# Question #e4ab2

##### 1 Answer

#### Explanation:

This looks like a job for the **Clausius - Clapeyron equation**, which looks like this

#color(blue)(ul(color(black)(ln(P_1/P_2) = (DeltaH_"vap")/R * (1/T_2 - 1/T_1))))#

Here

Now, you know that at an unknown atmospheric pressure

#T_2 = 110^@"C" + 273.15 = "383.15 K"#

Moreover, you know that at *normal pressure*, i.e. at

#T_1 = 100^@"C" + 273.15 = "373.15 K"#

As you know, a liquid *boils* when its vapor pressure is **equal** to the atmospheric pressure, so you can use the two atmospheric pressures in the Clausius - Clapeyron equation as

So, plug in your values into the equation to get

#ln("1 atm"/P_2) = ("40,660" color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("mol"^(-1)))))/(8.3145 color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("mol"^(-1)))) color(red)(cancel(color(black)("K"^(-1))))) * (1/(383.15 color(red)(cancel(color(black)("K")))) - 1/(273.15color(red)(cancel(color(black)("K")))))#

#ln("1 atm"/P_2) = -0.342041625#

This will be equivalent to

#e^ln("1 atm"/P_2) = e^(-0.342041625)#

which will get you

#"1 atm"/P_2 = 0.71032#

You can thus say that

#P_2 = "1 atm"/0.71032 = color(darkgreen)(ul(color(black)("1.4 atm")))#

Now, does this result make sense?

As atmospheric pressure **increases**, the molecules of water will need *more energy* in order to escape from the liquid, which implies that the temperature of the water will be **higher** at the new boiling point.

So in your case, a higher boiling temperature implies a higher a higher value for the vapor pressure of water at its boiling point, i.e. a higher atmospheric pressure.