7 replies to this topic

[jaffrey]

Hi can I make 2 points that of understanding that I totally made up myself and hope someone can justify whether or not they are valid.

1. The bubble point of a binary mixture is when the partial pressure of the MVC is equal to the external pressure and hence is first to vapourise as pure MVC at the initial point of evaporation.

2. The vapour pressure measures the tendancy of a component to evaporate from the mixture at that specific temperature. Therefore that is why when going way beyond the boiling point (rising temperature) of the MVC means vapour pressure rapidly falls while the LVC and it's boiling point which has not yet been reached has greater vapour pressure and so start evaporating at a greater rate to MVC. This forms the basis of a T-X-Y diargram.

thanks

[katmar]

No, you need to revise Dalton's law of partial pressures and Raoult's law too. For a binary mixture the bubble point is when the combined partial pressures equal the external pressure. It is wrong to think that the MVC comes off as a pure product at the bubble point.

[jaffrey]

No, you need to revise Dalton's law of partial pressures and Raoult's law too. For a binary mixture the bubble point is when the combined partial pressures equal the external pressure. It is wrong to think that the MVC comes off as a pure product at the bubble point.

ah ok thanks Katmar for pointing this out to me. So at the dew point when the vapour mixture starts to condense will it be when the combined vapour pressures of the 2 components are lower than the external pressure?

also am I right in thinking that there is a direct link between vapour pressure of a component at a temperature (t) and the rate of evaporation?

thanks

[djack77494]

Your statement, "So at the dew point when the vapour mixture starts to condense will it be when the combined vapour pressures of the 2 components are lower than the external pressure?" is a bit off. At the dew point, the combined partial pressures of your constituentshas dropped to just the point where it equals the external pressure.

The answer to your question, "am I right in thinking that there is a direct link between vapour pressure of a component at a temperature (t) and the rate of evaporation?" is, "No". Typically your have some source of heat/energy flowing into your system. The instantaneous rate of vaporization, assuming the liquid mixture to be saturated, is the heat input divided by the latent heat of vaporization. So the rate of vaporization (which you not quite correctly refer to as evaporation) correlates to the latent heat. The mixture's vapor pressure will equal the external pressure during this process.

[jaffrey]

Your statement, "So at the dew point when the vapour mixture starts to condense will it be when the combined vapour pressures of the 2 components are lower than the external pressure?" is a bit off. At the dew point, the combined partial pressures of your constituentshas dropped to just the point where it equals the external pressure.

The answer to your question, "am I right in thinking that there is a direct link between vapour pressure of a component at a temperature (t) and the rate of evaporation?" is, "No". Typically your have some source of heat/energy flowing into your system. The instantaneous rate of vaporization, assuming the liquid mixture to be saturated, is the heat input divided by the latent heat of vaporization. So the rate of vaporization (which you not quite correctly refer to as evaporation) correlates to the latent heat. The mixture's vapor pressure will equal the external pressure during this process.

thanks djack

so concerning flash distillation will the distillate ever be more concentrated in the LVC than MVC as tempertures increase. Hence this is why I thought maybe the LVC evaporates at a faster rate than MVC. Than it may well be the case that LVC is more concentrated?

thanks

[djack77494]

so concerning flash distillation will the distillate ever be more concentrated in the LVC than MVC as tempertures increase. Hence this is why I thought maybe the LVC evaporates at a faster rate than MVC. Than it may well be the case that LVC is more concentrated?

If you warm a mixture and then allow the warmed mixture to flash, you will get a separation of the constituents. The vapor phase will be richer in light components while the liquid will be enriched in the heavier compounds. I don't accept your statement the " the LVC evaporates at a faster rate than MVC" because I don't feel that this is a valid description of the phenomenum. Better to say that the LVC has a higher partial pressure due to its higher vapor pressure (or fugacity). Because of this, the lighter compounds exert a greater partial pressure than their liquid mole fractions would suggest. Since the vapor composition of a component is its partial pressure divided by the total pressure, the vapor phase is enriched in light components.

BTW, if you flash at a lower temperature, you will get an even lighter mix of compounds in the vapor phase but you will also get less vapor in total.

[jaffrey]

this is a general question but why are the terms "heavy" and "light" keys used to refer to LVC and MVC respectively? I dont think its linked to their densities is it?

[djack77494]

this is a general question but why are the terms "heavy" and "light" keys used to refer to LVC and MVC respectively? I dont think its linked to their densities is it?

For any given class of compounds, such as hydrocarbons or alkanes or whatever, those compounds with higher molecular weight (i.e. heavier) exert a lower vapor pressure than those with a lower molecular weight. This is a common phenomenum that applies across many families of compounds. Coincidentally, higher molecular weight compounds tend also to have higher densities.