P340: hermodynamics and Statistical Physics, Exam#2 Any answer without showing your work will be counted as zero 1. (15 points) he equation of state for the an der Waals gas (n = 1 mole) is (a) Find ( C / ) ( p + a ) ( b) = R. 2 (b) Find C p C as a function of and. (c) Assuming that C is independent of, express the entropy as a function of C, and. 2. (15 points) A room is to be kept at an indoor temperature of H = 294 K. he outdoor temperature is L. Heat, that leak through the windows and doors at a rate of Q leak = a, must be replenished by a room heater at the same rate. (a) We assume that the electric radiator convert electric power Ẇ el into heat with 100% efficiency. Find Ẇel as a function of a, H, and L. (b) Now we consider the case of heating the room with an ideal electric heat pump that uses Ẇ sup of electric power to extract heat at a rate of Q L at the outdoor temperature L and convert it into heat at a rate of Q H = Q L + Ẇhp at the indoor temperature. In general the heat derived from heat pump is given by Ẇ hp = (1 b)ẇsup, where the coefficient b represents the mechanical power inefficiency of the heat pump. Express power Ẇ sup as a function of a, b, H, and L. (c) Determine the outdoor temperature range where the heat pump is more economical than the electric radiator for fixed H = 294 K, and b = 0.75 (75% loss). 3. (20 points) he equation of state and the internal energy (Stefan s law) for the black body radiation are given by p = 1 3 a 4, U(, ) = a 4, where a = 8π 5 k 4 /(15h 3 c 3 ) = 7.55 10 16 Jm 3 K 4 is a constant, is the temperature, is the volume, and p is the pressure. (a) Show that the equation of state for an adiabatic process is p 4/3 = K, where K is a constant. (b) Describe the 4-steps (isothermal, adiabatic, isothermal, and adiabatic) Carnot engine for the black-body radiation in the (p ) plane (use 2 = 10 4 K, 1 = 5 10 3 K), and calculate the efficiency of the black-body Carnot engine. (c) Find C v and κ S. (d) Determine the entropy and other thermodynamics potentials, i.e. S(, ), U(S, ), H(S, p), F(, ), and G(, p). i
4. (10 points) A Diesel cycle for an ideal gas with γ = C p /C v operating between 1, 2 and 3, where 1 / 2 and 3 / 2 are respectively called the compression ratio and the cut-off ratio. Find the efficiency of the engine in terms of γ, compression ratio, and cut-off ratio. 5. (10 points) An ideal mixing corresponds to mixing of molecules having the same size, same interactions. It is applicable to mixings of ideal gases, liquids, etc. Consider N A blue and N B green marbles of the same size and weight. hey are mixed in a box. What is the entropy of mixing? Please expressed it in terms of x = N A /(N A + N B ). 6. (10 points) Using the able of thermodynamics parameters at temperature 0 = 298 K and p 0 = 1 bar below. Find the Gibbs free energy per mole (or chemical potential) for water and steam as a function of temperature. Evaluate the chemical potentials at = 350 K and = 400 K, at a constant pressure p 0 = 1 bar. Here, we can assume constant C p for both steam and liquid. See able 1. 7. (10 points) In a hydrogen fuel cell, the steps of the chemical reactions are at electrode : H 2 + 2OH 2H 2 O + 2e, at + electrode : 1 2 O 2 + H 2 O + 2e 2OH. Calculate the voltage of the fuel cell between these two electrodes. See able 1. 8. (10 points) A biological cell is a water solution of sugar, amino acids, and other molecules. ypically, it contains 200 water molecules for each molecule of something else. Sea water has a salinity of 3.5%, i.e. if you boil away a kilogram of seawater you will find 35 g of NaCl left in the pot. Each mole of NaCl is 58.44 g. If you drop a biological cell into sea water, what will be the osmotic pressure when the equilibrium is reached at = 300 K? We assume that the cell wall is semipermeable that only water molecules can pass through the cell wall. able 1: hermodynamic Properties of some substances at 0 = 298K and p 0 = 1 bar Substance G (kj) S (J/K) C p (J/K) H 2 O (l) 237.13 69.91 75.29 H 2 O (g) 228.57 188.83 33.58 H 2 (g) 0 130.68 28.82 O 2 (g) 0 205.14 29.38 ii
