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) \(\Delta E \approx 37.68\,\text{kJ}\).

Step by step solution

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) \(\Delta E \approx 37.68\,\text{kJ}\)

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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

Use the appropriate table to calculate \(\Delta H^{\circ}\) for (a) the reaction between copper(II) oxide and carbon monoxide to give copper metal and carbon dioxide. (b) the decomposition of one mole of methyl alcohol (CH \(_{3} \mathrm{OH}\) ) to methane and oxygen gases.

Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is widely used as a fuel for disposable lighters. When one mole of butane is burned in oxygen, carbon dioxide and steam are formed and \(2658.3 \mathrm{~kJ}\) of heat is evolved. (a) Write a thermochemical equation for the reaction. (b) Using Table \(8.3\), calculate the standard heat of formation of butane.

Determine whether the statements given below are true or false. Consider specific heat. (a) Specific heat represents the amount of heat required to raise the temperature of one gram of a substance by \(1^{\circ} \mathrm{C}\). (b) Specific heat is the amount of heat flowing into the system. (c) When 20 J of heat is added to equal masses of different materials at \(25^{\circ} \mathrm{C}\), the final temperature for all these materials will be the same. (d) Heat is measured in \({ }^{\circ} \mathrm{C}\).

Draw a cylinder with a movable piston containing six molecules of a liquid. A pressure of 1 atm is exerted on the piston. Next draw the same cylinder after the liquid has been vaporized. A pressure of one atmosphere is still exerted on the piston. Is work done on the system or by the system?

When ammonia reacts with dinitrogen oxide gas ( \(\Delta H_{\mathrm{f}}^{\circ}=82.05 \mathrm{~kJ} / \mathrm{mol}\) ), liquid water and nitrogen gas are formed. How much heat is liberated or absorbed by the reaction that produces \(345 \mathrm{~mL}\) of nitrogen gas at \(25^{\circ} \mathrm{C}\) and \(717 \mathrm{~mm} \mathrm{Hg}\) ?

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