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Chapter 6: Problem 49

For the process \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) at \(298 \mathrm{~K}\) and \(1.0 \mathrm{~atm}\), \(\Delta H\) is more positive than \(\Delta E\) by \(2.5 \mathrm{~kJ} / \mathrm{mol}\). What does the \(2.5 \mathrm{~kJ} / \mathrm{mol}\) quantity represent?

Short Answer

Expert verified
The 2.5 kJ/mol quantity represents the work done due to the pressure-volume change (P∆V) during the process of converting liquid water to water vapor at 298 K and 1.0 atm. This additional energy needed for the expansion is what makes ∆H more positive than ∆E by 2.5 kJ/mol.

Step by step solution

01

Write down the given information

We are given the following information: ∆H > ∆E by 2.5 kJ/mol The process: H₂O(l) → H₂O(g) Conditions: 298 K and 1.0 atm
02

Recall the enthalpy and internal energy equation

The relationship between the enthalpy change (∆H), the internal energy change (∆E), and the work done by the system (P∆V) is given by the equation: ∆H = ∆E + P∆V
03

Express the given information in terms of the equation

We know that ∆H > ∆E by 2.5 kJ/mol, which means that the right side of the equation is more positive by 2.5 kJ/mol. We can write this difference as: ∆H - ∆E = P∆V Since we aim to determine the meaning of the 2.5 kJ/mol quantity, we can express the equation as: 2.5 kJ/mol = P∆V
04

Interpret the result

The 2.5 kJ/mol quantity represents the work done due to the pressure-volume change (P∆V) during the process of converting liquid water to water vapor at 298 K and 1.0 atm. In other words, when liquid water changes to water vapor, it requires 2.5 kJ/mol of energy to do the work of expansion against the external pressure (in this case, 1.0 atm) so that the water molecules occupy a greater volume in the gaseous state. This additional energy needed for the expansion is what makes ∆H more positive than ∆E by 2.5 kJ/mol.

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Most popular questions from this chapter

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at \(1.00 \mathrm{~atm}\) and \(100 .{ }^{\circ} \mathrm{C}\) occupies a volume of \(30.6 \mathrm{~L}\). When 1 mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(1.00\) atm and \(100 .{ }^{\circ} \mathrm{C}, 40.66 \mathrm{~kJ}\) of heat is released. If the density of \(\mathrm{H}_{2} \mathrm{O}(l)\) at this temperature and pressure is \(0.996 \mathrm{~g} / \mathrm{cm}^{3}\), calculate \(\Delta E\) for the condensation of 1 mole of water at \(1.00 \mathrm{~atm}\) and \(100 .{ }^{\circ} \mathrm{C}\).

Consider the substances in Table 6.1. Which substance requires the largest amount of energy to raise the temperature of \(25.0 \mathrm{~g}\) of the substance from \(15.0^{\circ} \mathrm{C}\) to \(37.0^{\circ} \mathrm{C} ?\) Calculate the energy. Which substance in Table \(6.1\) has the largest temperature change when \(550 . \mathrm{g}\) of the substance absorbs \(10.7 \mathrm{~kJ}\) of energy? Calculate the temperature change.

It takes \(585 \mathrm{~J}\) of energy to raise the temperature of \(125.6 \mathrm{~g}\) mercury from \(20.0^{\circ} \mathrm{C}\) to \(53.5^{\circ} \mathrm{C}\). Calculate the specific heat capacity and the molar heat capacity of mercury.

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm}\). In the process, there is a heat gain by the system of \(350 . \mathrm{J} .\) c. A piston expands against \(1.00\) atm of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

A cubic piece of uranium metal (specific heat capacity \(=\) \(0.117 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) at \(200.0^{\circ} \mathrm{C}\) is dropped into \(1.00 \mathrm{~L}\) deuterium oxide ("heavy water," specific heat capacity \(=4.211 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) at \(25.5^{\circ} \mathrm{C}\). The final temperature of the uranium and deuterium oxide mixture is \(28.5^{\circ} \mathrm{C}\). Given the densities of uranium \((19.05\) \(\mathrm{g} / \mathrm{cm}^{3}\) ) and deuterium oxide \((1.11 \mathrm{~g} / \mathrm{mL})\), what is the edge length of the cube of uranium?
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