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Molecular motion is a manifestation of heat. Even at absolute zero, molecules in their ground state, have a zero-point energy ($\ne 0$), in the form of molecular vibrations. As temperature rises, rotational and other vibrational modes get excited. Translational motion arises as the solids melt into liquids or sublime into gases. (For materials, like Helium, ...


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Translational motion just means linear 'velocity'. That is, motion 'across-ways' (ie: in x-, y- and z- directions). So the question can be rephrased as 'why do molecules have velocities'? Well, if they didn't, everything would be realtively motionless, there would be nothing happening in the universe and you wouldn't be able to ask the question and we ...


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The specific heat of a molecule depends on the number of degrees of freedom the molecule has. There are several degrees of freedom available: translation (3), rotation (3), vibration (depends on the number of bonds in a molecule) and electronic modes. Now, for something that is monatomic, you have 3 translational modes (x,y,z directions), zero rotation ...


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You ask "what would we see", meaning you want to know what kind of radiation is given off. (i) sample from the surface looks like the surface of the Sun. i.e. it emits close-to-blackbody radiation at a temperature of 5800K.Except it now can't be a blackbody since you've taken it out of its environment. A black body is both perfect absorber and a perfect ...


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An ideal gas is treated as a set of $N$ indistinguishable particles with no interactions, which means the partition function of a single particle is simply, $$Z = \int d^3p \, d^3 q \, \exp \left\{ -\beta\frac{p^2}{2m}\right\}$$ and the partition function of the entire system is $Z^N/N!$. From this, we can obtain the ideal gas law, $$PV = N k_B T$$ Of ...


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Let's make some order: you say "As time passes the packets spread". Why should they spread? How do you imagine that these packets look like? Do you believe that initially the wave-packets are very small, and increase in width in time? No reason for that. As an example, assume that the gas consists in very light particles, and the gas is very cool, s.t. the ...


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I remember doing that same question to a professor of mine (a long time agooooo), and his answer was that we could imagine that each collision is like an observation, and the wave collapses back into a localized packet. The answer was not very convincing, but in any case, remember that physics is not mathematics, and you can get very accurate physiocal ...


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@Jim But as cylinder A has twice as much gas as B, $$n_{a}= 2n_{b}$$ By $$PV=nRT$$ , $$ T_{A}= \frac{T_{B}}{2} $$ $$T_{A} < T_{B} $$


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Yes it does depend on temperature. $PV = nRT$ so $V \propto T$ or volume is proportional to temperature in Kelvin. To go from 0 degrees Centrigrade to 1 degree Centrigrade the temperature in Kelvin changes from 273 K to 274 K (approximately) so the volume changes from $V$ to ${274 \over 273} V$ which is equal to $(1 + {1 \over 273}) V$ - so there is ...


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This is a key sentence (p.24 [1]): "Pseudocritical properties of gases can be estimated with gas composition and mixing rules or from correlations based on gas specific gravity." (I added bolding) so the supercompressability $Z$ factor (also called the deviation factor) can be defined by the following methods below. Method 1: gas composition values ...


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The gas is expanded adiabatically and then isothermally. Thus the temperature it has at the end of adiabatic expansion stays the same even after the isothermal process. Ideal Gas equation after adiabatic expansion: $p_aV_a=nRT_a$, where index "a" shows after. You do not have $V_a, T_a$ in this equation. However, another equation you can write down is the ...



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