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Key concepts from last class • The internal energy, U, is the total energy of the system: It is the sum of the kinetic and potential energies contributed by all the atoms, ions and molecules present in the system. It depends on temperature and in general on pressure. • When interactions between molecules can be neglected (ideal gas), the only contributions to the internal energy are the kinetic energy of the molecules. • Temperature is a measure of the average kinetic energy. • For a monoatomic ideal gas, the only contribution is the translational kinetic energy, which gives • For a polyatomic ideal gas, is also a function of temperature, but the value depends on the number of atoms and bonds. • We can change the internal energy by adding or subtracting energy (heat or work): U = w + q. Key equations / conclusions We know how to calculate work done by/on the system. It depends on the external pressure and the volume change : We know how to calculate heat for a process that involves changing the temperature at either constant pressure or constant volume : , . , .∆ (Cm,V is constant) , . , .∆ (Cm,p is constant) We know ∆ and because w = 0 for a process at constant volume, ∆ We also know that, for an ideal gas, U depends only on temperature. As a consequence, if Tf = Ti, then U = 0 Internal Energy 145 Because for a monoatomic ideal gas all the energy is translational kinetic energy: It makes sense that Um does not depend on the volume in the case of an ideal gas. Changing the volume changes the average distance between molecules. Because ideal gases are assumed not to experience any repulsive or attractive interactions, their internal energy is independent of the average distance between molecules. T = constant Same internal energy if interactions are negligible (ideal behavior) Internal Energy 146 In the case of a real gas, attractive forces dominate when molecules are closer together. (we discussed this when talking about Van der Waals gases) T = constant Now the three states have different internal energy. The internal energy is lowered by attractive forces. Important : The internal energy depends only on temperature for an ideal gas, but not for a real gas. In the case of a real gas, the density of the gas matters because it affects the potential energy of the system. Internal Energy 147 Because for a monoatomic ideal gas all the energy is translational kinetic energy: For polyatomic ideal gases (those that obey PV = nRT), we need to take into account vibrational and rotational modes, but the internal energy still depends on temperature only. For a real gas, the internal energy is lowered when attractive forces dominate (molecules closer together). Therefore, Um depends not only on T, but also on the density of the gas. State and Path Variables 148 Previously, we said… Heat and work cannot be stored: they are transient quantities that only apply to a system that undergoes a change in its state: There is no change in heat, because there is no initial and final heats. Instead, we’ll talk about the heat (and/or work) involved in changing the system from an initial to a final state. Note: We could talk about a change in temperature (volume, pressure, etc). Initial Vinitial Tinitial Pinitial qinitial Final Vfinal Tfinal Pfinal q 20°C >20°C 150°C <150°C qfinal Hum..Some variables are used to describe the system in a particular state while others are used to describe the change in state. (more to come!) State and Path Variables 149 Consider a system consisting of a pure substance (e.g. a pure liquid). Specifying the pressure, volume and temperature is sufficient to specify many other properties of the liquid (e.g. its density, refractive index, surface tension, etc). These properties are called state functions: they depend only on the state of the system, but not on how the system arrived to that state. If we measure the temperature of a cup of hot water, we don’t care if the water was heated on a hot plate or in strong sunlight. Initial Vinitial Tinitial Pinitial qinitial Final Vfinal Tfinal Pfinal q 20°C >20°C 150°C <150°C qfinal State and Path Variables 150 Internal energy (U) is a state function U is a state function because it is the sum of the kinetic and potential energies of all the molecules and atoms that make up the system. You don’t need to know how the energy was transferred to the system to be able to define it. Uinitial Vinitial Tinitial Pinitial qinitial Initial Final Ufinal Vfinal Tfinal Pfinal q 20°C >20°C 150°C <150°C qfinal State and Path Variables 151 And we also said… So w 0 after all… What if we do the one in the figure? Will it be the same? I guess it makes sense… you need to do work to do all this stuff, even if you end up in the same place you started I’ll try when I get home, but it looks like it not only depends on the initial and final states, but also on the particular path you follow Initial = final In this tutorial, we saw that the work performed by the system depends on the particular path we follow to go from the initial and final state. Therefore, specifying the initial and final states is not enough. We also need to know the exact path. State and Path Variables 152 Heat and Work are path variables. We can’t talk about the “initial heat” and “final heat”, because heat is not a property that describes a system. Instead, it depends on the path we followed to take the system from an initial state (characterized by a set of state functions), to a final state (characterized by other values of the state functions). Uinitial Vinitial Tinitial Pinitial Initial winitial 150°C w Ufinal Vfinal Tfinal Pfinal q 20°C qinitial Final >20°C <150°C qfinal wfinal State and Path Variables The Path shown in the figure is cyclic because the initial and final states are the same. This tells us that p, T, V, U, etc are all equal to zero. However, q and w are in principle NOT zero. Note that we use to specify a change in a state function. There is no q because there is no “initial q” and “final q” 153 Initial = final Internal Energy 154 ∆ initial final You can follow different paths to go from 1 to 6. The change in internal energy is always the same: U = U6 - U1. However, the amount of work and heat will depend on the path Internal Energy 155 ∆ What about this path? initial final You can use this to your advantage: You can use the path that is easiest to calculate, or even create another one. However, if you need to calculate q and/or w you are in trouble… gas 1 bar 1.1 bar 1 bar from atmosphere + 0.1 bar from weight. Frictionless piston, can move up or down 1 bar Frictionless piston, can move up or down 156 w 1.1 bar Stoppers don’t allow the piston to move (up or down) q 1.1 bar T , V 1.1 bar q P , T P.V = n.R.T Internal Energy 157 An important result ∆ We already saw that w = 0 if there’s no change in volume. Therefore, if you perform a process that does not change the volume of the system, the change in internal energy is simply the heat ∆ It means “keeping volume constant ” We already defined the molar heat capacity at constant volume, Cm,V : , , . , .∆ (Cm,V is constant) Internal Energy 158 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) ∆ 1 1 2 You may have information you don’t need 3p0 3p0 3p0 V0, 3p0 V0, 3p0 V0, p0 3 How can I reduce the pressure while keeping the volume constant? 2 Internal Energy 159 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) ∆ 1 1 2 3p0 3p0 V0, 3p0 V0, 3p0 3 3p0 q 2 V0, p0 How can I reduce the pressure while keeping the volume constant? We need to cool down the gas. Internal Energy 160 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) ∆ 1 3p0 3p0 3p0 V0, 3p0 V0, 3p0 V0, p0 2 0 if volume does not change 1 2 3 How can I reduce the pressure while keeping the volume constant? We need to cool down the gas. We need to cool down the gas. T2 < T1 we are lowering the internal energy of the gas, < 0, w = 0 Internal Energy 161 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) ∆ 1 3p0 3p0 3p0 V0, 3p0 V0, 3p0 V0, p0 2 0 if volume does not change 3 1 2 , .∆ How can I reduce the pressure while keeping the volume constant? We need to cool down the gas. We need to cool down the gas. T2 < T1 we are lowering the internal energy of the gas, < 0, w = 0 Internal Energy 162 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 3 1 3 2 3 You may have information you don’t need Internal Energy 163 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 3 1 3 3 2 1 2 3p0 3p0 3p0 p0 V0, 3p0 V0, 3p0 V0, p0 V0, p0 3 p0 3V0, p0 Internal Energy 164 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 3 1 3 3 2 1 3p0 2 3p0 3p0 3 p0 p0 w w=0 V0, 3p0 V0, 3p0 q V0, p0 V0, p0 q 3V0, p0 Internal Energy 165 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) No (or minimal) math answers: 3 1 3 w1-2 = 0 (no change in volume), w2-3 = -2p0V0 (constant pext) • q1-2 = n.Cv.(T2-T1) < 0 (constant volume, we have the Cv value) • T3 = T1 (P1V1 = P3V3) • 3 2 • 1 U1-3 = 0 (ideal gas, T does not change) q2-3 = n.CP.