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## Pressure in the Vicinity of a Lunar Astronaut Space Suit due to Outgassing of Coolant Water

Problem:
The space suits worn by Lunar astronauts are cooled by the sublimation of water through specially designed, porous surfaces on the suit. The rate of water loss is approximately one pound per hour. If

A = .06m2

is the approximate area of the porous space suit surfaces, predict the gas pressure in the immediate vicinity of the suit. Assume a temperature of 273°K.

Solution:
The rate of water loss is approximately one pound per hour. This value corresponds to a mass loss of

 1.4 x 10-4 kg/sec 1
 or a particle loss of dN/dt = 5 x 1021 /sec 2

where N = the total number of water particles available. If we assume that the temperature T = 273°K, and set the velocity v of the sublimating water particles equal to (2kT/m)1/2 (with m = 3 x 10-26 kg for water), then v = 500 m/sec approximately.

Now if n is the number density of the water vapor at the surface where sublimation from the suit into space is occurring, then

 nv = (dN/dt)/A 3
 or nA = (dN/dt)/v 3a.

where A = the area from which sublimation is occurring. Substituting for dN/dt and v in the second expression, we find that

 nA = 1019 /m 4

Setting A = .06m2, we find

 n = 2 x 1020/m3 5

and the pressure p = (2 x 1020/m3)/(2 x 1025/m3) atm = 2 x 10-5 atm, (where n0 =2 x 1025/m3 is Loschmidt's number, ie., the number density of an ideal gas at STP).

[Nota Bene: This calculated value of pressure may be used in estimating the probability of Paschen breakdown occurring in the immediate vicinity of an astronaut who might be working near an exposed high voltage if the Paschen curve of water vapor under the specified conditions is known.]

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