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

Answer:

Water will boil at $76^{0}\textrm{C}$.

Explanation:

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

                                         $ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]$

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:

                          $ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]$

                       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 $76^{0}\textrm{C}$