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IZhO 2020, Теория — Задача 2. Phase states and phase transitions

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XVI International Zhautykov Olympiad/Theoretical Competition · 10 януари 2020 г. · 10 т.

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THEORETICAL COMPETITION

January 10, 2020

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The time available for the theoretical competition is 4 hours. There are three questions.
Use only the pen provided.
You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.
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At the end of the exam, arrange all sheets for each problem in the following order:
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SOLUTIONS TO THE PROBLEMS OF THE THEORETICAL COMPETITION

SOLUTIONS TO THE PROBLEMS OF THE THEORETICAL COMPETITION

Attention. Points in grading are not divided!

Условие

At a given pressure, the transition from one phase state of matter to another always occurs at a strictly defined temperature, and the transition itself is called a phase transition. For example, ice at atmospheric pressure melts at $0\ {}^{\circ}\mathrm{C}$ , so that when the heat is supplied, the temperature of the mixture of ice and water remains unchanged until all the ice turns into water.

In all the subtasks proposed below, consider that the specific volume of the liquid phase is negligible compared to the specific volume of saturated vapor, which can be considered an ideal gas. Assume as well that the heat capacity of liquid water is independent of temperature.

Useful physical constants

Gas constant $R = 8{,}31\ J/(mol\cdot K)$ ;

molar mass of air $\mu_{air} = 29{,}0\ g/mol$ ;

acceleration of gravity $g = 9{,}81\ m/s^2$ .

Normal conditions:

pressure $P_0 = 1\ atm = 760\ mm\,Hg = 101325\ Pa$ ;

temperature $T_0 = 273{,}15\ K = 0\ {}^{\circ}\mathrm{C}$ .

Properties of water (H $_2$ O)

Molar mass of water $\mu_w = 18{,}0\ g/mol$ ;

water density $\rho_w = 1{,}00\ g/sm^3$ ;

ice density $\rho_i = 0{,}920\ g/sm^3$ ;

melting point of ice at normal pressure $t_m = 0{,}00\ {}^{\circ}\mathrm{C}$ ;

boiling point of water at normal pressure $t_b = 100{,}0\ {}^{\circ}\mathrm{C}$ ;

specific heat of water $c_w = 4{,}20\ J/(g\cdot K)$ ;

specific heat of melting ice $q_i = 334\ J/g$ ;

specific heat of water vaporization (at 100 ℃) $r_w = 2259\ J/g$ ;

Poisson's adiabatic exponent for water vapor $\gamma = C_P/C_V = 4/3$ .

Specific heat of phase transition

If a transition from one phase state to another is associated with the release or absorption of a certain amount of heat, called the transition heat, then such a transition is called the first-order phase transition. In this case, the transition heat $q$ for a unit mass is called the specific heat of the phase transition (melting, evaporation, sublimation).

Since the phase transition occurs at constant pressure, according to the first law of thermodynamics, the heat $q$ is spent on changing the internal energy $u$ and on performing the work $A$ against constant external pressure:

q = u_2 - u_1 + A,

where $u_1, u_2$ stand for the specific internal energies of the first and second phases, respectively.

During melting (crystallization), due to a small difference in the densities of the liquid and solid phases, the volume change as a result of the phase transition is small, therefore, the work $A$ can be neglected in comparison with the change in internal energy.

2.1 Evaluate how much of the evaporation heat of water at $t_b = 100\ {}^{\circ}\mathrm{C}$ is spent on changing the internal energy. Express your answer in %. [1 т.]

2.2 Evaluate the specific heat of water vaporization at room temperature $t = 20{,}0\ {}^{\circ}\mathrm{C}$ . [1 т.]

In the following, the specific heat of vaporization of all liquids is considered to be temperature independent.

