Question

The vapor pressure of a substance describes how readily molecules at the surface of the substance enter the gaseous phase. At the boiling point of a liquid, the liquid's vapor pressure is equal to or greater than the atmospheric pressure exerted on the surface of the liquid. Since the atmospheric pressure at higher elevations is lower than at sea level, the boiling point of water decreases as the elevation increases. The atmospheric pressure at sea level is 760 mmHg. This pressure decreases by 19.8 mmHg for every 1000-ft increase in elevation.Elevation Pressure0 m 760 mmHg1000 m 695 mmHg2000 m 630 mmHgThe boiling point of water decreases 0.05 ?C for every 1 mmHg drop in atmospheric pressure.A) What is the boiling point of water at an elevation of 7000m ?

Answer 1

Answer:

84.75°C is the boiling point of water at an elevation of 7000 meter.

Explanation:

Rate of change of pressure = 19.8 mmHg/1000 ft

1 foot = $\frac{1}{3.28}$ meter

$19.8 mmHg/1000 ft=\frac{19.8}{1000}\times 3.28 mmHg/m=0.065 mmHg$

Pressure change for every 1 m = 0.065 mmHg × 1= 0.065 mmHg

Elevation      Pressure

     0 m       760 mmHg

1000 m       695 mmHg

2000 m       630 mmHg

Pressure drop at the elevation of 7000 m: $7000\times 0.065 mmHg=455 mmHg$

Pressure at 7000 m = 760 mmHg - 455mmHg = 305 mmHg

The boiling point of water decreases 0.05°C for every 1 mmHg drop in atmospheric pressure.

At 7000 meter elevation the boiling of water will be :

$0.05^oC\times 305=15.25^oC$

Boiling point of water at 7000 meter elevation :

$100.0^oC-15.25^oC =84.75^oC$

84.75°C is the boiling point of water at an elevation of 7000 meter.