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The reason that in your sample set water seems to have a low value of $a$ despite having hydrogen bonding is because most of the molecules that you chose were of much higher molecular mass. You should compare water with some gas of approximately equal molecular mass, then you'll see that value of $a$ in water is quite high. Your other observations are ...


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A deep dive for the sceptics Since this is a question about which there seems to be a lot of misunderstanding, it seems worthwhile to take a deeper look, even though the given answers are perfectly satisfactory. First, some terminology clarification. When people say something like “the kinetic energy of molecules in a liquid is the same as in a gas at ...


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How can I compare the van der Waals constant $a$ (the liquefication constant) for different compounds based on my knowledge of bonding in the compound? Actually, as you indicated in the question, this comparison is not realistic since $a$ is unique to a particular compound/atom. As Chet Miller pointed out $a$ is depend on the each molecule/atom's critical ...


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In terms of the critical temperature and critical pressure, $$a=\frac{27R^2T_c^2}{64P_c}$$


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[OP] the average kinetic energy of evaporating water molecules You have to specify whether you are talking about the kinetic energy just before the water molecule breaks the hydrogen bonds to its neighbors or just afterwards. A millisecond before or after the event, of course, the average kinetic energy will be determined by the bulk temperature. One way to ...


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At gas/liquid phase equilibrium, average kinetic energy of evaporating molecules, i.e. those just passed to a gas phase, is equal to average kinetic energy of condensing molecules. The latter is then approximately proportional to $T$. If these average values were not equal, the system would not be in thermal equilibrium. Feedback to comments: @theorist In ...


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Since joule-thomson effect is isoenthalpic process, so everything is done at the cost of internal energy. Taking $a$ as the term involving intermolecular forces of attraction and $b$ as the excluded volume, we get: $$T_i = \frac{2a}{Rb}$$ As per above equation inversion temperature is the function of intermolecular forces of attraction. When the temperature ...


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You need to be more precise with the terms you use, you are mixing up rates of effusion and times of effusion. This is a trick question, gases with lower molar mass will always effuse faster as per Graham's law


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In a Joule expansion, which is an irreversible adiabatic expansion against a vacuum, $q = 0 \text{ and } p_{ext} = 0$. Thus, since the only type of work in a Joule expansion is $pV$-work: $$\Delta \text{U} = q + w = w = -p_{ext} \Delta V = 0 \text{, always.}$$ And since $\Delta \text{H} = \Delta \text{U}+ \Delta pV$: $$\Delta \text{H} = \Delta pV \text {, ...


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In the Joule experiment, $\Delta H$ is not zero for a real gas. $\Delta H=\Delta U+\Delta (PV)$, and even if $\Delta U$ is zero, there is no physical reason to expect that $\Delta (PV)$ would be zero in this process.


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Forget about the ideal gas. The definition of the compressibility factor is just $$Z=\frac{PV_m}{RT}$$


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