(T3-T2) > 0 (but we do not have the Cp value) 2 3p0 3p0 3p0 V0, 3p0 V0, 3p0 V0, p0 q 3 p0 p0 w V0, p0 q 3V0, p0 Internal Energy 166 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) No (or minimal) math answers: 3 1 3 2 3 • w1-2 = 0 (no change in volume), w2-3 = -2p0V0 (constant pext) • q1-2 = n.Cv.(T2-T1) < 0 (constant volume, we have the Cv value) • T3 = T1 (P1V1 = P3V3) • U1-3 = 0 (ideal gas, T does not change) q2-3 = n.CP.(T3-T2) > 0 (but we do not have the Cp value) U1-3 = 0 = q1-3 + w1-3 q1-3 = -w1-3 w1-3 = w1-2 + w2-3 = -2p0V0 (we didn’t need the Cv) q1-3 = -w1-3 = 2p0V0 Internal Energy 167 A cool result on the side… q1-2 = n.Cv.(T2-T1) q2-3 = n.CP.(T3-T2) = n.CP.(T1-T2) q1-2, w1-2 1 2 q1-3 = 2p0V0 q2-3, w2-3 3 q1-2 q2-3 (T3 = T1) q1-3 n.Cv.(T2-T1) + n.CP.(T1-T2) = 2p0V0 -n.Cv.(T1-T2) + n.CP.(T1-T2) = 2p0V0 n(CP – Cv).(T1-T2) = 2p0V0 3p0V0 = nRT1 p0V0 = nRT2 n(CP – Cv). 2p0V0 /(nR) = 2p0V0 (T1-T2) = (2p0V0 )/(nR) n(CP – Cv). 2p0V0 /(nR) = 2p0V0 (CP – Cv) / R = 1 (CP – Cv) = R (ideal gas) Internal Energy 168 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 3 1 9 2 Internal Energy 169 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 1 3p0 3p0 3 1 9 2 V0, 3p0 3V0, 3p0 How can I increase the volume while keeping the external pressure constant? 2 Internal Energy 170 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 1 3 1 3p0 3p0 w 9 2 V0, 3p0 3V0, 3p0 q How can I increase the volume while keeping the external pressure constant? U = q + w ∆ 0 temperatureincreased , 0, 0 2 Internal Energy 171 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 1 3 1 9 2 3p0 2 3p0 w V0, 3p0 3V0, 3p0 q U = q + w Easy: -6p0V0 , .∆ But we have the Cv! What do we do? Internal Energy 172 ∆ 1 U1-3 + U3-2 0 (no change in temperature, ideal gas) 2 ∆ , .∆ U3-2 U1-3 3 Careful: this is true for an ideal gas, because U depends only on temperature. 0, and we are in deep trouble… Otherwise, ∆ U1-3 + U3-2 = , . ∆ (even if the process is not done at constant volume) Internal Energy 173 Example: Calculate U, q and w for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 1 3 1 2 3p0 3p0 w 9 2 V0, 3p0 q 3V0, 3p0 U = q + w Easy: -6p0V0 U - w = q , .∆ Describing the state of a system: m, V, n, T, P… Units! Equations of state: The ideal gas, the van der Waals gas. Types of Energy: kinetic, potential. Units. Work: mechanical work against a constant or variable force. Thermodynamics : definitions Energy exchanges: work and heat. pV work Heat: Heat capacity. Internal energy: molecular interpretation Internal energy: First principle Heat capacity Enthalpy: Definition, Reversible processes. 174 Internal Energy 175 Monoatomic ideal gas U: kinetic (translational, rotational, vibrational) + potential Ideal gas Temperature is a measure of the average kinetic energy. When you increase the temperature of a system, the energy you supply is “stored” in these motions. The more ‘places’ to store energy, the more energy you need to supply to increase the temperature by a given amount (e.g. 1 C) the higher Cm,v and the Cm,p , .∆ (if , is constant) The more ways that a molecule has of taking up energy, the higher the heat capacity. Constant Volume Heat Capacity 176 R = 8.31 J/(K.mol) Gas Cv (J/(K.mol)) CP (J/(K.mol)) CP-CV (J/(K.mol)) Helium (He) 12.46 20.77 8.31 Nitrogen (N2) 20.8 29.1 8.3 Methane (CH4) 27.4 35.8 . 8.4 Ethane (C2H6) 44.4 52.9 . . 8.5 Gas Cv (J/(K.mol)) CP (J/(K.mol)) CP-CV (J/(K.mol)) Helium (He) 1.5 R= 3/2 R 2.5 R = 5/2 R 8.31 = R Nitrogen (N2) 2.5 R = 5/2 R 3.5 R = 7/2 R 8.3 = R Methane (CH4) 3.3 R 4.3. R 8.4 = 1.01R Ethane (C2H6) 5.3 R 6.3. R . 8.5 = 1.02 R Heat Capacity 177 At constant pressure, as you increase the temperature, the volume will increase pushing the piston up. That means you are performing work, and part of the heat from the burner is wasted (you need more energy to produce the same increase in temperature). Changes in volume are negligible for solids and liquids. Constant volume: All the heat you transfer is used to increase the kinetic energy (i.e. temperature) of the gas From “The molecules of Life” by Kuriyan et al. Heat Capacity 178 = , , +R , , For a gas that obeys pV= nRT Liquids and solids (hard to compress) If you are interested in a proof check any p.chem. book Constant Volume Heat Capacity 179 Let’s get crazy.. How crazy? Let’s look at a protein. Lysozyme, for example. It has 129 amino acids. You are out of control girl… , , ∆ Constant Volume Heat Capacity At pH 4.5, the heat capacity is about 4 kcal/mol/K This is 16 kJ/mol/K, or about 2,000 times R 180 Constant Volume Heat Capacity Let’s get crazy.. Let’s look at a protein. Lysozyme, for example. It has 129 amino acids. 