You are in a mountain range with atmospheric air pressure of 520 , and you wish to boil some eggs. What is the approximate boiling point of the water at this air pressure?
Express your answer as an integer and include the appropriate units.
88.6 *C
There may be an easier way to do this (a graph but you didn't say you had one). You can use the Clausius-Clapeyron equation and solve for T2.
ln (p2/p1) = (dHvap/R)(1/T1 - 1/T2)
p1 = 760; T1 = 100
p2 = 520; T2 = ?
dHvap is heat vaporization in J. You'll need to look that up. T2 will be the boiling point of the water at 520 mm Hg pressure.
To determine the approximate boiling point of water at a given atmospheric pressure, we can refer to the vapor pressure-temperature relationship of water. This relationship is known as the Clausius-Clapeyron equation, which states that the natural logarithm of the vapor pressure of a substance is inversely proportional to its temperature.
To calculate the boiling point, we need to know the air pressure and its relationship to the standard atmospheric pressure at sea level, which is approximately 1013.25 hPa (hectopascals).
Let's convert the given atmospheric pressure from 520 to 520/1013.25 times the standard atmospheric pressure:
Conversion factor = 520/1013.25 = 0.513
Therefore, the adjusted atmospheric pressure at the mountain range is approximately 0.513 times the standard atmospheric pressure.
Next, we need to determine the boiling point of water at the adjusted pressure. At standard atmospheric pressure, water boils at 100 degrees Celsius. Since we are at an adjusted pressure, we can use the Clausius-Clapeyron equation to estimate the boiling point.
The equation is as follows:
ln(P1/P2) = -ΔHvap/R * (1/T1 - 1/T2)
Where:
P1 and P2 are the two pressures (standard atmospheric and adjusted atmospheric pressure)
ΔHvap is the enthalpy of vaporization (40.7 kJ/mol for water)
R is the ideal gas constant (8.314 J/(mol * K))
T1 and T2 are the two temperatures (standard boiling point and the unknown boiling point)
Rearranging the equation to solve for T2:
T2 = ΔHvap / (R * (-ln(P2/P1)) + 1/T1)
Plugging in the values:
T1 = 100 degrees Celsius = 373.15 Kelvin
P1 = standard atmospheric pressure = 1013.25 hPa
P2 = adjusted atmospheric pressure = 0.513 * 1013.25 hPa
Calculating T2:
T2 = (40.7 kJ/mol) / (8.314 J/(mol * K) * (-ln(0.513 * 1013.25 / 1013.25)) + (1/373.15)
Simplifying the equation:
T2 = (40.7 kJ/mol) / (-8.314 J/mol * (ln(0.513)) + (1/373.15)
Using a calculator:
T2 ≈ 99.2 degrees Celsius
Therefore, the approximate boiling point of water at an atmospheric pressure of 520 hPa is 99 degrees Celsius.