On the saturated vapor pressure problem of the Antoine equation
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Recently, I have been using the Antoine equation to develop an Excel formula for calculating saturated vapor pressures. I encountered a problem: when using the Antoine equation to calculate the saturated vapor pressure of toluene at 30 degrees Celsius, the value obtained is 5 KPa. My understanding is that this pressure should be the absolute pressure. But if it is absolute pressure, then the corresponding gauge pressure is -96 KPa (gauge pressure = absolute pressure – atmospheric pressure). In actual production, a certain amount of toluene is kept in the reaction equipment. After the device is sealed, the gauge pressure on it should be -96 KPa according to the calculation formula, but in reality it is not. Did I misunderstand something about this question? So what should the pressure gauge reading on the device be theoretically at this point?
The saturated vapor pressure calculated using Antoine’s equation refers to the absolute pressure relative to the liquid, excluding atmospheric pressure. So in your example, the saturated vapor pressure of toluene at 30 degrees Celsius is 5 KPa, which does not include atmospheric pressure. If you want to know the gauge pressure reading inside the device, you need to take atmospheric pressure into account. Gauge pressure can be obtained by subtracting atmospheric pressure from absolute pressure. Under these conditions, assuming the atmospheric pressure is 101.3 KPa, the gauge pressure reading on the device should be 5 KPa – 101.3 KPa = -96.3 KPa. In fact, there may be some errors in the pressure displayed on the instrument; as a result, there can be a difference between the theoretically calculated value and the actual measured value. This may be the reason why the actual situation you observe differs from what you expected. .
Something doesn’t seem right – what does -96KPa mean? It’s not even certain that a vacuum pump can achieve such a pressure level
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