Is Boiling Point an Intensive Property?

Boiling point is indeed an intensive property, which means it does not depend on the amount of the substance present. This property is a crucial characteristic of a substance and remains constant under the same conditions of temperature and pressure, regardless of the quantity of the substance. In this comprehensive blog post, we will delve into the technical details and specific aspects of boiling point as an intensive property, providing a valuable resource for physics students.

Understanding Intensive Properties

Intensive properties are physical or chemical properties of a substance that do not depend on the amount or size of the system. These properties are characteristic of the substance itself and are independent of the quantity of the material. Some examples of intensive properties include:

1. Temperature: The temperature of a substance is an intensive property, as it does not change with the amount of the substance present.
2. Density: The density of a substance is an intensive property, as it is a measure of the mass per unit volume and does not depend on the total mass or volume of the substance.
3. Melting point: The melting point of a substance is an intensive property, as it is the temperature at which the solid and liquid phases of the substance coexist in equilibrium.

In contrast, extensive properties are those that depend on the amount or size of the system, such as mass, volume, and total energy.

Boiling Point as an Intensive Property

The boiling point of a substance is the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid, and bubbles of vapor form inside the liquid. This temperature is an intensive property because it is a characteristic of the substance itself and does not depend on the quantity of the substance.

Factors Affecting Boiling Point

The boiling point of a substance is influenced by several factors, including:

1. Pressure: The boiling point of a substance is directly related to the pressure exerted on the liquid. As the pressure increases, the boiling point also increases, and vice versa. This relationship is described by the Clausius-Clapeyron equation:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

where P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively, ΔHvap is the enthalpy of vaporization, and R is the universal gas constant.

1. Intermolecular forces: The strength of the intermolecular forces between the molecules of a substance affects its boiling point. Substances with stronger intermolecular forces, such as hydrogen bonding, generally have higher boiling points.

2. Molecular mass: The molecular mass of a substance also influences its boiling point. Substances with higher molecular masses tend to have higher boiling points, as they require more energy to overcome the intermolecular forces and transition to the gaseous state.

Examples of Boiling Point as an Intensive Property

1. Water: The boiling point of water is 100°C (212°F) at a pressure of 1 atmosphere (101.325 kPa). This boiling point is an intensive property, as it remains the same regardless of the amount of water present.

2. Ethanol: The boiling point of ethanol (C2H5OH) is 78.3°C (172.9°F) at a pressure of 1 atmosphere. This boiling point is an intensive property and does not change with the quantity of ethanol.

3. Nitrogen: The boiling point of nitrogen (N2) is -195.8°C (-320.4°F) at a pressure of 1 atmosphere. This boiling point is an intensive property and is a characteristic of the nitrogen molecule.

Numerical Problems

1. Problem: The boiling point of water at a pressure of 2 atm is approximately 121°C. Calculate the boiling point of water at a pressure of 3 atm.

Solution:
Using the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
where:
– P1 = 2 atm
– P2 = 3 atm
– T1 = 121°C (394.15 K)
– ΔHvap = 40.65 kJ/mol
– R = 8.314 J/mol·K

Solving for T2:
ln(3/2) = (40650 J/mol) / (8.314 J/mol·K) * (1/394.15 K - 1/T2)
T2 = 127.6°C
Therefore, the boiling point of water at a pressure of 3 atm is approximately 127.6°C.

1. Problem: The boiling point of ethanol at a pressure of 1 atm is 78.3°C. Calculate the boiling point of ethanol at a pressure of 0.5 atm.

Solution:
Using the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
where:
– P1 = 1 atm
– P2 = 0.5 atm
– T1 = 78.3°C (351.45 K)
– ΔHvap = 38.56 kJ/mol
– R = 8.314 J/mol·K

Solving for T2:
ln(0.5/1) = (38560 J/mol) / (8.314 J/mol·K) * (1/351.45 K - 1/T2)
T2 = 72.6°C
Therefore, the boiling point of ethanol at a pressure of 0.5 atm is approximately 72.6°C.

Figures and Data Points

To further illustrate the concept of boiling point as an intensive property, let’s consider the following figure and data points:

The graph shows the relationship between the boiling point and the pressure for various substances. The key data points are:

Substance	Boiling Point at 1 atm (°C)
Water	100
Ethanol	78.3
Nitrogen	-195.8
Oxygen	-183.0
Hydrogen	-252.8

As you can see, the boiling point of each substance is an intensive property, as it remains constant regardless of the quantity of the substance. The boiling point is a characteristic of the substance itself and is influenced by factors such as intermolecular forces and molecular mass, as discussed earlier.

Conclusion

In summary, boiling point is an intensive property, which means it does not depend on the amount of the substance present. The boiling point of a substance is a characteristic property that remains constant under the same conditions of temperature and pressure, regardless of the quantity of the substance. Understanding the concept of boiling point as an intensive property is crucial for physics students, as it helps them to accurately identify and analyze the properties of different substances.

References:

• Intensive and Extensive Properties
• Clausius-Clapeyron Equation
• Boiling Point and Intermolecular Forces