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1

If you know the phase diagram of the gas, and can calculate the exact pressure the gas should have given the volume outside of the molten salt, you can use the measured pressure/temperature to determine any difference in pressure. Then you can calculate the number of moles of gas that must be absorbed to generate the pressure difference. If the gas is ...


5

It helps to rewrite $$\text{rate} \propto \frac{PA}{\sqrt{TM}}$$ by assuming the ideal gas law holds, as follows: $$\text{rate} \propto \frac{RA}{V_m}\sqrt{\frac{T}{M}}$$ where $V_m$ is the molar volume or inverse of molar particle density. Written this way it is clear that if the particles occupy the same volume, increasing their temperature increases ...


0

The seminal Van der Waals equation of state P=( RT/V-b)- a/V^2 where v = 1/ρ is molar volume. It can be rearranged by expanding 1/(v - b) into a Taylor’s series: Z= 1+ (b- a/RT)1/V + (b^2/v^2) +( b^3/v^3)+........ The second virial coefficient has roughly the correct behavior, as it decreases monotonically when temperature is lowered. The third and higher ...


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The compression factor $Z$ is defined as the ratio of the real molar volume with respect to the theoretical ideal $$Z = \frac{V_m}{V_m^{ideal}}$$ where if $Z$ = 1, the gas behaves ideally. If we want to get some sort of understanding for how this function behaves when you change the Pressures and Temperatures of your system, we need to substitute into ...


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Why thermodynamical equations are just for gases? They are not. The equation $\Delta U = q + P\Delta V$ applies to any phase (gas, liquid, solid...) when only pV work is done. In the particular form of the equation you present, the pressure is in addition constant during the work. Gases are (1) an easy way to introduce thermodynamics concepts because ...


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