calculate the boiling point of water if the atmospheric is 630 mm hg
Use the Clausius-Clapeyron equation.
P1 is 630mm Hg and T1 is the unknown.
P2 is 760 mm Hg and T2 is 100 C.
Remember T must be used in kelvin.
98.6 degree
To calculate the boiling point of water at a given atmospheric pressure, you can use the Clausius-Clapeyron equation. This equation relates the boiling point of a substance to its vapor pressure at a different temperature.
The Clausius-Clapeyron equation is given as:
ln(P1/P2) = -ΔHvap/R * [(1/T1) - (1/T2)]
where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
ΔHvap = enthalpy of vaporization for the substance
R = gas constant (8.314 J/(mol·K))
In our case, we want to find the boiling point of water (T2) when the atmospheric pressure is 630 mm Hg. We know that the vapor pressure of water at its boiling point is equal to the atmospheric pressure.
Let's assume the normal boiling point of water at 1 atm (760 mm Hg) is 100°C (373 K). We can plug these values into the equation and solve for T2.
ln(P1/P2) = -ΔHvap/R * [(1/T1) - (1/T2)]
ln(760/630) = -ΔHvap/8.314 * [(1/373) - (1/T2)]
Now, we can rearrange the equation to solve for T2:
ln(760/630) * 8.314 = (1/373) - (1/T2)
ln(760/630) * 8.314 + 1/373 = 1/T2
Substituting the values and simplifying:
(0.2075) * 8.314 + (1/373) = 1/T2
1.721 + (1/373) = 1/T2
1.721 + 0.0027 = 1/T2
1.7237 = 1/T2
Now, we can solve for T2:
T2 = 1/1.7237
T2 ≈ 0.58
Therefore, the boiling point of water at an atmospheric pressure of 630 mm Hg is approximately 0.58°C.
To calculate the boiling point of water at a given atmospheric pressure, you can make use of the Clausius-Clapeyron equation. This equation establishes the relationship between the vapor pressure of a substance and its boiling point:
ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and T1 are the initial vapor pressure and boiling point of the substance (known values)
P2 is the desired vapor pressure (in this case, the atmospheric pressure)
T2 is the desired boiling point (the value you want to find)
ΔHvap is the enthalpy of vaporization of the substance (for water, it is 40.7 kJ/mol)
R is the ideal gas constant (8.314 J/(mol·K))
In this case, the initial vapor pressure (P1) is the known vapor pressure of water at its normal boiling point, which is 760 mmHg.
Let's solve the equation for T2, keeping in mind that the unit for pressure should be consistent (mmHg or atm) and that the units for R and ΔHvap should also match:
ln(P2/760 mmHg) = (40.7 kJ/mol / (8.314 J/(mol·K))) * (1/T2 - 1/373.15 K)
Now, substitute the given atmospheric pressure, P2 (630 mmHg in this case), into the equation:
ln(630 mmHg/760 mmHg) = (40.7 kJ/mol / (8.314 J/(mol·K))) * (1/T2 - 1/373.15 K)
Simplifying further:
ln(0.8289) = 4.9/T2 - 4.9/373.15
Now, rearrange the equation to isolate T2:
4.9/T2 = ln(0.8289) + 4.9/373.15
T2 = 4.9 / (ln(0.8289) + 4.9/373.15)
Using a calculator, you can determine the value of T2:
T2 ≈ 99.8°C
Therefore, at an atmospheric pressure of 630 mmHg, the boiling point of water is approximately 99.8°C.