... 2 The Gibbs free energy change is zero in the case of reversible processes carried out at constant temperature and pressure, but that isn't the case if these conditions are not observed. As a demonstration consider an isothermal reversible expansion of an ideal gas. Since the temperature is constant, the free energy change is given by $$\Delta G = \Delta H - T ... 2 Internal energy of any gas is given by$$U=U_\text{trans} + U_\text{rotational} + U_\text{vibrational} + U_\text{intermolecular} + U_\text{electronic} + U_\text{relativistic} + U_\text{bonds}$$Last three aren't affected by ordinary heating. And U_\text{intermolecular} for ideal gas is zero. That's why for ideal gas U is only function of temperature. But ... 2 The temperature and the volume of the inner ear are constant. When your ears pop during descent, air from the cabin goes into the ear, increasing the pressure. The law is the following:$$n / P = const$$You can derive this from the ideal gas law. It has no special name. 2 To carry out the B transformation, you have to heat a lot. If you don't heat, the volume will of course increase, but the pressure will decrease. B is a process difficult to carry out, because it is not easy to heat enough a gas who is inflated to as to maintain the inner pressure. Then, at the end of B, the temperature is very high. You block the position ... 2 Your analysis is correct in terms of number density. But let's see how it plays out in terms of molar density. Let n be the number of moles and A be Avagadro's number. Then N=nA. If we substitute this into your first equation, then I get$$P=\frac{n(Ak)T}{V-nAb}-\frac{aA^2n^2}{V^2}$$But, since Ak=R, we obtain:$$P=\frac{nRT}{V-nb'}-\frac{a'n^2}{V^2}$$where ... 2 We know from the first Newton motion law, that the net force acting on an object in rest must be zero. The forces acting on the piston are gravity and gas pressure:$$\vec F_g + \vec F_\mathrm{p,down} + \vec F_\mathrm{p,up}=\vec 0 \tag{1}$$If$V$is the given bottom gas volume,$V_0$is the total gas volume,$n\$ is the molar amount of each of gases, the gas ...

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