Question

The molar enthalpy of vaporization of water at \(373 \mathrm{K}\) and 1.00 atm is \(40.7 \mathrm{kJ} / \mathrm{mol} .\) What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Hint: Assume that water vapor is an ideal gas.)

Short answer

The fraction of energy used to change the internal energy of the water is approximately 1.08, and the fraction of energy used to do work against the atmosphere is approximately -0.08. The negative sign indicates that work is done against the atmospheric-pressure volume-extension.

Step-by-step solution

Ideal Gas Law

Recall the ideal gas law, given as: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. In this problem, we have the temperature and pressure, as well as the number of moles (since we are working with molar enthalpy).

Enthalpy Change

The molar enthalpy of vaporization is the energy required to vaporize 1 mole of water. In this case, this enthalpy change, ΔH, is given as \(40.7 \frac{kJ}{mol}\). The change in enthalpy is related to the change in internal energy (ΔU) and the work done against the atmosphere (W) by: ΔH = ΔU + W

Work Done Against the Atmosphere

In this problem, the work done against the atmosphere is the work done to push the gas out against the atmospheric pressure, given as: W = -PΔV Where ΔV is the change in volume during vaporization. Since we are dealing with 1 mole of water vapor, we can write the ΔV as: ΔV = RT / P

Calculate the Work

Now we can calculate the work done against the atmosphere. Using the ideal gas constant R in \(J/(mol.K)\): R = 8.314 \(\frac{J}{mol \cdot K}\) W = -PΔV = -P(RT/P) = -RT W = -(8.314 J/mol*K)(373 K) = -3100 J/mol

Calculate the Change in Internal Energy

Now that we have the work done against the atmosphere, we can find the change in internal energy by using the enthalpy change equation: ΔH = ΔU + W ΔU = ΔH - W Convert ΔH from \(kJ/mol\) to \(J/mol\): ΔH = \(40.7 kJ/mol * 1000 J/kJ = 40700 J/mol\) ΔU = \(40,700 J/mol - (-3,100 J/mol) = 43,800 J/mol\)

Determine the Fraction of Energy for ΔU and W

Now that we have both the ΔU and W values, we can determine the fraction of the energy used for each. Total energy is: Total Energy = ΔH = 40,700 J/mol Fraction of energy for ΔU: Fraction for ΔU = ΔU / Total Energy = 43,800 J/mol / 40,700 J/mol ≈ 1.08 Fraction of energy for W: Fraction for W = W / Total Energy = -3,100 J/mol / 40,700 J/mol ≈ -0.08 The fraction of energy used to change the internal energy of the water is approximately 1.08, and the fraction of energy used to do work against the atmosphere is approximately -0.08. The answer should be interpreted as follows: for each unit of energy used to vaporize water, 1.08 units is used to increase the internal energy and -0.08 units is used to do work against the atmosphere. The negative sign indicates that work is done against the atmospheric-pressure volume-extension.

Key concepts

Ideal Gas Law

The Ideal Gas Law is fundamental to understanding how gases behave under various conditions of pressure, volume, and temperature. It is represented by the equation PV = nRT, where P represents the pressure of the gas, V is the volume it occupies, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin.

When dealing with the vaporization of water, as in the textbook exercise, the Ideal Gas Law allows us to determine the change in volume (ΔV) that occurs when water turns into vapor. This is essential since the ΔV contributes to the work done against the atmosphere during the phase change. It's important to remember that real gases only obey the Ideal Gas Law under certain conditions, typically at high temperatures and low pressures, where they approximate ideal behavior. In this problem, by assuming water vapor behaves as an ideal gas, we can simplify the calculations and directly apply the law.

Internal Energy Change

The Internal Energy Change (ΔU) during a phase transition, like the vaporization of water, is a quantity that reflects how much energy is absorbed or released by a system, excluding any work done on or by the system. Essentially, it represents the change in energy at a microscopic level, including kinetic and potential energy of the molecules.

According to the first law of thermodynamics, the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings. Hence, in our textbook exercise, the relationship ΔH = ΔU + W is used, where ΔH is the enthalpy change, which in this case is the molar enthalpy of vaporization of water, and W represents work done. By rearranging this equation, we can solve for ΔU. It's crucial to understand that the internal energy is a state function, depending only on the state of the system and not on how the system arrived at that state.

Work Done Against Atmosphere

The concept of Work Done Against the Atmosphere is closely linked to thermodynamics and refers to the work performed by a system when it pushes its surroundings (in this case, the atmosphere) to make space for its own expansion or when it contracts and does work on the system. In the context of gases, it is represented by the formula W = -PΔV.

The negative sign indicates that work done by the system is considered negative, as the system loses energy. When water vaporizes, it expands and does work on the atmosphere, which is equivalent to the product of the external pressure and the change in volume. Understanding this concept is crucial when calculating the total energy requirements for a process like vaporization. In the provided exercise, the work done is part of the total molar enthalpy of vaporization and is necessary to understand the energy distribution during the transformation of water from liquid to gas.