Question

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured to be 314.0 mm Hg. At what temperature (in °C) will water boil at base camp ? The vapor pressure of water at 373 K is 760.0 mm Hg. (ΔH°vap for H2O = 40.7 kJ/mol and R = 8.314 J/mol K).

a. 344°C.
b. 70.8°C .
c. 2.91E-3°C .
d. 79.8°C.
e. 57.8°C.

Answer 1

Water will boil at
`76^(0)\textrm{C}`.

Step-by-step explanation:

According to clausius-clapeyron equation for liquid-vapor equilibrium:

`ln((P_(2))/(P_(1)))=(-\Delta H_(vap)^(0))/(R)[(1)/(T_(2))-(1)/(T_(1))]`

where,
`P_(1)` and
`P_(2)` are vapor pressures of liquid at
`T_(1)` (in kelvin) and
`T_(2)` (in kelvin) temperatures respectively.

Here,
`P_(1)` = 760.0 mm Hg,
`T_(1)` = 373 K,
`P_(2)` = 314.0 mm Hg

Plug-in all the given values in the above equation:

`ln((314.0)/(760.0))=(-40.7* 10^(3)(J)/(mol))/(8.314(J)/(mol.K))* [(1)/(T_(2))-(1)/(373K)]`

or,
`T_(2)=349 K`

So,
`T_(2)=349K=(349-273)^(0)\textrm{C}=76^(0)\textrm{C}`

Hence, at base camp, water will boil at
`76^(0)\textrm{C}`