Respuesta :
Answer:
Hi there!
the correct answer to this question is: A
Explanation:
The correct answer is 230 mmHg. Hence, when liquid water boils at 50°C, the water vapor pressure is (760/2) mmHg or 230 mm Hg
Answer: The vapor pressure of the water when it boils at 50°C is 99.91 mmHg
Explanation:
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{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 mmHg
[tex]P_2[/tex] = final pressure = ?
[tex]\Delta H[/tex] = Enthalpy of vaporization = 40.65 kJ/mol = 40650 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]50^oC=[50+273]K=323K[/tex]
Putting values in above equation, we get:
[tex]\ln(\frac{P_2}{760})=\frac{40650J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{323}]\\\\\ln \frac{P_2}{760}=-2.029mmHg\\\\\frac{P_2}{760}=e^{-2.029}=0.1315mmHg\\\\P_2=0.1315\times 760=99.91mmHg[/tex]
Hence, the vapor pressure of the water when it boils at 50°C is 99.91 mmHg