181 How crazy? You are out of control girl… The more ways that a molecule has of taking up energy, the higher the heat capacity. Measuring changes in the Cp of proteins is a useful way of following denaturation and other conformational changes. Constant Volume Heat Capacity 182 , From “The Molecules of Life” by J. Kuriyan et al. Constant Volume Heat Capacity 183 This area is the amount of heat you need to unfold one mole of protein at constant pressure , , . This is the enthalpy of unfolding. From “The Molecules of Life” by J. Kuriyan et al. Describing the state of a system: m, V, n, T, P… Units! Equations of state: The ideal gas, the van der Waals gas. Types of Energy: kinetic, potential. Units. Work: mechanical work against a constant or variable force. Thermodynamics : definitions Energy exchanges: work and heat. pV work Heat: Heat capacity. Internal energy: molecular interpretation Internal energy: First principle Heat capacity Enthalpy: Definition, Reversible processes. 184 A new state function: Enthalpy 185 By definition: H = U + pV It can be shown that the change in enthalpy of a system can be identified with the heat transferred to it at constant pressure, and in the absence of work other than pV*: H = qp This is interesting… q is a path variable, but qp = H and qV = U are sate functions * This means that the system performs only pV work, and does not do other types of work, such as electrical work. Ok… A new state function: Enthalpy the change in enthalpy of a system can be identified with the heat transferred to it at constant pressure 186 Remember the “selfish” sign convention: • negative heat means the system releases heat to the surroundings. H = qp • positive heat: the system absorbs heat from the surroundings. More definitions • An exothermic process releases energy as heat to the surroundings. • An endothermic process absorbs energy as heat from the surroundings. A new state function: Enthalpy the change in enthalpy of a system can be identified with the heat transferred to it at constant pressure H = qp 187 Remember the “selfish” sign convention: • negative heat means the system releases heat to the surroundings. • positive heat: the system absorbs heat from the surroundings. H > 0 Endothermic process H < 0 Exothermic process • Cooking an egg • mixing water and strong acids • Melting ice cubes • Combustion (burning something • Producing sugar by photosynthesis in the presence of oxygen) A new state function: Enthalpy H > 0 Endothermic process 188 H < 0 Exothermic process Will the final temperature be greater or smaller than 25°C? 25° NH4NO3 (s) + H2O(l) → NH4+(aq) + NO3- (aq) water 25° A new state function: Enthalpy willincrease willnotchange 189 willdecrease Thereisnowaytotell Will the final temperature be greater or smaller than 25°C? 25° NH4NO3 (s) + H2O(l) → NH4+(aq) + NO3- (aq) water 25° A new state function: Enthalpy H > 0 Endothermic process 190 H < 0 Exothermic process A new state function: Enthalpy 193 Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the final temperature of the resulting solution? Coffee cup calorimetry Insulated coffee cup Stirring rod thermometer This is a constant pressure calorimeter because the cork lid will not allow the pressure to build up inside the cup (that is, the pressure inside will be the same as outside, which is the atmospheric pressure) • This is a process at constant pressure, so q = ΔH • The cup is insulated, so there’s no heat being transferred from the inside of the cup to the outside • The reaction is endothermic, so it will absorb heat, which will result in a decrease in temperature (because the cup is insulated) A new state function: Enthalpy 194 Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the final temperature of the resulting solution? Define the system clearly: Everything inside the coffee cup. The heat absorbed by the reaction (q1 > 0) is taken from the solution (q2 < 0). The cup is insulated, so there’s no heat being transferred from the inside of the cup to the outside: q =0 q 1 + q2 = 0 A new state function: Enthalpy 195 Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the final temperature of the resulting solution? 80g/mol The heat absorbed by the reaction (q1 > 0) is taken from the solution (q2 < 0). q1 + q2 = 0 How much heat does the reaction absorb? .