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For the vaporization of one mole of water at \(100^{\circ} \mathrm{C}\) determine (a) \(\Delta H\) (Table 8.3) (b) \(\Delta P V\) (in kilojoules) (c) \(\Delta E\)

Short Answer

Expert verified
Answer: (a) \(\Delta H = 40.7\,\text{kJ}\), (b) \(\Delta PV \approx 3.02\,\text{kJ}\), and (c) \(\Delta E \approx 37.68\,\text{kJ}\).

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01

Determine \(\Delta H\)

According to Table 8.3, the enthalpy of vaporization for one mole of water at \(100^\circ C\) is given as \(\Delta H_\text{vap} = 40.7 \,\text{kJ}\,\text{mol}^{-1}\). Therefore, for the vaporization of one mole of water at \(100^{\circ} \mathrm{C}\), the change in enthalpy is \(\Delta H = 40.7 \, \text{kJ}\).
02

Calculate \(\Delta PV\)

To calculate the change in pressure volume, we use the ideal gas law, \(PV=nRT\). At \(100^\circ C\), water vapor is in equilibrium with liquid water, so the pressure is equal to the vapor pressure of water at this temperature. The vapor pressure of water at \(100^\circ C\) is \(1\,\text{atm}\). Remember to convert the temperature to Kelvin: \(T = 100 + 273.15 = 373.15\,\text{K}\). For one mole, \(n=1\). Using the universal gas constant, \(R = 0.0821\,\frac{\text{L}\,\text{atm}}{\text{K}\,\text{mol}}\). The change in pressure volume is given by: \(\Delta PV = (1\,\text{atm})(1\,\text{mol})(0.0821\,\frac{\text{L}\,\text{atm}}{\text{K}\,\text{mol}})(373.15\,\text{K})\). To convert the units of the \(\Delta PV\) to kilojoules, use the conversion factor: \(\frac{1\,\text{L}\,\text{atm}}{0.1013\,\text{kJ}}\). The resulting \(\Delta PV\) in kilojoules is: \(\Delta PV = (1\,\text{atm})(1\,\text{mol})(0.0821\,\frac{\text{L}\,\text{atm}}{\text{K}\,\text{mol}})(373.15\,\text{K})\times \frac{1\,\text{L}\,\text{atm}}{0.1013\,\text{kJ}} \approx 3.02 \,\text{kJ}\).
03

Calculate \(\Delta E\)

To calculate the change in internal energy, use the relationship \(\Delta E = \Delta H - \Delta PV\). With the given values, this becomes: \(\Delta E = 40.7\,\text{kJ} - 3.02\,\text{kJ} \approx 37.68\,\text{kJ}\). In conclusion, for the vaporization of one mole of water at \(100^{\circ} \mathrm{C}\), we found that: (a) \(\Delta H = 40.7\,\text{kJ}\) (b) \(\Delta PV \approx 3.02\,\text{kJ}\) (c) \(\Delta E \approx 37.68\,\text{kJ}\)

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics
Thermodynamics is a core subject in physics and chemistry that deals with heat, work, temperature, and the laws governing their interactions. It describes how thermal energy is converted to and from other forms of energy and how it affects matter. The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed.

When we talk about the vaporization of water, we're looking at an endothermic process, where heat is absorbed to convert liquid water into a gas. This is where the concept of enthalpy (∆H) shines, as it quantifies the total heat content of a system at constant pressure. The enthalpy of vaporization is defined as the amount of heat required to turn one mole of a substance from its liquid phase into vapor without changing its temperature.

In the example given, the enthalpy of vaporization provides insight into the energy needed for the phase transition of water at its boiling point. This ties into the broader thermodynamic view, where understanding energy changes is critical in predicting the behavior of a system under various conditions.
Ideal Gas Law
The ideal gas law is a fundamental equation in thermodynamics that describes the relationship between pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas, under the assumption that the gas behaves ideally. This relationship is mathematically expressed as PV = nRT, where R is the ideal gas constant.

In the context of the problem, after water vaporizes, it behaves as a gas and thus, the ideal gas law can be applied to determine the change in the product of pressure and volume (∆PV) during the phase change. It's important to note that the ideal gas law assumes no interactions between gas molecules and that they occupy no space, which is a good approximation for gases at high temperatures and low pressures.