The Clausius–Clapeyron relation

When the pressure changes, the temperature of the first-order phase transition changes as well, i.e. the phase transition occurs at a strictly defined dependence $P(T)$ between the pressure $P$ and the temperature $T$ of the matter under investigation. This dependence, depicted on the (T,P)-plane, is called the (T,P)-phase diagram, and the P(T) curve itself is called the phase equilibrium curve. The Clapeyron-Clausius relation gives the slope of the phase equilibrium curve P(T) in the following form:

\frac{dP}{dT} = \frac{q}{T(v_2 - v_1)},

where $q$ denotes the specific heat of transition from phase 1 with the specific volume $v_1$ to phase 2 with the specific volume $v_2$ .

2.3 Assuming that the pressure of the saturated water vapor at the temperature $t_b = 100\ {}^{\circ}\mathrm{C}$ is known, obtain an explicit dependence of the pressure of the saturated water vapor on its temperature $P(T)$ . [0,4 т.]

2.4 Evaluate the boiling point of water at the highest peak in Kazakhstan – Khan-Tengri mountain. The height of the Khan-Tengri mountain peak is $h \approx 7000\ m$ above the sea level. The altitude air temperature should be considered constant and equal to $t_0 = 0\ {}^{\circ}\mathrm{C}$ . [1 т.]

2.5 At what pressure (in atmospheres) the ice melts at the temperature of $t = -\ 1{,}00\ {}^{\circ}\mathrm{C}$ ? [0,6 т.]

2.6 It is known that ice crystals begin to break down if a force is applied along any direction of the crystal to create a pressure $P > P_{cr} \sim 1000\ atm$ . Therefore, snow in frosty weather "crunches" when walking. Estimate the maximum air temperature $t_{max}$ at which the snow still “crunches” when walking. [0,6 т.]

2.7 One mole of the saturated water vapor occupies a vessel and has the temperature of $t_b = 100\ {}^{0}\mathrm{C}$ . The vapor heats up and at the same time its volume changes such that it remains saturated at all times. Find the molar heat capacity of vapor in such a process. [1 т.]

Border boiling

Border boiling is boiling at the interface between two immiscible liquids. The border boiling point may vary significantly from the volume boiling points of each liquid.

Tetrachloromethane or hydrogen tetrachloride is a heavy (density $\rho = 1{,}60\ g/sm^3$ ) transparent liquid with a molar mass $\mu = 153{,}8\ g/mol$ . Under normal atmospheric pressure, carbon tetrachloride boils at a temperature of $t = 76{,}65\ {}^{\circ}\mathrm{C}$ , while it practically does not dissolve in water. A vessel with a volume of $V = 100\ ml$ is half-filled with the carbon tetrachloride, and the same (by volume) amount of water is poured over it. In this case, a clear water-carbon tetrachloride border is formed. When the vessel is uniformly heated in a water bath, the border boiling at the liquid interface begins at the temperature of $t^* = 66{,}0\ {}^{\circ}\mathrm{C}$ , which is significantly lower than the volume boiling temperature of each liquid.

2.8 Calculate the specific heat of evaporation of carbon tetrachloride, if it is known that the pressure of the saturated water vapor at the border boiling point is $P_w(t^*) = 196\ mm\,Hg$ . [1,2 т.]

2.9 Find the mass of liquid remaining in the vessel by the time the other liquid is completely boiled away at such border boiling. [1 т.]

Consider another pair of immiscible liquids, water and fluoroketone.

Fluoroketone, sometimes called “dry water”, is used to extinguish fires in libraries, museums, and offices because it does not wet paper. It is a heavy (density $\rho = 1{,}72\ g/sm^3$ ) transparent liquid with a molar mass $\mu = 316\ g/mol$ , which practically does not dissolve in water. The boiling point of fluoroketone at atmospheric pressure is $t_f = 49{,}2\ {}^{\circ}\mathrm{C}$ , its specific heat of vaporization is $r = 95{,}0\ J/g$ . If water is poured over the fluoroketone into the vessel, a clear water-fluoroketone border is also formed.

2.10 Estimate the boiling point $t_x$ of liquids at the water-fluoroketone border if the saturated vapor pressure of water is known at the volume boiling point of fluoroketone to be $P_w(t_f) = 89{,}0\ mm\,Hg$ . [2,2 т.]