P340: hermodynamics and Statistical Physics, Exam#2, Solution 1. he equation of state for the an der Waals gas (n = 1 mole) is ( p + a ) ( b) = R. 2 (a) Show that C is independent of, i.e. ( C / ) = 0. (b) Find C p C as a function of p and. (c) Assuming that C is independent of, show that the entropy of the an der Waals gas is S = C ln + R ln( b) + constant. hus p = R v b a v 2, du = ( ) u d + v ( ) u v = ( ) p p = a v v 2 ( ) u dv = C v d + a v v 2dv (a) Since du is an exact differential, we find (b) ( ) Cv i.e. C v is a function of only. C p = ( ) S, p C p C v = vβ2 κ C p C v = R [ = R (v b) 2 = ( (a/v 2 ) ) v = 0, ( ) 2 ( ) 2 β p vκ = κ ] [ ] 1. R (v b) 2 2 a v 3 In the limit that a 0 and b 0, we find C p C v R. (c) o evaluate entropy, we begin with ds = 1 (du + pd ) = 1 ( ) v p = [( ) ( ) ] U U d + d + pd. ( ) p 2 ( ) p v, 1
We know ( U/ ) = C. Now we evaluate ( ) U = ( ) p p = R b p = a 2. Substituting the result in the entropy equation, we find ds = C d + R b d. If C is independent of, we can integrate the equation to obtain where S 0 is an integration constant. S = C ln + R ln( b) + S 0, 2. A room is to be kept at an indoor temperature of H = 294 K. he outdoor temperature is L. Heat, that leak through the windows and doors at a rate of Q leak = a, must be replenished by a room heater at the same rate. (a) We assume that the electric radiator convert electric power Ẇ el into heat with 100% efficiency. Find Ẇel as a function of a, H, and L. (b) Now we consider the case of heating the room with an ideal electric heat pump that uses Ẇ sup of electric power to extract heat at a rate of Q L at the outdoor temperature L and convert it into heat at a rate of Q H = Q L + Ẇhp at the indoor temperature. In general the heat derived from heat pump is given by Ẇ hp = (1 b)ẇsup, where the coefficient b represents the mechanical power inefficiency of the heat pump. Express power Ẇ sup as a function of a, b, H, and L. (c) Determine the outdoor temperature range where the heat pump is more economical than the electric radiator for fixed H = 294 K, and b = 0.75 (75% loss). (a) Ẇ el = a = a( H L ). (b) o supply constant temperature, we need Q H = a = a( H L. Since Q H = Q L + Ẇhp = (cop + 1)Ẇhp = [ H /( H L )]Ẇhp, we find Ẇ sup = 1 1 bẇhp = H L (1 b) H Q H = a ( H L ) 2 (1 b) H. (c) Setting Ẇsup Ẇel, we find L H b his means that L b H = 221 K or above 52 C, the heat pump can provide more heat for heating the house than an electric space heater. 2
3. he equation of state and the internal energy (Stefan s law) for the black body radiation are given by p = 1 3 a 4, U(, ) = a 4, where a = 8π 5 k 4 /(15h 3 c 3 ) = 7.55 10 16 Jm 3 K 4 is a constant, is the temperature, is the volume, and p is the pressure. (a) Show that the equation of state for an adiabatic process is p 4/3 = K, where K is a constant. (b) Describe the 4-steps (isothermal, adiabatic, isothermal, and adiabatic) Carnot engine for the black-body radiation in the (p ) plane, and calculate the efficiency of the black-body Carnot engine (c) Find C v and κ S, and determine the entropy and other thermodynamics potentials, i.e. S(, ), U(S, ), H(S, p), F(, ), and G(, p). (a) From the first law of thermodynamics, we find ds = du + Pd = (4a 3 d + a 4 d ) + 1 3 a 4 d. For an adiabatic process, ds = 0, we find d + d 3 = d[ln( 1/3 )] = 0, or 1/3 = constant, or 1 3 a 4 4/3 = p 4/3 = constant. (b) A cycle of a Carnot engine is divided into 4 processes: i. isothermal: Q 2 = du + Pd = 4 3 a 4 2 ( 2 1 ) ii. adiabatic: dq = 0, 2 1/3 2 = 1 1/3 3 iii. isothermal: Q 1 = du + Pd = 4 3 a 4 1 ( 4 3 ) iv. adiabatic: dq = 0, 1 1/3 4 = 2 1/3 1 W = Q 2 + Q 1 = 4 3 a( 2 1 )[ 3 1 3 3 2 1], η = W Q 2 = 2 1 2 = 1 1 2. (c) Using definition of each physical quantities, we obtain: C = ( ) U = 4a 3, κ S = 1 ( ) p U(S, ) = a 4 = a 3 S ( 3S 4a = 3 4p, S(, ) = 4 3 a 3, ) 4/3 = 34/3 4 4/3 a 1/3S4/3 1/3, H(p, S) = U + p = 4 3 a 4 = 31/4 a 1/4p1/4 S, F(, ) = U S = 1 3 a 4, G(, p) = H S = 0
