Z Let's say I have hot water vapor gas at 1 atm pressure and I want to cool it to condense the liquid. I assume that this is where the mean free path is as big as the molecule itself (0.27 nm diameter). To what temperature should I cool the gas? The equation for mean free path (2) kB T V2-T-d2.P is: 1 = b. The answer to part a is far below the real temperature of 100 °C. Why could the result from pt. a be so wrong?

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Let's say I have hot water vapor gas at 1 atm pressure and I want to cool it to condense
the liquid. I assume that this is where the mean free path is as big as the molecule itself (0.27
nm diameter). To what temperature should I cool the gas? The equation for mean free path (1)
is: A
kg T
V2-T-d2.P'
b. The answer to part a is far below the real temperature of 100 °C. Why could the result from
pt. a be so wrong?
Transcribed Image Text:Let's say I have hot water vapor gas at 1 atm pressure and I want to cool it to condense the liquid. I assume that this is where the mean free path is as big as the molecule itself (0.27 nm diameter). To what temperature should I cool the gas? The equation for mean free path (1) is: A kg T V2-T-d2.P' b. The answer to part a is far below the real temperature of 100 °C. Why could the result from pt. a be so wrong?
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