Writer Quick Response Code in Java SURFACE EFFECTS IN CONDENSATION
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If we compress a gas isothermally, it is supposed to start condensing at a point 0, as shown in the P- V diagram in Fig. 2.6. If we compress the system further, the pressure is supposed to remain constant. Actually the pressure will sometimes follow the dotted line shown in Fig. 2.6; along this dotted line the system is not
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Fig. 2.6 Supersaturation and supercooling.
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in stable equilibrium, however, because the slightest jar will abruptly reduce the pressure to the correct vapor pressure. Similarly, if a liquid is expanded beyond the point 0 I, it will sometimes follow the dotted curve shown, but this too would not be a situation of stable equilibrium. These phenomena are respectively known as supersaturation and supercooling. They owe their existence to surface effects, which we have previously ignored. We give a qualitative discussion of the surface effects responsible for supersaturation. Vapor pressure as we have defined it is the pressure at which a gas can coexist in equilibrium with an infinitely large body of its own liquid. We now denote it by Poo(T). On the other hand, the pressure at which a gas can coexist in equilibrium with a finite droplet (of radius r) of its own liquid is not the vapor pressure Poo(T) but a higher pressure Pr(T). The difference between Poo(T) and P/T) is due to the surface tension of the droplet. Before we try to give a qualitative description of the mechanism of condensation, we calculate Pr(T). Suppose a droplet of liquid is placed in an external medium that exerts a pressure P on the droplet. Then the work done by the droplet on expansion is empirically given by dW= PdV- ada (2.15) where da is the increase in the surface area of the droplet and a the coefficient of surface tension. The first law now takes the form
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dU = dQ - P dV + ada
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Integrating this, we obtain for the internal energy of a droplet of radius r the expressIOn
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where U oo is the internal energy per unit volume of an infinite droplet. Correspondingly the Gibbs potential takes the form
1 7Tr 3
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Consider a droplet of radius r in equilibrium with a gas of temperature T and pressure P. For given T and P, r must be such that the total Gibbs potential of the entire system is at a minimum. This condition determines a relation between P and r for a given T. Let the mass of the droplet be M 1 and the mass
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of the gas be M 2 The total Gibbs potential is of the form 2 Glolal = M 2g 2 + M i g 1 + 4'1Tar (2.18) where g2 and gl are respectively the chemical potential of an infinite body of gas and liquid. We now imagine the radius of the drop changed slightly by evaporation, so that 8Ml = -8M2 . The equilibrium condition is
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= 0 = 8M1 (
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(2.20) where p is the mass density of the droplet, we have 2a g2 - g1 = P;
as the condition for equilibrium. Differentiating both sides with respect to the pressure P of the gas at constant temperature, and remembering the Maxwell relation (8g18Ph = lip, we obtain 1 1 p' -
20 ( 8r ) pr 2 8P T
2a ( 8 P ) p2 r 8P T
where p' is the mass density of the gas. We assume that the gas is sufficiently dilute to be considered as ideal gas. Hence (2.23) where m is the mass of a gas atom. Further, lip may be neglected compared to lip', and (8pI8P)::=: O. Therefore
( ~) 8P
m 2aP
Integrating both sides of this equation we find P as a function of r for a given temperature:
Pr(T) = Poo(T)exp ( pkT-;
20m 1 )
which is the expression we seek. A graph of Pr(T) is shown in Fig. 2.7. N ow we can give a qualitative description of what happens when a gas starts to condense. According to (2.25) only liquid droplets of a given radius r can exist in equilibrium with the gas at a given T and P. The droplets that are too large find the external pressure too high. They attempt to reduce the external pressure by gathering vapor, but this makes them grow still larger. The droplets that are too small, on the other hand, find the pressure too low, and tend to evaporate, but this makes them still smaller, and they eventually disappear. Thus, unless all