Caesium (m.p. 29°C, b.p. 686°C) was introduced into a container and heated to 500°C. When a hole of...

Question

Caesium (m.p. 29°C, b.p. 686°C) was introduced into a container and heated to 500°C. When a hole of diameter 0.50 mm was opened in the container for 100 s, a mass loss of 385 mg was measured. Calculate the vapour pressure of liquid caesium at 500 K.

Method The pressure of vapour is constant inside the container despite the effusion of atoms because the hot liquid metal replenishes the vapour. The rate of effusion is therefore constant, and given by eqn XXXXXXXXXXTo express the rate in terms of mass, multiply the number of atoms that escape by the mass of each atom.

(21.16)

Robert · Accepted Answer

The mass loss Δm in an interval Δt is related to the collision flux by: x 
    Δm = (Z
w
A
0
) Δt m 
                Where A
0
 is the area of the hole and m is the mass of one Caesium atom.
                         
                            
         
   
Or,
Or,

Caesium (m.p. 29°C, b.p. 686°C) was introduced into a container and heated to 500°C. When a hole of diameter 0.50 mm was opened in the container for 100 s, a mass loss of 385 mg was measured....

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