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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.

Respuesta :

Answer:

Water will boil at [tex]76^{0}\textrm{C}[/tex].

Explanation:

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

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

where, [tex]P_{1}[/tex] and [tex]P_{2}[/tex] are vapor pressures of liquid at [tex]T_{1}[/tex] (in kelvin) and [tex]T_{2}[/tex] (in kelvin) temperatures respectively.

Here, [tex]P_{1}[/tex] = 760.0 mm Hg, [tex]T_{1}[/tex] = 373 K, [tex]P_{2}[/tex] = 314.0 mm Hg

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

                          [tex]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}][/tex]

                       or, [tex]T_{2}=349 K[/tex]

So, [tex]T_{2}=349K=(349-273)^{0}\textrm{C}=76^{0}\textrm{C}[/tex]

Hence, at base camp, water will boil at [tex]76^{0}\textrm{C}[/tex]

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