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1) You are correct that the thermodynamic temperature is a measure only of the translational kinetic energy. Intramolecular vibrations do not contribute to the temperature. 2) You are also correct that rotational and vibrational KE can be converted to translational KE in a collision (as long as total momentum and kinetic energy of the system are conserved)....

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It has already been suggested to use liquid nitrogen to cool oxygen and that is probably your best bet. You could research something called a cryocooler that can be used to condense air or pure oxygen but it is very costly and would not be worth it unless you plan on using a continuous supply liquid oxygen and/or nitrogen. If you're just going to do few ...

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Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid? No, but they will have the same temperature. Solids don't have any translational energy, the atoms time-averaged positions are constant. If you want to link the temperature to the ...

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I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different? The total kinetic energy is different. The average kinetic energy per particle is the same. Through this question, I want to ask whether temperature ...

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The short answer: If 2 samples of water of the same temperature is considered,the average kinetic energy of water molecules of samples A nad B are the same. But, within the same water sample, water molecules have different kinetic energy with a particular statistical distribution. Even if I had set by some magic wand the same energy for all water ...

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The temperature of water is implicitly covered by actual and saturated vapour content, or equivalent vapour pressures. There is deterministic relation between relative humidity, air temperature and temperature of water, supposed to be in dynamic equilibrium. In such a state, the cooling effect of water evaporation is balanced by the thermal transfer from ...

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From the open system (control volume) version of the first law of thermodynamics, between cross sections x and y, $$\dot{Q}-\dot{W}_s-\dot{m}\Delta h=0$$where $\dot{Q}$ is the rate of heat addition to the control volume, $\dot{W}_s$ is the rate of doing shaft work, $\dot{m}$ is the mass flow rate, and $\Delta h$ is the change in specific enthalpy between ...

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For an adiabatic system like a piston where $\delta Q = 0$, using the first law of thermodynamics gives you the following expression: $$\mathrm dU = \delta Q + \delta W$$ $$\mathrm dU = - p\,\mathrm dV$$ This expression is pretty much useless however, in that you can't integrate it, since $T$, $V$, and $p$ are all constantly changing interdependently in ...

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