Respuesta :
Answer: The vapor pressure of water in the air is 27.58 torr
Explanation:
We are given:
Temperature of water in air = 85.0°F
Converting the temperature from degree Fahrenheit to degree Celsius is:
[tex]^oF=\frac{9}{5}^oC+32[/tex]
where,
[tex]^oF[/tex] = temperature in Fahrenheit
[tex]^oC[/tex] = temperature in centigrade
So,
[tex]85.0^oF=\frac{9}{5}^oC+32\\\\T(^oC)=29.44^oC[/tex]
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = initial pressure which is the pressure at normal boiling point = 760 torr
[tex]P_2[/tex] = final pressure = ?
[tex]\Delta H_{vap}[/tex] = Enthalpy of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature = [tex]100^oC=[100+273]K=373K[/tex]
[tex]T_2[/tex] = final temperature = [tex]29.44^oC=[29.44+273]=302.44K[/tex]
Putting values in above equation, we get:
[tex]\ln(\frac{P_2}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{302.44}]\\\\P_2=35.72torr[/tex]
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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
- For water:
Molar mass of water = 18 g/mol
[tex]\text{Moles of water}=\frac{68g}{18g/mol}=3.78mol[/tex]
- For air:
Average molar mass of air = 29 g/mol
[tex]\text{Moles of air}=\frac{32g}{29g/mol}=1.103mol[/tex]
Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
[tex]\chi_{water}=\frac{n_{water}}{n_{water}+n_{air}}\\\\\chi_{water}=\frac{3.78}{3.78+1.103}=0.774[/tex]
To calculate the mole fraction of substance, we use the equation given by Raoult's law, which is:
[tex]p_{A}=p_T\times \chi_{A}[/tex]
where,
[tex]p_A[/tex] = vapor pressure of water = ?
[tex]p_T[/tex] = total pressure = 35.72 torr
[tex]\chi_A[/tex] = mole fraction of water = 0.774
Putting values in above equation, we get:
[tex]p_{water}=35.72\times 0.772\\\\p_{water}=27.58torr[/tex]
Hence, the vapor pressure of water in the air is 27.58 torr
