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Lecture 9 CH131 Summer 1 2020 6/9/2020 1:47 PM
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Lecture 9 CH131 Summer 1Tuesday, June 9, 2020
• Phase diagramsChapter 11: Solutions• Vapor pressure lowering, 𝑃 𝑥 𝑃°• Boiling point elevation, 𝑇 𝑇° 𝐾 𝑚• Freezing point depression, 𝑇° 𝑇 𝐾 𝑚• Osmotic pressure, Π 𝑅𝑇𝑀 ;
Ch12: Thermodynamic processes and thermochemistry• Thermodynamic concepts• Heat and workNext: Continue ch12
Phase diagramsLines of 𝑃 versus 𝑇 at which different phases
are present at the same time.
That is, values of 𝑃 and 𝑇 at which different phasesare in equilibrium.
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Phase diagramsLecture 9 CH131 Summer 1 2020
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What about the intersection marked “A”Lecture 9 CH131 Summer 1 2020
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Triple pointLiquid tert-butyl alcohol, C CH3 3OH, can boil and freeze at the same time... http://goo.gl/4K1SR
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Supercritical fluidSupercritical transition of liquid Cl2 … http://goo.gl/xo2jU
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Changes of phaseCO2 gas at 1 atm, 0℃
is heated to 1 atm, 60℃; then compressed to 75 atm, 60℃;
then cooled to 75 atm, 0℃; then expanded to 65 atm, 0℃.
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TP CO2 gas at 1 atm, 0℃ is heated to 1 atm, 60℃;
then compressed to 75 atm, ℃; then cooled to 75 atm, 0℃;
then expanded to 65 atm, 0℃. At this point, the CO2 will be a …
1. gas2. supercritical fluid3. liquid4. solid
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TP CO2 gas at 1 atm, 0℃ is heated to 1 atm, 60℃;
then compressed to 75 atm, ℃; then cooled to 75 atm, 0℃;
then expanded to 65 atm, 0℃. At this point, the CO2 will haveundergone a phase transition …
1. once2. twice3. three times4. None of the above
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Colligative propertiesNon-volatile solute negligible vapor pressure …
• lowers vapor pressure of solvent: 𝑃 𝑥 𝑃° , ∆𝑃 𝑃° 𝑃 𝑥 𝑃°𝑥 is the mole fraction of the solvent: 𝑥
Pure solvent means 𝑥 1 and so 𝑃 𝑃°𝑥 is the mole fraction of the solute: 𝑥 1 𝑥
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Effect of solute on vapor pressureThe vapor pressure of water at 32℃ is 4.76 kPa. Calculate the vapor pressure after 68.0 g of ethylene glycol 𝑀 62.07 is dissolved in 500. g of water at 32℃.
Answer: 4. 58 kPa
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The more particles, the greater the effectSugar in water, 1 mol of particles per mol of sugar, so 𝑖 1
NaCl in water, 2 mol of particles per mol of NaCl, so 𝑖 2
Na2SO4 in water, 3 mol of particles per mol of Na2SO4, so 𝑖 3
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Effect of solute on vapor pressureThe vapor pressure of water at 32℃ is 4.76 kPa. Calculate the vapor pressure after 68.0 g of Na PO 𝑀 163.94 is dissolved in 500. g of water at 32℃.
Answer: 4.49 kPa not 4.69 kPa
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Colligative propertiesNon-volatile solute negligible vapor pressure …
• lowers vapor pressure of solvent: 𝑃 𝑥 𝑃° , ∆𝑃 𝑃° 𝑃 𝑥 𝑃°
• raises boiling point of solvent: 𝑇 𝑇° 𝑖 𝑚solute 𝐾b 𝑚 𝐾
• lowers freezing point of solvent: 𝑇° 𝑇 𝑖 𝑚solute 𝐾 𝑚 𝐾
𝑚solute is known a molality, the moles of the solute per kg of solvent.
Be careful to distinguish this from
molarity M, the moles of solute per liter of solution.
