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Chapter 6: Problem 40

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 \(100 .{ }^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The change in internal energy \(\Delta E\) for the condensation of 1 mole of water at 1 atm and 100°C is -37.555 kJ.

Step by step solution

01

Convert the given information to the appropriate units.

We are given the volume of H₂O(g) as 30.6 L and the density of H₂O(l) as 0.996 g/cm³. We need to convert these to the appropriate units: 1. Convert the volume of H₂O(g) from L to cm³: \(30.6 \mathrm{~L} \times 1000 \mathrm{~cm^{3}/L} = 30600 \mathrm{~cm^{3}}\) 2. Convert the density of H₂O(l) to kg/m³: \(0.996 \mathrm{~g/cm^{3}} \times 1000 \mathrm{~kg/m^{3}/(g/cm^{3})} = 996 \mathrm{~kg/m^{3}}\)
02

Find the volume of H₂O(l).

For 1 mole of H₂O(l), the mass is \(1 \times 18.0153 \mathrm{~g/mol} = 18.0153 \mathrm{~g}\). Thus, we can find the volume of H₂O(l) at the given conditions using the density: $$V_{H_{2}O(l)} = \frac{m}{\rho} = \frac{18.0153 \mathrm{~g}}{0.996 \mathrm{~g/cm^{3}}} = 18.073 \mathrm{~cm^{3}}$$
03

Calculate the change in volume ΔV.

Now we will find the change in volume during the condensation process: $$\Delta V = V_{H_{2}O(l)} - V_{H_{2}O(g)} = 18.073 \mathrm{~cm^{3}} - 30600 \mathrm{~cm^{3}} = -30581.927 \mathrm{~cm^{3}}$$
04

Calculate the work done on the system w.

We can find the work done on the system by using the constant pressure: $$w = -P \Delta V$$ We are given the pressure P as 1 atm, and we should convert it to the SI unit of pressure (Pa): $$P = 1 \mathrm{~atm} \times 101325 \mathrm{~Pa/atm} = 101325 \mathrm{~Pa}$$ Now we can calculate the work done: $$w = -101325 \mathrm{~Pa} \times (-30581.927 \mathrm{~cm^{3}}) \times \frac{1 \mathrm{~J}}{100 \mathrm{~Pa \cdot cm^{3}}} = 3105.06 \mathrm{~J} = 3.105 \mathrm{~kJ}$$
05

Calculate the change in internal energy ΔE.

Finally, we can find ΔE using the First Law of Thermodynamics: $$\Delta E = q + w$$ We are given the heat released as q = -40.66 kJ. Therefore, $$\Delta E = (-40.66 \mathrm{~kJ}) + (3.105 \mathrm{~kJ}) = -37.555 \mathrm{~kJ}$$ Thus, the change in internal energy for the condensation of 1 mole of water at 1 atm and 100°C is -37.555 kJ.

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Most popular questions from this chapter

In a coffee-cup calorimeter, \(100.0 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{NaOH}\) and \(100.0 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\). After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\). Assuming that all the solutions have a density of \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), calcu- late the enthalpy change for the neutralization of \(\mathrm{HCl}\) by \(\mathrm{NaOH}\). Assume that no heat is lost to the surroundings or to the calorimeter.

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \text { (unbalanced) } $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{~kJ}\). If \(2.50 \times 10^{2} \mathrm{~mL} \mathrm{~N}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50 \mathrm{~atm}\) and \(4.50 \times\) \(10^{2} \mathrm{~mL} \mathrm{O}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50\) atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains \(1.00 \mathrm{~kg}\) water and has a total heat capacity of \(10.84 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), what is the heat capacity of the calorimeter components?

For the following reactions at constant pressure, predict if \(\Delta H>\Delta E, \Delta H<\Delta E\), or \(\Delta H=\Delta E .\) a. \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) b. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) c. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{~kJ}, w=+88 \mathrm{~kJ}\) b. \(q=+82 \mathrm{~kJ}, w=-47 \mathrm{~kJ}\) c. \(q=+47 \mathrm{~kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?
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