Отговори

Решение

Problem 2. Phase states and phase transitions (10,0 points)

Specific heat of phase transition

2.1 The work of steam against constant external pressure during evaporation of a unit water mass is found as

A = P(v_2 - v_1). \qquad (1)

Since $v_2 \gg v_1$ , we can neglect the specific volume of liquid $v_1$ in comparison with the specific volume of vapor $v_2$ . Then, considering water vapor as an ideal gas with the equation of state

Pv = \frac{RT}{\mu_w} \qquad (2)

the work sought is obtained as

A = P(v_2 - v_1) \approx \frac{RT_b}{\mu_w}. \qquad (3)

Thus, the ratio of work to the total heat of evaporation at $T = 373\ K$ is determined by the expression

\frac{A}{r_{\text{в}}} = \frac{RT_b}{\mu_w r_w}, \qquad (4)

and the rest of the heat goes to the increase of the internal energy of the system $\Delta u = r_w - A$ , i.e.

\frac{\Delta u}{r_w} = 1 - \frac{RT_b}{\mu_w r_w} = 92{,}4\ \%. \qquad (5)

2.2 The evaporation of one mole of water at a temperature $T$ consumes heat

\mu_w r(T) = U_2(T) - U_1(T) + PV_2 = U_2(T) - U_1(T) + RT. \qquad (6)

A similar expression for the temperature $T_b = 373\ K$ has the form

\mu_w r_w = U_2(T_b) - U_1(T_b) + RT_b\ . \qquad (7)

Subtracting equation (7) from equation (6), we obtain for the change in the molar heat of evaporation

\mu_w\Delta r = \Delta U_2 - \Delta U_1 + R\Delta T = C_P\Delta T - \mu_w c_w\Delta T = \mu_w\left(\frac{C_P}{\mu_w} - c_w\right)\Delta T, \qquad (8)

where $\Delta T = T - T_b$ and $\Delta r = r(T) - r_w$ .

Given that for water vapor, the molar heat capacity at constant pressure is

C_P = 4R, \qquad (9)

we obtain the specific heat of water evaporation

r(T) = r_w - \left(c_w - \frac{4R}{\mu_w}\right)(T - T_b) = 2447\ J/g. \qquad (10)

It is interesting to note that the heat of evaporation is increased by $\Delta r/r_w \approx 8\ \%$ .

The Clausius–Clapeyron relation

2.3 Neglecting the specific volume of water compared to the volume of vapor, we apply the Clapeyron-Clausius equation to the vaporization in the form

\frac{dP}{dT} = \frac{r}{Tv} = \frac{\mu_w r_w}{RT^2}P \qquad (11)

\frac{dP}{P} = \frac{\mu_w r_w dT}{RT^2}. \qquad (12)

Integrating this expression at $r = r_w = const$ gives rise to

P = P_0\exp\left(\frac{\mu_w r_w}{R}\left(\frac{1}{T_b} - \frac{1}{T}\right)\right). \qquad (13)

2.4 As it follows from equation (13) the explicit dependence of the boiling point of water on external pressure has the form

T = \frac{T_b}{1 - \frac{RT_b}{\mu_w r_w}\ln\frac{P}{P_0}}\ . \qquad (14)

According to the barometric formula for an isothermal atmosphere, we have

P = P_0\exp\left(-\frac{\mu_{air}gh}{RT_0}\right). \qquad (15)

Substituting this expression into formula (14), we obtain the dependence of the boiling temperature on height and the numerical value of the boiling temperature of water at the altitude of $h = 7\ km$

T = \frac{T_b}{1 + \frac{T_b\mu_{air}gh}{T_0\ \mu_w\ r_w}} = 349{,}6\ K = 76{,}6\ {}^{\circ}\mathrm{C}. \qquad (16)

2.5 It follows from the Clapeyron-Clausius relation the following holds at the vicinity of 0 ℃

\frac{dP}{dT} = \frac{r_w}{T_0\left(\frac{1}{\rho_w} - \frac{1}{\rho_i}\right)}. \qquad (17)

Therefore, we obtain that in order to lower the melting temperature of ice by $1\ {}^{0}\mathrm{C}$ , the pressure should be increased by

\Delta P = \frac{dP}{dT}\Delta T = 139\ atm, \qquad (18)

so that the pressure should be equal to $P = 140\ atm$ .

