216k views
1 vote
The ambient temperature is 85.0°F and the humidity of the surrounding air is reported to be 68.0%. Using the Clausius-Clapeyron equation and the boiling point of water as 100.0°C at 760 torr, calculate the vapor pressure (in torr) of water in the air. Use 40.7 kJ/mol as the ∆Hvap of water.

1 Answer

0 votes

Answer: The vapor pressure of water in the air is 27.58 torr

Step-by-step explanation:

We are given:

Temperature of water in air = 85.0°F

Converting the temperature from degree Fahrenheit to degree Celsius is:


^oF=(9)/(5)^oC+32

where,


^oF = temperature in Fahrenheit


^oC = temperature in centigrade

So,


85.0^oF=(9)/(5)^oC+32\\\\T(^oC)=29.44^oC

To calculate the final pressure, we use the Clausius-Clayperon equation, which is:


\ln((P_2)/(P_1))=(\Delta H)/(R)[(1)/(T_1)-(1)/(T_2)]

where,


P_1 = initial pressure which is the pressure at normal boiling point = 760 torr


P_2 = final pressure = ?


\Delta H_(vap) = Enthalpy of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K


T_1 = initial temperature =
100^oC=[100+273]K=373K


T_2 = final temperature =
29.44^oC=[29.44+273]=302.44K

Putting values in above equation, we get:


\ln((P_2)/(760))=(40700J/mol)/(8.314J/mol.K)[(1)/(373)-(1)/(302.44)]\\\\P_2=35.72torr

We are given:

68.0 % of water in the air

This means that 68 grams of water is present in the air

Mass of air = 100 - 68 = 32 g

To calculate the number of moles, we use the equation:


\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

  • For water:

Molar mass of water = 18 g/mol


\text{Moles of water}=(68g)/(18g/mol)=3.78mol

  • For air:

Average molar mass of air = 29 g/mol


\text{Moles of air}=(32g)/(29g/mol)=1.103mol

Mole fraction of a substance is given by:


\chi_A=(n_A)/(n_A+n_B)


\chi_(water)=(n_(water))/(n_(water)+n_(air))\\\\\chi_(water)=(3.78)/(3.78+1.103)=0.774

To calculate the mole fraction of substance, we use the equation given by Raoult's law, which is:


p_(A)=p_T* \chi_(A)

where,


p_A = vapor pressure of water = ?


p_T = total pressure = 35.72 torr


\chi_A = mole fraction of water = 0.774

Putting values in above equation, we get:


p_(water)=35.72* 0.772\\\\p_(water)=27.58torr

Hence, the vapor pressure of water in the air is 27.58 torr

User Alecov
by
6.6k points