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# chapter11 - 496 Chapter 11 Liquids Solids and Materials...

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Chapter 11: Liquids, Solids, and Materials 496 End-of-Chapter Solutions for Chapter 11 Summary Problem Part 1 Answer: (a) 190 mm Hg (b) weaker (c) 30 °C (d) 32 kJ Strategy and Explanation: Given Figure 11.5, determine the vapor pressure of diethyl ether at a given temperature, determine the relative strength of intermolecular attractions between diethyl ether and ethanol, determine the temperature of diethyl ether with a given vapor pressure. Given the normal boiling point and vapor pressure of diethyl ether at a given temperature, calculate the enthalpy of vaporization. (a) Use the graph shown in Figure 11.5, draw a vertical line from 0 °C on the x-axis up to the diethyl ether curve, then a horizontal line from the curve to the y-axis. The vapor pressure of diethyl ether at 0 °C is approximately 190 mm Hg. (b) The liquid with a higher vapor pressure has weaker intermolecular attractions, so, looking at Figure 11.5, diethyl ether has weaker intermolecular attractions than ethanol. This is consistent with the molecular structures: ethanol has an –OH group and liquid molecules of ethanol can experience strong hydrogen bonding interactions; whereas, diethyl ether has the O atom bonded to C atoms, and therefore, the strongest intermolecular force is dipole-dipole forces. (c) Use the graph shown in Figure 11.5, draw a horizontal line from 600 mm Hg on the y-axis to the diethyl ether curve, then a vertical line down to the temperature axis. The temperature is approximately 30 °C. (d) There is no need to use the graph shown in Figure 11.5 to answer this question. Use the two-set Clausius-Claypeyron equation, Section 11.2 (page 494): ln P 2 P 1 " # \$ % & = () H vap R 1 T 2 ( 1 T 1 " # \$ % & with R is given in Table 10.4. The vapor pressure above the liquid at the normal boiling point is 760.0 mm Hg. Assume the second temperature is 20. °C, since typical precision on temperature readings are ± 1 °C. T 1 = 34.6°C + 273.15 = 307.8 K T 2 = 20.°C + 273.15 = 293 K ln 410 mm Hg 760.0 mm Hg " # \$ % & = () H vap 8.314 J K * mol " # \$ % & 1 kJ 1000 J " # \$ % & 1 293 K ( 1 307.8 K " # \$ % & ln 0.54 ( ) = "# H vap 0.008314 kJ mol \$ % & ( ) 0.00341 " 0.003249 ( ) = "# H vap 0.008314 kJ mol \$ % & ( ) 0.00016 ( ) " 0.62 = "# H vap 0.019 mol kJ \$ % & ( ) " 32 kJ mol = # H vap Part 2 Answer: (a) gas (b) liquid (c) –120.5 °C (d) 0.225 atm (e) Strategy and Explanation: Given a phase diagram, determine the phase of a substance under various conditions, the temperature of a liquid at a given a vapor pressure, the vapor pressure of a substance in a given phase and at a given temperature; identify which of two phases is more dense. Phase diagrams are described in Section 11.3. (a) Look up the point on the given phase diagram where the pressure is 0.75 atm and the temperature is room temperature (approximately 20-25 °C). This point occurs in the part of the phase diagram represented by the gas phase.

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Chapter 11: Liquids, Solids, and Materials 497 (b) Look up the point on the given phase diagram where the pressure is 0.75 atm and the temperature is – 114 °C.
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chapter11 - 496 Chapter 11 Liquids Solids and Materials...

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