2.6 In order for ice crystals to break when walking, and not to melt under the influence of pressure $P_{cr}$ , the outdoor temperature should be lower than

t_{max} = \frac{P_{cr}}{(dP/dT)} \approx -7{,}21\ {}^{\circ}\mathrm{C}\ , \qquad (19)

in which the derivative $(dP/dT)$ is determined by formula (17).

2.7 Since for one mole of vapor PV = RT, then

d(PV) = PdV + VdP = RdT, \qquad (20)

thus, the elementary work of the vapor when changing its volume is derived as

PdV = RdT - VdP. \qquad (21)

From the first law of thermodynamics it follows that the heat supplied to the vapor has the form

\delta Q = dU + \delta A = C_V dT + RdT - VdP = C_P dT - VdP\ . \qquad (22)

Given that from the Clapeyron-Clausius relation $dP/dT = r_w\mu_w/(T_b V)$ ,, we obtain the heat capacity of the vapor

C = \frac{\delta Q}{dT} = C_P - \frac{VdP}{dT} = C_P - \frac{\mu_w r_w}{T_b} = -75{,}7\ J/(K\cdot mol). \qquad (23)

Thus, the heat from the vapor must be removed so that it does not overheat as a result of expansion. It is interesting to note that the specific heat in this process turned out to be almost equal to the specific heat of water with a minus sign $c = C_P/\mu_w - r_w/T_b = -4{,}21\ J/(g\cdot K)$ .

Border boiling

2.8 A liquid boils when bubbles are formed inside such that the pressure of its saturated vapor reaches the atmospheric pressure P0. At the liquids border, the total vapor pressure in the bubbles formed upon boiling is the sum of the partial pressures of the saturated vapor of carbon tetrachloride and water at t*

P_0 = P(t^*) + P_w(t^*). \qquad (24)

It follows that the saturated vapor pressure of carbon tetrachloride at a boiling point is found as

P^* = P(t^*) = P_0 - P_w(t^*). \qquad (25)

From the Clapeyron-Clausius relation for carbon tetrachloride it follows that

\frac{dP}{P} = \frac{\mu r dT}{RT^2}. \qquad (26)

After integrating from the lower bound $T = t + 273{,}15 = 349{,}8\ K$ to the higher bound $T^* = t^* + 273{,}15 = 339{,}15\ K$ results in the following formula

ln(P_0/P^*) = r\mu\Delta T/RTT^*, \qquad (27)

which means the heat of vaporization of carbon tetrachloride is obtained as

r = \frac{RTT^*\ln(P_0/P^*)}{\mu(T - T^*)} \approx 180\ \mathrm{J/g}. \qquad (28)

For reference: the experimental value is $r = 195\ J/g$ .

2.9 The ratio of evaporation rates from the border layer is obviously equal to the ratio of the masses of vapor of tetrachloromethane and water in the bubbles formed during boiling, which, in turn, is equal to the ratio of the densities of the vapors found as

\frac{m}{m_w} = \frac{\rho}{\rho_w} = \frac{P^*\mu}{P_w(t^*)\mu_w} \approx 25. \qquad (29)

Thus, carbon tetrachloride evaporates 25 times faster (by weight) than water. This means that by the time of evaporation of carbon tetrachloride, the amount of water that finally evaporates is written as

\Delta m = \frac{\rho V}{2}\frac{m_w}{m} = 3{,}25\ g\ . \qquad (30)

Accordingly, the amount of water remaining after evaporation of all carbon tetrachloride is derived as

M_w = \rho_w V/2 - \Delta m = 46{,}7\ g. \qquad (31)

2.10 Let border boiling occur at a certain temperature $t_x$ , then the saturated vapor pressure of fluoroketone $P$ and the saturated vapor pressure of water $P_w$ at this temperature should equal the external atmospheric pressure, i.e.