Figure 1: he Carnot cycle example with 2 = 10 4 K, 1 = 5 10 3 K, 2 = 1 m 3, and 3 = 2 m 3 is shown in this example. For the photon gas, isothermal process is equivalent to an isobaric process. 4. (10 points) A Diesel cycle for an ideal gas with γ = C p /C v operating between 1, 2 and 3, where 1 / 2 and 3 / 2 are respectively called the compression ratio and the cut-off ratio. Find the efficiency of the engine in terms of γ, compression ratio, and cut-off ratio. he Diesel cycle is described by 1 2 adiabatic, 2 3 isobaric, 3 4 adiabatic, and 4 1 constant volume. hus Q h = C p ( 3 2 ), Q c = C v ( 4 1 ), e = 1 Q c = 1 1 4 1 = 1 1 1 (p 4 p 1 ) Q h γ 3 2 γ p 2 ( 3 2 ), where we applied the ideal gas law in the last equation, and use γ = C p /C. Now, the adiabatic processes provide the following identities: p 4 γ 1 = p 2 γ 3, p 1 γ 1 = p 2 γ 2. hus we find p 4 p 1 p 2 = e = 1 1 γ ( ) γ ( ) γ 3 2 1 1 1 (p 4 p 1 ) p 2 ( 3 2 ) = 1 1 γ ( 2 1 ) γ ( 3 / 2 ) γ 1 ( 3 / 2 ) 1 5. (10 points) An ideal mixing corresponds to mixing of molecules having the same size, same interactions. It is applicable to mixings of ideal gases, liquids, etc. Consider N A blue and N B green marbles of the same size and weight. hey are mixed in a box. What is the entropy of mixing? Please expressed it in terms of x = N A /(N A + N B ). 4
he number of possible states for the mixing is Ω = N! N A!N B! hus the mixing entropy is S = k ln Ω = Nk(x ln(x) + (1 x) ln(1 x), where x = N A /N. 6. (10 points) Using the able of thermodynamics parameters at temperature 0 = 298 K and p 0 = 1 bar below. Find the Gibbs free energy per mole (or chemical potential) for water and steam as a function of temperature. Evaluate the chemical potentials at = 350 K and = 400 K, at a constant pressure p 0 = 1 bar. Here, we can assume constant C p for both steam and liquid. See able 1. he Gibbs free energy at constant pressure is given by G() = G( 0 ) 0 S()d where G is the Gibbs free energy relative to the Gibbs free energy of H 2 and 1 2 O 2 at 0 and p 0. he entropy is given by hus S() = S( 0 ) + 0 C p () d = S 0 + C p ln 0. G() = G 0 S 0 ( 0 ) C p [ ln ] ( 0 ) 0 Using the able 1, we can calculate the Gibbs free energy vs temperature. Figure 2 shows the Gibbs free energies of H 2 O liquid and gas phases. hey cross at the boiling point. heir values are 241.1 kj (l), 238.5 kj (g) at = 350 K, and 245.5 kj (l), 248.4 kj (g) at = 400 K respectively 7. (10 points) In a hydrogen fuel cell, the steps of the chemical reactions are at electrode : H 2 + 2OH 2H 2 O + 2e, at + electrode : 1 2 O 2 + H 2 O + 2e 2OH. Calculate the voltage of the fuel cell between these two electrodes. See able 1. he available Gibbs free energy in this reaction is 237.13 kj (see able 1). Each reaction transports 2 electrons across the electrodes, and thus the voltage available is = G Q = 237.13 10 3 J/mol 2 (1.602 10 19 ) (6.022 10 23 ) 5 = 1.23 olts
Figure 2: he Gibbs free energy vs the temperature at p 0 = 1 bar. (Problem #6). At the boiling point, the Gibbs functions of water and steam are equal. 8. (10 points) A biological cell is a water solution of sugar, amino acids, and other molecules. ypically, it contains 200 water molecules for each molecule of something else. Sea water has a salinity of 3.5%, i.e. if you boil away a kilogram of seawater you will find 35 g of NaCl left in the pot. Each mole of NaCl is 58.44 g. If you drop a biological cell into sea water, what will be the osmotic pressure when the equilibrium is reached at = 300 K? We assume that the cell wall is semipermeable that only water molecules can pass through the cell wall. he chemical potentials are equal for both solutions. hus µ 0 (, p 1 ) N B1k N A p 2 p 1 = n B2 n B1 R = µ 0 (, p 2 ) N B2k N A Since NaCl is ionized, each molecule counts as two solutes, we find n B2 = 2 3.5 104 g = 2 599 mol/m3 58.44 g/mol m3 n B1 = 1.0 mol 200 mol 18 g/mol 10 6 m 3 /g = 278 mol/m3. hs osmotic pressure is p 2 p 1 = 22.9 bar. 6