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The main assumptions that are relevant here is that the size of the individual molecules is negligible compared to the average distance between them, and that the intermolecular potentials can be ignored. The first is invalidated if the molecules are too close to one another, i.e. if the number density of particles is too high. The latter is invalidated by ...

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The reasons for nonideal behavior are different for the two scenarios you describe, and as such they do not cancel each other out. The ideal gas law is an approximation that works well when the gas molecules are very far apart and don't collide very often. they are close together at low temperatures and high pressures and collide more often and with greater ...

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Any equation can be written as a function equal to zero - it is just the matter of little algebra to move all the terms to one side of the equation. It sometimes make it less transparent for a physicist, but it is a standard mathematical convention, and for a good reason. When two bodies are in equilibrium with each other, it often means that they have same ...

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We can always define $f$ so that the constraint is zero. If we start with a constraint $g_{AB}(A_i; C_i) = X$, just define $f_{AB}(A_i; C_i) = g_{AB}(A_i; C_i) - X$. The point is that there's some conservation law at work. In the kind of situation stat mech textbooks usually think about, $A_i, C_i$ will be macroscopic quantities. However, that doesn't need ...

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Pressure, like temperature, is an intensive thermodynamic property. It does not depend on the mass. Suppose you have a room filled with air (which can typically be considered an ideal gas), at 1 atmosphere pressure and room temperature of 20 C. If you divide the room into two equal parts by a wall, the pressure and temperature in each half will be the same, ...

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Since you seem to consider a condition where the temperature is not a factor, I am assuming isothermal conditions i.e., $T=$ constant. CASE I: $V=$ constant $$\frac{m}{P}=\frac{MV}{RT}\Rightarrow P\propto m$$ More the gas stuffed inside the container, the more the pressure. CASE II: $P=$ constant. $$\frac{m}{V}=\frac{MP}{RT}\Rightarrow V\propto m$$ More the ...

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The water vapor does approximately follow the ideal gas law at 100 C. But the number of moles of water vapor has changed, and this needs to be taken into account.

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Saturated steam is NOT an ideal gas. It is water vapor that is in equilibrium with the liquid that it is in contact with. This means that any heat input to the system will vaporize some of the water at 100 deg C, and any heat removal from the system will condense some of the steam at 100 deg C. This type of system follows the Antoine equation rather than ...

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$\frac{3}{2}kT$ is only translational kinetic energy. If it is diatomic or polyatomic molecule, there is also rotational kinetic energy and the formula is $$<K>=\dfrac{f}{2}kT$$ where $f$ is degree of freedom

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Feynman is correct. The reduced partial pressure of water vapour in the drier air reduces the rate that vapour condenses onto the liquid, but it doesn't affect the rate that the liquid evaporates. But certainly, the net evaporation is faster because of the reduced condensation. If we replace the top of the container, eventually the partial pressure of the ...

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The SI unit of the specific heat at constant pressure is Joule per Kelvin and kilogram, J/(K-kg). However, notice that an increment of the temperature of one degree Celsius is the same as an increment of one Kelvin, that is the same as joule per degree Celsius per kilogram (J/°C/kg)

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Why does adiabatic contraction bring Carnot engine back to initial state? The short answer is the work done on the gas during the adiabatic compression has to increase the internal energy of the gas by the same amount that the adiabatic expansion reduced the internal energy of the gas, so that the total change in internal energy of the cycle is zero. Keep ...

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