P_0 = P(t_x) + P_w(t_x). \qquad (32)

Thus, the saturated vapor pressure of fluoroketone at the border boiling point decreases by the value of the saturated vapor pressure of water at this temperature

P(t_x) = P_0 - P_w(t_x). \qquad (33)

From the Clapeyron – Clausius equation (in the approximation of small liquid volume and vapor ideality) it follows that the slope of the phase equilibrium line P(T) at the volume boiling point of fluoroketone reads as

\alpha_f = \frac{\mathrm{d}P}{\mathrm{d}T} = \frac{r\mu P_0}{RT_f^2}. \qquad (34)

For water at the same temperature, a similar derivative is more than 6 times less

\alpha_\mathrm{w} = \frac{\mathrm{d}P}{\mathrm{d}T} = \frac{\mu_w r_w P_w(t_f)}{RT_f^2}. \qquad (35)

Since $\alpha_f/\alpha_\mathrm{w} \approx 6{,}30$ , the decrease in pressure and, correspondingly, in the boiling point are both small relative to the same values for fluoroketone, therefore, we can use the linear approximation near $t_f$

P_0 - P(t_x) = \alpha_f\Delta T = P_\mathrm{w}(t_x) = P_\mathrm{w}(t_f) - \alpha_\mathrm{w}\Delta T, \qquad (36)

where $\Delta T = T_f - T_x$ , wherefrom the lowering of the boiling point is found as

\Delta T = \frac{P_\mathrm{w}(\mathrm{t}_f)}{(\alpha_f + \alpha_\mathrm{w})}\ . \qquad (37)

Finally, the temperature for the border boiling is obtained as

t_x = t_f - \Delta T = 46{,}3\ {}^{\circ}\mathrm{C}. \qquad (38)

For reference: the experimental value is $t_x = (46 \pm 1)\ {}^{\circ}\mathrm{C}$ .

