3. The saturation vapor pressure over ice is given by the equatiorn where a, b a

Question

3. The saturation vapor pressure over ice is given by the equatiorn where a, b are constants and T is the temperature in units of [K]. Starting from the Clausius-Clapeyron equation, determine the values of a and b; for the latent heat of vaporization use L 2.834 × 106 J kg-1 and for the specific gas constant for water vapor use RW = 461.7 J kg-1 K-1. Show your calculations 4. We observe saturation vapor pressure values for water vapor at-5°C,-10 and-15°C we get 408.43, 255.81, and 163.93 Pa, respectively. Compute the errors that we would make by using the equation derived in Problem 3. instead of using the measurements

Explanation / Answer

Solution :- Given data , L = 2.843 ×10^6 j kg^-1

Rw = 461.7kg^-1K^-1

Clausius - Clapeyron equation is given by

dP/dT = L/T(delta)v , where (delta)v is the change in specific taking it as 1

log10es = a - b/T

T = aT -b/log10es

T - aT = -b/log10es

T(1 - a) = -b/log10es

T = -b/log10es x (1 - a)

Substituting the value of T in the above equation we get,

dP/d(-b/log10es x ( 1 - a) = L/[-b/ log10es (1 - a) ]

dP = L/[b/log10es (1 - a)] d(b/log10es (1 -a ))

integrating both sides, we get

P = L = 2.834 x 106 J/kg , since P = L, therefore,

Therefore, the saturation vapor pressure is equal to the latent heat of vaporisation,

log10es = a - b/T

log10{2.834 x 106) = a - b/T

6.45 = a - b/T

since T = -b/log10(es) x ( 1 -a )

6 .45 = a - b/[b/log10es (1 -a )]

6.45 = a - log10es ( 1 -a)

6.45 = a - 6.45 - 6.45a

12.90 = - 6.44a

a = 12.90/-6.44

a = -2.0031

6.45 = -2.0031 - b/T

b/T = 8.4531

taking T as 1

b = 8.4531.

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