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TP A non-volatile solute lowers the vapor pressure of the solvent. This in turn means the boiling point of the solvent must increase. Why? Because …
1. higher temperature is necessary to evaporate the solute2. the solute particles stick to the solvent particles,
analogous to van der Waals 𝑎3. at the normal boiling point the vapor pressure of the
solvent will be too low4. the solute vapor pressure is so low5. Some other reason
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Effect of solute on vapor pressure (Fig 16.4)Lecture 9 CH131 Summer 1 2020
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Effect of solute on vapor pressure (Fig 16.4)Lecture 9 CH131 Summer 1 2020
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Freezing point exampleCalculate the freezing point of 3.00 kg of water to which has been added 525 g of ethylene glycol, OHC2H4OH 𝑀 62.07 , 𝐾 1.86 K kg/molAnswer: –5.24 oC
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Freezing point example againCalculate the freezing point of 3.00 kg of water to which has been added 525 g of sodium chloride, NaCl 𝑀 58.44 , 𝐾 1.86 K kg/molAnswer: –11.1 oC
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TP The vapor pressure of water at 32℃ is 4.76 kPa. A glass of water is sealed in a 1.00 L container filled with air at 32℃. After the water comes to equilibrium with the air in the container,
1 the total pressure is 100.00 kPa, 2 there is 500. g of liquid water in the glass, 3 and the partial pressure of water vapor in the container is …
1. less that 4.76 kPa2. 4.76 kPa3. more than 4.76 kPa4. Further information required
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TP Then, 35.0 g of ethylene glycol if dissolved in the liquid water. After the water returns to equilibrium with its vapor, the mass of the liquid water …
1. will have decreased2. will be unchanged3. will have increased4. Further information required
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Effect of solute on vapor pressureThe vapor pressure of water at 32℃ is 4.76 kPa. A glass of water is sealed in a 1.00 L container filled with air at 32℃. After the water comes to equilibrium with the air in the container,
1 the total pressure is 100.00 kPa, 2 there is 500. g of liquid water in the glass, 3 and the partial pressure of water vapor in the container is 4.76 kPa.
Then, 35.0 g of ethylene glycol is dissolved in the liquid water. Ignore any resulting change in volume of the liquid water.
Calculate change in the mass of the liquid water after it has returned to equilibrium with its vapor.
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Effect of solute on vapor pressure𝑇 32.0℃,𝑃 100 kPa,𝑉 1.00 L,𝑃° 4.76 kPa500. g H O 𝑙 , 35.0 g ethylene glycol 𝑀 62.07
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Effect of solute on vapor pressure𝑇 32.0℃,𝑃 100 kPa,𝑉 1.00 L,𝑃° 4.76 kPa500. g H O 𝑙 , 35.0 g ethylene glycol 𝑀 62.07𝑥 0.980,∆𝑃 0.0948 kPa
∆𝑛 3.43 10 mol,∆𝑚 0.000616 g
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Differing vapor pressures → osmotic pressureWe have seen that solutes cause condensation of water vapor.
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Differing vapor pressures → osmotic pressureThe same differing vapor pressure leads to osmotic pressure.
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Differing vapor pressures → osmotic pressureThe same differing vapor pressure leads to osmotic pressure.
The flow stops when the pressure Π 𝐷𝑔ℎ due to the column height difference ℎ is equal to the vapor pressure difference Π 𝑃° 𝑃 .
This difference is called osmotic pressure and is given by
𝑃° 𝑃 Π 𝑖𝑐 𝑅𝑇 𝑀 𝑅𝑇
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Recipe for osmotic pressure calculationsOsmotic pressure Π 𝑖𝑐 𝑅𝑇 𝑀 𝑅𝑇.