	Content	Points
2.1	Formula (1): $A = P(v_2 - v_1)$	0,2	1,0
	Formula (2): $Pv = \frac{RT}{\mu_w}$	0,2
	Formula (4): $\frac{A}{r_{\text{в}}} = \frac{RT_b}{\mu_w r_w}$	0,2
	Formula (5): $\frac{\Delta u}{r_w} = 1 - \frac{RT_b}{\mu_w r_w}$	0,2
	Numerical value in formula (5): $92{,}4\ \%$	0,2
2.2	Formula (6): $\mu_w r(T) = U_2 - U_1 + PV_2 = U_2 - U_1 + RT$	0,2	1,0
	Formula (7): $\mu_w r_w = U_2(T_b) - U_1(T_b) + RT_b$	0,2
	Formula (9): $C_P = 4R$	0,2
	Formula (10): $r(T) = r_w - \left(c_w - \frac{4R}{\mu_w}\right)(T - T_b)$	0,2
	Numerical value in formula (10): $2447\ J/g$	0,2
2.3	Formula (11): $\frac{dP}{dT} = \frac{r}{Tv} = \frac{\mu_w r_w}{RT^2}P$	0,2	0,4
	Formula (13): $P = P_0\exp\left(\frac{\mu_w r_w}{R}\left(\frac{1}{T_b} - \frac{1}{T}\right)\right)$	0,2
2.4	Formula (14): $T = \frac{T_b}{1 - \frac{RT_b}{\mu_w r_w}\ln\frac{P}{P_0}}$	0,2	1,0
	Formula (15): $P = P_0\exp\left(-\frac{\mu_{air}gh}{RT_0}\right)$	0,4
	Formula (16): $T = \frac{T_b}{1 + \frac{T_b\mu_{air}gh}{T_0\ \mu_w\ r_w}}$	0,2
	Numerical value in formula (16): 76,6 ℃	0,2
2.5	Formula (17): $\frac{dP}{dT} = \frac{q_i}{T_0\left(\frac{1}{\rho_w} - \frac{1}{\rho_i}\right)}$	0,2	0,6
	Formula (18): $\Delta P = \frac{dP}{dT}\Delta T$	0,2
	Numerical value in formula (17): $\Delta P = 139\ atm$ or $P = 140\ atm$	0,2
2.6	Formula (19): $t_{max} = \frac{P_{cr}}{(dP/dT)}$	0,4	0,6
	Numerical value in formula (19): $t_{max} \approx -7{,}21\ {}^{\circ}\mathrm{C}$	0,2
2.7	Formula (21): $PdV = RdT - VdP$	0,2	1,0
	Formula (22): $\delta Q = dU + \delta A = C_V dT + RdT - VdP = C_P dT - VdP$	0,2
	Formula (23): $C = \frac{\delta Q}{dT} = C_P - \frac{VdP}{dT} = C_P - \frac{\mu_w r_w}{T_b}$	0,4
	Numerical value in formula (23): $C = -75{,}7\ J/(K\cdot mol)$	0,2
2.8	Formula (24): $P_0 = P(t^) + P_w(t^)$	0,4	1,2
	Formula (26): $\frac{dP}{P} = \frac{\mu r dT}{RT^2}$	0,2
	Formula (27): $\ln(P_0/P^) = r\mu\Delta T/RTT^$	0,2
	Formula (28): $r = \frac{RTT^\ln(P_0/P^)}{\mu(T - T^*)}$	0,2
	Numerical value in formula (28): $r \approx 180\ J/g.$	0,2
2.9	Formula (29): $\frac{m}{m_w} = \frac{\rho}{\rho_w} = \frac{P^\mu}{P_w(t^)\mu_w}$	0,4	1,0
	Formula (30): $\Delta m = \frac{\rho V}{2}\frac{m_w}{m}$	0,2
	Formula (31): $M_w = \rho_w V/2 - \Delta m$	0,2
	Numerical value in formula (31): $M_w = 46{,}7\ g$	0,2
2.10	Formula (32): $P_0 = P(t_x) + P_w(t_x)$	0.2	2,2
	Formula (34): $\alpha_f = \frac{\mathrm{d}P}{\mathrm{d}T} = \frac{r\mu P_0}{RT_f^2}$	0,3
	Formula (35): $\alpha_\mathrm{w} = \frac{\mathrm{d}P}{\mathrm{d}T} = \frac{r_w\mu_w P_\mathrm{w}(t_f)}{RT_f^2}$	0,3
	Estimation: $\alpha_f/\alpha_\mathrm{w} \approx 6{,}30$	0,3
	Formula (36): $P_0 - P(t_x) = \alpha_f\Delta T = P_\mathrm{w}(t_x) = P_\mathrm{w}(t_f) - \alpha_\mathrm{w}\Delta T$	0,3
	Formula (37): $\Delta T = \frac{P_\mathrm{w}(\mathrm{t}_f)}{(\alpha_f + \alpha_\mathrm{w})}$	0,4
	Formula (38): $t_x = t_f - \Delta T$	0,2
	Numerical value in formula (38): $t_x = 46{,}3\ {}^{\circ}\mathrm{C}$	0,2
Total			10,0

Mathematical hints for the theoretical problems

Mathematical hints for the theoretical problems

The following integrals may be useful:

\int\frac{dx}{ax+b} = \frac{1}{a}\ln|ax+b|.

\int x^n dx = \frac{x^{n+1}}{n+1},\ \text{where } n \text{ is integer}

(1+x)^{\gamma} \approx 1 + \gamma x + \frac{\gamma(\gamma-1)}{2}x^2,\ \textit{for } x \ll 1 \textit{ and any } \gamma

Оригинал в Архива: IZhO-2020-Theory_eng.pdf · официални решения: IZhO-2020-Theory_eng_sol.pdf