Question

What is the approximate vapor pressure when the liquid water boils at about 50°C?

A. O 230 mmHg

B. 5 mmHg

C. O 760 mmHg

D. O 380 mmHg

Answer 1

Hi there!

the correct answer to this question is: A

Step-by-step 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 2

Answer: The vapor pressure of the water when it boils at 50°C is 99.91 mmHg

Step-by-step explanation:

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

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

where,

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

`P_2` = final pressure = ?

`\Delta H` = Enthalpy of vaporization = 40.65 kJ/mol = 40650 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 =
`50^oC=[50+273]K=323K`

Putting values in above equation, we get:

`\ln((P_2)/(760))=(40650J/mol)/(8.314J/mol.K)[(1)/(373)-(1)/(323)]\\\\\ln (P_2)/(760)=-2.029mmHg\\\\(P_2)/(760)=e^(-2.029)=0.1315mmHg\\\\P_2=0.1315* 760=99.91mmHg`

Hence, the vapor pressure of the water when it boils at 50°C is 99.91 mmHg