Additionally, ∆PV provides a bridge between the thermodynamic property changes of a system since the work done during a phase change at constant pressure can be estimated using this value. It is key for understanding other thermodynamic processes at play during the vaporization of water.
Internal Energy
Internal energy is the total energy contained within a system due to the kinetic and potential energies of its particles. In thermodynamics, changes in internal energy (∆E) are of particular interest because they represent how the energy of a system changes in response to heat transfer and work done on or by the system.

The change in internal energy can be calculated from the enthalpy of vaporization and the work associated with the change in pressure and volume (∆PV), using the relationship ∆E = ∆H - ∆PV. This equation is derived from the first law of thermodynamics and is crucial in understanding how energy is conserved during a phase change.

Applying this concept to our current scenario, where water is vaporizing, enables us to quantify the change in the kinetic and potential energies of the molecules as they transition from a liquid to a gas phase. The subtraction of the work term (∆PV) from the enthalpy (∆H) gives us the net change in internal energy (∆E), offering us a sober look into the intrinsic energy dynamics of the phase change.

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Most popular questions from this chapter

Given $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H^{\circ}=3351.4 \mathrm{~kJ} $$ (a) What is the heat of formation of aluminum oxide? (b) What is \(\Delta H^{\circ}\) for the formation of \(12.50 \mathrm{~g}\) of aluminum oxide?

Calcium carbide, \(\mathrm{CaC}_{2}\), is the raw material for the production of acetylene (used in welding torches). Calcium carbide is produced by reacting calcium oxide with carbon, producing carbon monoxide as a byproduct. When one mole of calcium carbide is formed, \(464.8 \mathrm{~kJ}\) is absorbed. (a) Write a thermochemical equation for this reaction. (b) Is the reaction exothermic or endothermic? (c) Draw an energy diagram showing the path of this reaction. (Figure \(8.4\) is an example of such an energy diagram.) (d) What is \(\Delta H\) when \(1.00 \mathrm{~g}\) of \(\mathrm{CaC}_{2}(\mathrm{~g})\) is formed? (e) How many grams of carbon are used up when \(20.00 \mathrm{~kJ}\) of heat is absorbed?

A sample of sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), is contaminated by sodium chloride. When the contaminated sample is burned in a bomb calorimeter, sodium chloride does not burn. What is the percentage of sucrose in the sample if a temperature increase of \(1.67^{\circ} \mathrm{C}\) is observed when \(3.000 \mathrm{~g}\) of the sample is burned in the calorimeter? Sucrose gives off \(5.64 \times 10^{3} \mathrm{~kJ} / \mathrm{mol}\) when burned. The heat capacity of the calorimeter is \(22.51 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\).

To produce silicon, used in semiconductors, from sand \(\left(\mathrm{SiO}_{2}\right)\), a reaction is used that can be broken down into three steps: $$ \begin{aligned} \mathrm{SiO}_{2}(s)+2 \mathrm{C}(s) \longrightarrow \mathrm{Si}(s)+2 \mathrm{CO}(g) & & \Delta H=689.9 \mathrm{~kJ} \\ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(g) & & \Delta H=-657.0 \mathrm{~kJ} \\ \mathrm{SiCl}_{4}(g)+2 \mathrm{Mg}(s) \longrightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s) & & \Delta H=-625.6 \mathrm{~kJ} \end{aligned} $$ (a) Write the thermochemical equation for the overall reaction for the formation of silicon from silicon dioxide; \(\mathrm{CO}\) and \(\mathrm{MgCl}_{2}\) are byproducts. (b) What is \(\Delta H\) for the formation of one mole of silicon? (c) Is the overall reaction exothermic?

When \(35.0 \mathrm{~mL}\) of \(1.43 \mathrm{M} \mathrm{NaOH}\) at \(22.0^{\circ} \mathrm{C}\) is neutralized by \(35.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) also at \(22.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, the temperature of the final solution rises to \(31.29^{\circ} \mathrm{C}\). Assume that the specific heat of all solutions is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), that the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and that volumes are additive. (a) Calculate \(q\) for the reaction. (b) Calculate \(q\) for the neutralization of one mole of \(\mathrm{NaOH}\).

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