P340: Thermodynamics and Statistical Physics, Exam#2 Any answer without showing your work will be counted as zero

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1 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 ) = 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 = 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

2 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 (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 (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 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) H 2 O (g) H 2 (g) O 2 (g) ii

3 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

4 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

5 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 ) = 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 ) 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 )[ ], η = W Q 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

6 Figure 1: he Carnot cycle example with 2 = 10 4 K, 1 = 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 = = (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 (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

7 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 kj (l), kj (g) at = 350 K, and kj (l), 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 kj (see able 1). Each reaction transports 2 electrons across the electrodes, and thus the voltage available is = G Q = J/mol 2 ( ) ( ) 5 = 1.23 olts

8 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 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 = g = mol/m 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