what is the boiling point of water at the top of mount everest where the atmospheric pressure is only 34% as strong

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?

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

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explanation: