Calculating Energy Change Phase Transition Water To Steam

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In thermodynamics, understanding energy changes during phase transitions is crucial. A common example is the conversion of water from its liquid state to steam. This process, occurring at a constant temperature and pressure, involves the absorption of heat energy. This article delves into the calculation of energy change when one mole of water at 100°C and 1 atm pressure is converted to steam at the same temperature. We will explore the concepts of enthalpy, heat of vaporization, and the application of the first law of thermodynamics to determine the energy change (ΔU) for this process.

Phase transitions are physical processes that involve the change of a substance from one state of matter to another. These transitions, such as melting, boiling, and sublimation, occur at specific temperatures and pressures. When water transforms from liquid to steam at 100°C (its boiling point at 1 atm pressure), it undergoes a phase transition known as vaporization. This process requires energy input to overcome the intermolecular forces holding the water molecules together in the liquid phase. The energy absorbed during this phase transition at constant pressure is known as the enthalpy change or heat of vaporization (ΔHvap).

Enthalpy (H) is a thermodynamic property of a system, defined as the sum of the system's internal energy (U) and the product of its pressure (P) and volume (V): H = U + PV. Enthalpy is a state function, meaning its value depends only on the initial and final states of the system, not the path taken. Changes in enthalpy (ΔH) are particularly useful for analyzing processes occurring at constant pressure, such as phase transitions under atmospheric conditions.

The heat of vaporization (ΔHvap) is a specific type of enthalpy change that represents the amount of heat required to convert one mole of a substance from liquid to gas at its boiling point. For water at 100°C and 1 atm, the heat of vaporization is a significant value, reflecting the strong hydrogen bonds between water molecules. This high heat of vaporization is crucial for many natural processes, including climate regulation and the transport of heat in living organisms. In this case, we are given that the amount of heat absorbed during the conversion of one mole of water to steam is 40670 Joules (J).

The first law of thermodynamics states that energy is conserved; it cannot be created or destroyed, but it can be transferred or converted from one form to another. Mathematically, the first law is expressed as: ΔU = Q - W, where:

  • ΔU is the change in internal energy of the system
  • Q is the heat added to the system
  • W is the work done by the system

In this scenario, the heat absorbed (Q) during the vaporization of one mole of water is given as 40670 J. The work done (W) by the system during this process is primarily due to the expansion of volume as water transitions from a liquid to a gaseous state. At constant pressure, the work done can be calculated using the formula: W = PΔV, where:

  • P is the pressure
  • ΔV is the change in volume

To calculate ΔV, we need to consider the molar volumes of water and steam at 100°C and 1 atm. The molar volume of liquid water is relatively small compared to that of steam. We can approximate the volume of steam using the ideal gas law: PV = nRT, where:

  • P is the pressure (1 atm)
  • V is the volume
  • n is the number of moles (1 mole)
  • R is the ideal gas constant (8.314 J/(mol·K))
  • T is the temperature in Kelvin (100°C = 373.15 K)

Rearranging the ideal gas law to solve for V, we get:

V = nRT / P

Substituting the values:

V = (1 mol) * (8.314 J/(mol·K)) * (373.15 K) / (1 atm)

Since 1 atm = 101325 Pa (Pascals), we need to convert the pressure to Pascals to maintain consistent units. Also, we need to express the volume in cubic meters (m³) since the ideal gas constant is in J/(mol·K), and 1 J = 1 Pa·m³.

P = 1 atm = 101325 Pa

V = (1 mol) * (8.314 J/(mol·K)) * (373.15 K) / 101325 Pa

V ≈ 0.0306 m³

This volume represents the molar volume of steam. The molar volume of liquid water at 100°C is approximately 1.88 x 10⁻⁵ m³, which is significantly smaller than the volume of steam. Therefore, the change in volume (ΔV) can be approximated as the volume of steam:

ΔV ≈ 0.0306 m³

Now we can calculate the work done:

W = PΔV = (101325 Pa) * (0.0306 m³)

W ≈ 3100 J

Finally, we can calculate the change in internal energy (ΔU) using the first law of thermodynamics:

ΔU = Q - W

ΔU = 40670 J - 3100 J

ΔU ≈ 37570 J

To provide a clearer understanding, let’s break down the calculation process step by step:

  1. Identify the given information:
    • Heat absorbed (Q) = 40670 J
    • Pressure (P) = 1 atm = 101325 Pa
    • Temperature (T) = 100°C = 373.15 K
    • Number of moles (n) = 1 mole
    • Ideal gas constant (R) = 8.314 J/(mol·K)
  2. Calculate the volume of steam (Vsteam) using the ideal gas law:
    • V = nRT / P
    • V = (1 mol * 8.314 J/(mol·K) * 373.15 K) / 101325 Pa
    • V ≈ 0.0306 m³
  3. Estimate the change in volume (ΔV):
    • The volume of liquid water is negligible compared to the volume of steam.
    • ΔV ≈ Vsteam ≈ 0.0306 m³
  4. Calculate the work done (W) by the system:
    • W = PΔV
    • W = 101325 Pa * 0.0306 m³
    • W ≈ 3100 J
  5. Calculate the change in internal energy (ΔU) using the first law of thermodynamics:
    • ΔU = Q - W
    • ΔU = 40670 J - 3100 J
    • ΔU ≈ 37570 J

Thus, the energy change (ΔU) for the conversion of one mole of water to steam at 100°C and 1 atm is approximately 37570 J.

The calculated energy change (ΔU) of approximately 37570 J represents the increase in the internal energy of the system as water transitions from the liquid to the gaseous phase. This energy is primarily used to overcome the intermolecular forces, specifically the hydrogen bonds, that hold water molecules together in the liquid state. The molecules in the gaseous phase have greater kinetic energy and are more spread out, resulting in a higher internal energy.

Understanding the energy changes associated with phase transitions is critical in various fields:

  • Meteorology: The evaporation of water is a key process in the Earth's climate system, influencing humidity, cloud formation, and precipitation. The energy absorbed during evaporation is later released during condensation, driving weather patterns.
  • Engineering: In industrial processes such as steam power generation, the heat of vaporization of water is utilized to convert water into high-pressure steam, which then drives turbines to produce electricity. The efficiency of these processes depends on accurate knowledge of the energy changes involved.
  • Chemistry: Chemical reactions often involve phase transitions, and understanding the associated energy changes is crucial for predicting reaction outcomes and optimizing reaction conditions.
  • Biology: The high heat of vaporization of water plays a crucial role in thermoregulation in living organisms. Evaporation of sweat, for example, helps to dissipate heat and cool the body.

When calculating energy changes during phase transitions, several common mistakes can occur. Being aware of these pitfalls can help ensure accurate results:

  1. Incorrect unit conversions: Ensure that all values are expressed in consistent units. For example, pressure should be in Pascals (Pa), volume in cubic meters (m³), and temperature in Kelvin (K) when using the ideal gas law. Failing to convert units properly can lead to significant errors.
  2. Neglecting the work done: The work done during expansion is a critical component of the energy change calculation. It’s essential to calculate the work done (W = PΔV) and subtract it from the heat absorbed (Q) to find the change in internal energy (ΔU). Omitting this step will result in an overestimation of ΔU.
  3. Approximating ΔV: While the volume of liquid water is much smaller than that of steam, it should not be completely ignored, especially in high-precision calculations. However, for most practical purposes, approximating ΔV as the volume of steam is reasonable.
  4. Misunderstanding the sign conventions: Heat absorbed by the system (endothermic process) is positive, while heat released by the system (exothermic process) is negative. Work done by the system is positive, while work done on the system is negative. Using the correct sign conventions is crucial for accurate calculations.
  5. Using incorrect values for constants: Ensure that the correct values for constants like the ideal gas constant (R = 8.314 J/(mol·K)) are used. Using an incorrect value will lead to errors in the calculations.

In summary, calculating the energy change during the phase transition of water from liquid to steam involves applying the first law of thermodynamics and understanding the concepts of enthalpy and heat of vaporization. For the conversion of one mole of water at 100°C and 1 atm to steam, the energy change (ΔU) is approximately 37570 J. This calculation highlights the significant amount of energy required to overcome intermolecular forces during phase transitions. Accurate determination of energy changes is essential in various scientific and engineering applications, ranging from meteorology to industrial processes. By understanding the principles and steps involved, and by avoiding common mistakes, we can confidently analyze and predict energy changes in thermodynamic processes. The intricate dance of energy during phase transitions underscores the fundamental principles governing the behavior of matter and its transformations.