∆ 5 80 Because the process is at constant pressure 25.69 1.61 We have the molar value (per mol), so we need to multiply by the number of moles. A new state function: Enthalpy 196 Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the final temperature of the resulting solution? 80g/mol The heat absorbed by the reaction (q1 > 0) is taken from the solution (q2 < 0). q1 + q2 = 0 How much heat does the reaction absorb? .∆ 5 80 25.69 1.61 This amount of heat comes from cooling down the solution We already saw that: . ∆ where m = 100 g (density of water is 1 g/mL), and c is 4.184 J.g-1.°C-1 1.61 . ∆ =0 A new state function: Enthalpy 197 Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the final temperature of the resulting solution? 80g/mol The heat absorbed by the reaction (q1 > 0) is taken from the solution (q2 < 0). q1 + q2 = 0 How much heat does the reaction absorb? .∆ 5 80 25.69 1.61 -1,610 This amount of heat comes from cooling down the solution 100 . 4.184 . . 23 = 19.2°C We already saw that: . ∆ where m = 100 g (density of water is 1 g/mL), and c is 4.184 J.g-1.°C-1 1.61 . ∆ =0 Energy and Enthalpy 198 OK, that is a lot of stuff. Let’s try a problem. We did most of the hard work when calculating U some slides ago Totally. We just need to incorporate two new equations: H = U + pV H = qp Internal Energy 199 Example: Calculate U, q and w and H for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 1 3 1 2 3p0 3p0 w 9 2 V0, 3p0 q 3V0, 3p0 U = q + w Easy: -6p0V0 U - w = q , . , .∆ Internal Energy 200 Example: Calculate U, q and w and H for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) 1 3 1 2 3p0 3p0 w 9 2 V0, 3p0 3V0, 3p0 q U = q + w Easy: -6p0V0 U - w = q H = qp , . , .∆ Energy and Enthalpy 201 Maybe something more challenging? Let’s see…. H = U + pV H = qp Internal Energy 202 Example: Calculate U, q and w and H for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) ∆ 3p0 3p0 3p0 V0, 3p0 V0, 3p0 V0, p0 0 if volume does not change 3 1 2 , . , .∆ How can I reduce the pressure while keeping the volume constant? We need to cool down the gas. We need to cool down the gas. T2 < T1 we are lowering the internal energy of the gas, < 0, w = 0 Internal Energy 203 Example: Calculate U, q and w and H for the process depicted in the figure. Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol) ∆ 3p0 3p0 3p0 V0, 3p0 V0, 3p0 V0, p0 0 if volume does not change 3 1 , . , .∆ H = U + pV 2 H = U + (pV) = U + (pfVf- piVi) = U + (p0V0- 3p0V0) = U - 2p0V0 Energy and Enthalpy 204 Still pretty easy… Maybe something with liquids? We did ALL the hard work when calculating U some slides ago H = U + pV H = qp Energy and Enthalpy 205 Energy and Enthalpy Initial state: 1 mol of H2O(l) at 0°C and 1 bar Final state: 1 mol of H2O(l) at 100°C and 10 bar Goal: calculate U and H 206 Energy and Enthalpy 207 Initial state: 1 mol of H2O(l) at 0°C and 1 bar Final state: 1 mol of H2O(l) at 100°C and 10 bar Goal: calculate U and H What do you mean “choose a path”? Can I choose anything? That is the beauty of state functions like U and H. Their changes do not depend on the path, just on the initial and final states Energy and Enthalpy 208 Initial state: 1 mol of H2O(l) at 0°C and 1 bar Final state: 1 mol of H2O(l) at 100°C and 10 bar Goal: calculate U and H We know how to deal with constant pressure, constant temperature and constant volume, so let’s imagine a path that combines reversible processes that keep one variable constant. Initial state: final state: T = 273K T = 373K P= 1 bar P= 10 bar HTotal UTotal Energy and Enthalpy: Dependence on p, V, and T 209 Initial state: 1 mol of H2O(l) at 0°C and 1 bar Final state: 1 mol of H2O(l) at 100°C and 10 bar Goal: calculate U and H We know how to deal with constant pressure, constant temperature and constant volume, so let’s imagine a path that combines reversible processes that keep one variable constant. Constant T Constant p Initial state: T = 273K P= 1 bar H1 U1 Intermediate state: T = 273K P= 10 bar H2 U2 HT = H1 +H2 UT = U1 + U2 final state: T = 373K P= 10 bar Energy and Enthalpy: Dependence on p, V, and T 210 Initial state: 1 mol of H2O(l) at 0°C and 1 bar Final state: 1 mol of H2O(l) at 100°C and 10 bar Goal: calculate U and H We should get the same result if hold p constant first and T constant second Constant p Initial state: T = 273K P= 1 bar H3 U3 Constant T Intermediate state: T = 373K P= 1 bar H4 U4 HT = H3 +H4 UT = U3 + U4 final state: T = 373K P= 10 bar Energy and Enthalpy: Dependence on p, V, and T Initial state: 1 mol of H2O(l) at 0°C and 1 bar Final state: 1 mol of H2O(l) at 100°C and 10 bar Goal: calculate U and H Watch the whole solution in the screencast 211