This question is incomplete, the complete is;
Water has a heat of vaporization (ΔHvap) of 44.01 kJ mol-1 and boils at 100 degrees C at sea level.
What is the boiling point of water at the top of Mount Everest where the atmospheric pressure is only 34% as strong as the pressure at sea level?
Answer: the boiling point of water at the top of mount Everest where the atmospheric pressure is only 34% as strong is 74 °C
Explanation:
Given that;
P1 = 1 atm,
T1 = 100°C = 373 K
P2 = P1 × 34% = 1×0.34 = 0.34 atm
T1 = ?
ΔH = 44.01 kJ mol⁻¹ = 44.01×10³ J/mol
R = 8.314 J/k.mol
now using the Clausius - Clapeyron equation;
p1_∫^p2 d.InP = ΔHvap/R T1_∫^T2. 1/T².dT
⇒ In(P2/P1) = ΔHvap/R (1/T1 - 1/T2)
so we substitute;
ln( 0.34/1 ) = ( 44.01×10³ / 8.314) × ( 1/373 - 1/T2)
-1.0788 = 5293.4808 × ( 0.00268 - 1/T2)
-1.0788 = 14.1865 - 5293.4808(1/T2)
5293.4808(1/T2) = 14.1865 + 1.0788
5293.4808(1/T2) = 15.2653
(1/T2) = 15.2653 / 5293.4808
(1/T2) = 0.0028837
T2 = 1 / 0.0028837
T2 = 346.8 K
WE convert to Celsius
t2 =346.8 K − 273.15 = 73.65 °C ≈ 74 °C
Therefore, the boiling point of water at the top of mount Everest where the atmospheric pressure is only 34% as strong is 74 °C
