Solutions to Problem Set 5

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1 Cornell University, Physics Department Fall 204 PHYS-334 Statistical Physics Prof. Itai Cohen Solutions to Problem Set 5 David C. sang, Woosong Choi 5. Refrigeration Reif 5.22: Refrigeration cycles have been developed for heating buildings. he procedure is to design a device which absorbs heat from the surrounding earth or air outside the house and then delivers heat at a higher temperature to the interior of the building. Such a device is called a heat pump. ) a) If a device is used in this way, operating between the outside absolute temperature o and an interior absolute temperature i, what would be the maximum number of kilowatt-hours of heat that could be supplied to the building for every kilowatt-hour of electrical energy needed to operate the device? b) Obtain a numerical answer for the case that the outside temperature is 0 C and the interior temperature is 25 C. a) he change in total entropy is We want to know what Q i W S = Q i i Q i W o 0 is, so we move terms around and get Q i W ) i o o hus for o < i, Q i W i i o b) Substituting the values i = 298K and i = 273K we get Q i W.9

2 5.2 Gas Engine Reif 5.26: A gasoline engine can be approximately represented by the idealized cyclic process abcd shown in the accompanying diagram of pressure p versus volume V of the gas in the cylinder. Here a b represents the adiabatic compression of the air-gasoline mixture, b c the rise in pressure at constant volume due to the explosion of the mixture, c d the adiabatic expansion of the mixture during which the engine performs useful work, and d a the final cooling down of the gas at constant volume. Assume this cycle to be carried out quasi-statically for a fixed amount of ideal gas having a constant specific heat. Denote the specific heat ratio by γ c p /c v. Calculate the efficiency η ratio of work performed to heat intake Q ) for this process, expressing your answer in terms of V,, and γ. First, for the heating and cooling processes we can write Q = νc v c b ) Q 2 = νc v d a ) Since over a cycle abcd the gas returns to its original state, Q = Q 2 +W. hus, the efficiency is η = W = Q Q 2 = Q 2 = d a Q Q Q c b For adiabatic proccess, pv γ is constant and for ideal gas pv = νr, V γ is also constant. hus we can relate a and d to b and c respectively. a = b V d = c V Substituting this in the equation of η we get η = c b c b V ) γ ) γ ) γ = ) γ V 2

3 5.3 riple Point I Reif 8.2: he vapor pressure p in millimeters of mercury) of solid ammonia is given by ln p = / and that of liquid ammonia by ln p = /. a) What is the temperature of the triple point? b) What are the latent heats of sublimation and vaporization at the triple point? c) What is the latent heat of melting at the triple point? a) riple point is the point where the phase transition between all three phases occur. For that to happen, the phase transition curve between solid-vapor and liquid-vapor must coincide at a point. Equating the equations we get ln p = = C C = C C = ) hus, C = 95.2K b) From the Clausius-Clapeyron equation Reif 8.5.0) dp d = L 2 V Assuming that the ammonia vapor obeys the ideal gas law and that the volume of the solid and the liquid is negligibly small compared to the volume of the gas, we can deduce ln p = l R + constant hus, from the given equations we can find that c) At the triple point, l sublimination = 3754R 3754K 8.3J/mol K J/mol l vaporization = 3063R 3063K 8.3J/mol K J/mol l solid liquid = S solid S liquid ) = S solid S gas + S gas S liquid ) = l sublimination l vaporization 3

4 hus, the latent heat of melting is l melting ) 0 4 J/mol = J/mol 5.4 Latent Heat Reif 8.7: he molar latent heat of transformation in going from phase to phase 2 at the temperature and the pressure p is l. What is the latent heat of the phase transformation at a slightly different temperature and corresponding presusre), i.e., what is dl/d )? Express your answer in terms of l and the molar specific heat c p, coefficient of expansion α, and molar volume v of each phase at the original temperature and pressure p. dl l = S 2 S ) d = ds2 d ds ) + S + S 2 ) d We also have ds = C ) p V d dp = V αdp + C p d ds d dl d = V2 α 2 l V + C p2 + V α l V C p ) + l = C p2 C p + l dp = V α d + C p Box V αl = V + C p ] [ + V α α 2 V 4

5 5.5 ie Lines Bowley and Sanchez.6: Suppose that the thermodynamic potential for a system with a conserved quantity φ is of the form Φφ) = Φ0) + cφ 2 φ 2 a) 2 What it the slope of the double tangent to the curve? Obtain φ and φ 2, the values of φ where the double tangent touches the curve. What are the spinodal limits, φ and φ 2, for this thermodynamic potential? Show that the spinodal limits lie between φ and φ 2. For what value of φ is the system in a metastable state? a) First, we will find the double tangent values φ and φ 2. For the two values of the double tangent touching the curve, it should satisfy ) ) Φ Φ = = Φφ 2) Φφ ) φ φ φ 2 φ From the first part of the equality, we get φ φ 2 4cφ φ 2 φ 2 a) = 4cφ 2φ 22 φ 2 a) φ φ 2)φ 2 + φ φ 2 + φ 22 φ 2 a) = 0 From this, φ = φ 2 or φ 2 + φ φ 2 + φ 22 φ 2 a = 0. But, we don t want φ = φ 2 since it would not lead us to a double tangent. Now, take the average of the first two derivatives in the equation it should still be the same as the right hand side of the equation, since all three terms are equal. hat is, 2 [ ) ) Φ Φ + phi φ phi φ 2 ] = Φφ 2) Φφ ) φ 2 φ 2cφ φ 2 φ 2 a) + 2cφ 2φ 2 2 φ 2 a) = c φ 2 φ 2 2φ 2 a ) φ 22 φ 22 2φ 2 a ) φ φ 2 2φ + φ 2)φ 2 φ φ 2 + φ 22 φ 2 a) = φ 2 φ 22 )φ 2 + φ 22 2φ 2 φ φ a) 2 Now, we since φ φ 2 we can divide φ 2 φ 22 on the right hand side by φ φ 2 and simplify φ + φ 2)φ 2 2φ φ 2 + φ 22 ) = φ + φ 2)φ φ 2) 2 = 0 hus, φ = φ 2. By φ 2 + φ φ 2 + φ 22 φ 2 a = 0 this becomes φ 2 = φ 2 a. φ,2 = ±φ a 5

6 b) Now we want to find the spinodal limits for this potential. Spinodal limits are the values where 2 Φ φ 2 = 0. 2 Φ φ 2 = 4cφ2 φ 2 a + 2φ 2 ) = 4c3φ 2 φ 2 a) = 0 Solutions to this equation is φ,2 = ± φ a 3 Since φ = ±φ a from a), we see that the spinodal limit values φ,2 lie between φ and φ 2. c) Metastable states are the states with the second derivative of the potential negative while the phase parameter remaining within the double tangent contact points. Since the second derivative is 2 Φ = 4c3φ 2 φ 2 φ a), for values 3φ 2 < φ 2 2 a we get negative second derivative. For φ values from φ a to φ a / 3 and from φ a / 3 to φ a, we get positive second derivative which means that the states are metastable. 6

Solutions to Problem Set 4

Cornell University, Physics Department Fall 2014 PHYS-3341 Statistical Physics Prof. Itai Cohen Solutions to Problem Set 4 David C. sang, Woosong Choi 4.1 Equilibrium Fluctuations (a From lecture we had