1. Use measured osmotic pressure, Π, and temperature, 𝑇, to evaluate concentration, 𝑐 Π/ 𝑖 𝑅 𝑇 , in mol/L.
2. Use cell volume to express concentration in terms of moles, 𝑛 𝑐 𝑉.
3. Use solute mass to calculate molar mass, 𝑀 mass/𝑛.
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Osmotic pressure Π = i csolute R T1.40 g of polyethylene 𝑖 1 dissolved in 100. mL of benzene generates an osmotic pressure of 0.248 kPa at 25 ℃. Calculate the molar mass of the polyethylene. 𝑅 8.3145 10
1. Calculate the concentration…
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Osmotic pressure Π = i csolute R T1.40 g of polyethylene 𝑖 1 dissolved in 100. mL of benzene generates an osmotic pressure of 0.248 kPa at 25 ℃. Calculate the molar mass of the polyethylene. 𝑅 8.3145 10
1. Calculate the concentration…1.00 10‒4 mol/L
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Osmotic pressure Π = i csolute R T1.40 g of polyethylene 𝑖 1 dissolved in 100. mL of benzene generates an osmotic pressure of 0.248 kPa at 25 ℃. Calculate the molar mass of the polyethylene. 𝑅 8.3145 10
1. Calculate the concentration…1.00 10‒4 mol/L
2. Calculate the moles…
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Osmotic pressure Π = i csolute R T1.40 g of polyethylene 𝑖 1 dissolved in 100. mL of benzene generates an osmotic pressure of 0.248 kPa at 25 ℃. Calculate the molar mass of the polyethylene. 𝑅 8.3145 10
1. Calculate the concentration…1.00 10‒4 mol/L
2. Calculate the moles…10‒5 mol
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Osmotic pressure Π = i csolute R T1.40 g of polyethylene 𝑖 1 dissolved in 100. mL of benzene generates an osmotic pressure of 0.248 kPa at 25 ℃. Calculate the molar mass of the polyethylene. 𝑅 8.3145 10
1. Calculate the concentration…1.00 10‒4 mol/L
2. Calculate the moles…10‒5 mol
3. Calculate the molar mass…
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Osmotic pressure Π = i csolute R T1.40 g of polyethylene 𝑖 1 dissolved in 100. mL of benzene generates an osmotic pressure of 0.248 kPa at 25 ℃. Calculate the molar mass of the polyethylene. 𝑅 8.3145 10
1. Calculate the concentration…1.00 10‒4 mol/L
2. Calculate the moles…10‒5 mol
3. Calculate the molar mass…1.40 10 5 g/mol
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TP 1.0 mg of a substance dissolved in 1.0 mL of water generates an osmotic pressure of 1.0 kPa, at 25 ℃. The molar mass of the solute 𝑖 1 is …𝑅 8.3145 10
1. 250 g/mol2. 500 g/mol3. 1000 g/mol4. 2500 g/mol5. 5000 g/mol6. Some other value
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Ch12: Thermodynamic processes and thermochemistry
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First law of thermodynamicsEnergy, 𝑈, is exchanged between system and surroundings as heat, 𝑞, and work, 𝑤,
Δ𝑈 𝑞 𝑤
Positive values increase energy of systemΔ𝑈 𝑞 𝑤
𝑞 0: heat flow into the system𝑤 0: work done on the system
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How do we know heat is present?Since 𝑞 𝑚𝑐∆𝑇 …
we can use temperature change of surroundings to monitor heat flow.
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How do we know heat is present?Temperature decrease in surroundings, ∆𝑇 0 means …
energy flow into system from surroundings …energy of system goes up …
and so, 𝑞 0.
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How do we know heat is present?Temperature increase in surroundings, ∆𝑇 0, means …
energy flow out of system into surroundings …energy of system goes down …
and so, 𝑞 0.
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TP When NaHCO3 𝑠 is dissolved in 200 mL of HCl 𝑎𝑞 , the temperature of the solution goes down. This means the chemical reaction between the NaHCO3 𝑠 and the HCl 𝑎𝑞 results in the chemical system …
1. giving off heat and so 𝑞 02. giving off heat and so 𝑞 03. absorbing heat and so 𝑞 04. absorbing heat and so 𝑞 0
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How do we know if work is present?Macroscopic movement, for example of a piston.
Work done by gas: force distance 𝐹/𝐴 Δ𝑥 𝐴 𝑃ext Δ𝑉Work done on gas: 𝑤 𝑃ext Δ𝑉
Expansion of gas pushes against 𝑃ext: gas expends energy, 𝑤 0
Compression of gas pushed on by 𝑃ext: gas gains energy, 𝑤 0
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TP When NaHCO3 𝑠 is dissolved in 200 mL of HCl 𝑎𝑞 , CO2 𝑔 bubbles form. This means the chemical reaction between the NaHCO3 𝑠 and the HCl 𝑎𝑞 results in the chemical system …
1. doing work and so 𝑤 02. doing work and so 𝑤 03. having work done on it and so 𝑤 04. having work done on it and so 𝑤 0
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