headspace vapor concentration 2

[handee]

This is pretty basic stuff, but I could use a sanity check if anyone has the time. I'm calculating vapor concetration in the headspace of a tank of solvent (MW=53.1, bp=78C, vp=83torr) at ambient temperature. At atm. pressure, the mole fraction should simply be 83/760, and using PV=nRT I can calculate the concentration in g/L (I get ~0.25g/L). But when I reduce the pressure, that value doesn't change...let's say I cut the operating pressure in half, half the number of moles/liter, but the vapor mole fraction doubles, and voila - the concentration in g/L stays the same.

Am I missing something? TIA

Fran McConville

 

[Montemayor]

Fran:

I think what you are missing is the basis of your calculation: Dalton's Law, which states

P(Total) = P(A) + P(B)

where,
P(Total) = System pressure; your initial 760 mmHg
P(A) = partial pressure of your solvent, 83 mmHg
P(B) = partial pressure of another gas in the vapor space; possibly Nitrogen (or air?)

From basic chemistry & physics, if your solvent is a pure liquid at ambient temperature (you failed to state that its vapor pressure of 83 mmHg is at ambient temperature, but we have to assume that it is) its vapor pressure = its partial pressure.

You also failed to state that the level of the tank remains constant (or, more importantly, that the vapor space above the liquid is constant). That means that you have an equation of state (PV = nRT) with 4 variables and you have 3 of them known: the temperature, pressure, and volume. The number of moles stays the same in the same volume; only the ratio of the moles between the contributing solvent and Nitrogen changes - according to their partial pressure). If you change the temperature (by increasing it, for example) you increase the solvents vapor pressure - and correspondingly, its partial pressure. The N2 partial pressure remains the same. In this example, you get more moles in the given vapor space. But take into consideration that the system pressure MUST increase in order to contain more moles in the vapor space.

However, if you decrease the system pressure - down to a partial vacuum - you get another effect. This is so because you failed to note that in order to achieve your partial vacuum you MUST decrease the initial number of N2 moles; otherwise you can't have a partial vacuum - by definition. The sum of the partial pressures still equals the total system pressure. The partial pressure of the solvent remains the same while the partial pressure of the N2 is reduced. But the total volume has remained constant.

The partial pressure of a vapor is that pressure exerted by the vapor is it were alone in the given, set volume. Since you haven't changed the vapor space volume and you haven't changed the temperature, the amount of solvent moles per the fixed, constant vapor space volume is still the same - however, the system pressure has been reduced.

I hope this helps explain Dalton's contribution to partial pressure phenomena.

 

[handee]

Montemayor, thanks for taking the time to post such a detailed reply.

I suppose Dalton's law was implied in the conditions I proposed: P(A)=83, P(B)=760-83. And yes I was also assuming that everything is constant except for the pressure.

You've certainly helped explain, and confirmed, my finding that the number of moles of solvent in the headspace does not change as the pressure is reduced. But what a counter-intuitive phenomenon...no?

Fran McConville