*** Please either answer to all parts or skip my question. I need to compare wit

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                            *** Please either answer to all parts or skip my question. I need to compare with my own answer***

Problem 2: Boiling temperature of water as a function of altitude Assume that the atmosphere is isothermal, and that it falls off exponentially with altitude as we found earlier this semester _mgh P = Poexp kpT where Po is the atmospheric pressure at sea level In problem #1 you showed that under normal conditions the vapor pressure of water is little effected by the presence of the inert atmospheric components. The vapor pressure of the water is given by where R kbNa and C is a constant. The molar heat of vaporization is relatively constant over the tempera ture range from 50 to 100°C and has the value L~50kJ/mole. a) Assume that when the vapor pressure of the water becomes equal to the atmospheric pressure, the water begins to boil. Show that the temperature at which the liquid boils as a function of the altitude, h, is given by: Tbo Ta L where m is the mass of a nitrogen molecule and Tbo is the boiling temperature at atmospheric pressure b) Use the chain rule to calculate the change in boiling temperature for small changes in altitude: dTb/dh (dTb/dP) (dP/dh). Show that the result of part (a) is equivalent to this result for small changes in altitude c) Use your result from part (a) to estimate the temperature at which water boils on the summit of Mount Whitney (4417 meters). What implications could this have for high altitude cooking?

Explanation / Answer

for isothermal atmoshphere

P = Poexp(-mgh/kbT)

alos, vapour pressure P = Cexp(-L/RTw)

a. for vapour pressure to be equal to the pressure at some altitude h

Po*exp(-mgh/kb*T) = Cexp(-L/RT)

also

Po = Cexp(-L/RTbo)

hence

Cexp(-L/RTbo)exp(-mgh/kb*T) = Cexp(-L/RT)

exp(-L/RTbo-mgh/kb*T) = exp(-L/RT)

L/RTbo+mgh/kb*T = L/RT

L/RTbo = L/RT - mgh/kb*T = (1/T)(L/R - mgh/kb)

R = N*kb

T = (L/R - mgh/kb)RTbo/L = Tbo(1 - mgh*N/L)

let 1 - mghN/L = Ta/Tbo

then

T = Tbo/(1 + Tbo mghN/Ta*L)

b. dTb/dh = (dTb/dP)(dP/dh) = -Tbo(Tbo mgN/Ta*L)/(1 + Tbo mghN/Ta*L)^2

dTb/dh = -Tbo^2(mgN)/TaL(1 + Tbo mghN/Ta*L)^2

c. h = 4417 m

T = 373.16(1 - 6.022*10^-23*28.97*1.6*10^-27*9.81*4417/50,000) = 364.11797513727482403584 K

hence

Tb = 90.95 C

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