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# Lecture 19 Overview Ch. 4-5 PowerPoint PPT Presentation

Lecture 19 Overview Ch. 4-5. List of topics Heat engines Phase transformations of pure substances, Clausius-Clapeyron Eq. van der Waals gases Thermodynamic potentials, chemical reactions. P. 2. 1. T H. 3. T C. V 1. V 2. V. Problem 1 (heat engine).

Lecture 19 Overview Ch. 4-5

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### Lecture 19 Overview Ch. 4-5

• List of topics

• Heat engines

• Phase transformations of pure substances, Clausius-Clapeyron Eq.

• van der Waals gases

• Thermodynamic potentials, chemical reactions

P

2

1

TH

3

TC

V1

V2

V

### Problem 1 (heat engine)

Working substance for the cycle shown in the Figure is 1 mole of ideal gas. Find the efficiency of this heat engine (in terms of TH and TC).

1 – 2

2 – 3

3 – 1

### Problem (vdW)

An insulating membrane (does not conduct heat) divides an insulated tank into two equal volumes. Each volume contains one mole of the same van der Waals gas (the constants a and b are known). The pressure in volume I is P1, in volume II – P2 . The piston has been removed. Find the pressure, explain your reasoning.

V, P1

V, P2

### Isothermal Process for a vdW gas

10 kg of water at 200C is converted to ice at - 100C by being put in contact with a reservoir at - 100C. This process takes place at constant pressure and the heat capacities at constant pressure of water and ice are 4180 and 2090 J/kg·K respectively. The heat of fusion of ice is 3.34·105 J/kg.

### Latent heat, dS for phase transformations

(a) Calculate the heat absorbed by the cold reservoir.

(b) Calculate the change in entropy of the closed system “reservoir + water/ice”.

The conversion consists of three processes: (a) water at 200C  water at 00C; (b) water at 00C  ice at 00C; (c) ice at 00C  ice at -100C:

(a) the heat absorbed by the cold reservoir

(b) the change in entropy of the sub-system “water/ice”:

cooling of water

forming ice

cooling of ice

The increase of entropy of the reservoir:

The total change of entropy of the whole system:

### Vapor equation

• For Hydrogen (H2) near its triple point (Ttr=14K), the latent heat of vaporization Lvap=1.01 kJ/mol. The liquid density is 71 kg·m-3, the solid density is 81 kg·m-3, and the melting temperature is given by Tm =13.99+P/3.3, where Tm and P measured in K and MPa respectively. Compute the latent heat of sublimation.

• The vapor pressure equation for H2: where P0 = 90 MPa .

• Compute the slope of the vapor pressure curve (dP/dT) for the solid H2 near the triple point, assuming that the H2 vapor can be treated as an ideal gas.

1. Near the triple point:

P

liquid

solid

Ptr

gas

T

Ttr

### Vapor equation (cont.)

2. At the solid-gas phase boundary:

Assuming that the H2 vapor can be treated as an ideal gas

### Phase transformations

The pressure-temperature phase diagram of carbon is shown below. For simplicity, assume that the molar volumes of graphite and diamond are independent of temperature and pressure at 5.3×10-6 and 3.4×10-6 m3, respectively. 1 kbar = 108  N/m2 .

(a) Determine the latent heat per mole of transformation at T = 1000 K.

(b) Sketch the graph of Gibbs free energy of carbon at the constant temperature T = 2000 K as a function of pressure between P = 50 and 70 kbar. Mark the transition pressure. Explain the changes, if any, of the slope of G.

(c) Sketch the graph of entropy per mole of the material at the constant pressure P = 90 kbar as a function of temperature between 4000 and 5500 K. State any assumptions you make and explain your graph.

(a)

(b)

G.

the slope of G changes at the phase transition because of the change in volume

63 kbar

P

(c)

S

4000

5500

T

### C.-C. Equation

The vapor pressure of solid ammonia is given by the relation:

where units –mm of Hg, T – absolute temperature.

The vapor pressure of liquid ammonia is given by the relation:

• What is the temperature of triple point?

• Compute the latent heat of vaporization at the triple point. (Assume that the vapor can be treated as an ideal gas, and the density of vapor is negligibly small compared to that of the liquid)

• The latent heat of sublimation at the triple point is 31.4 kJ/mol. What is the latent heat of melting at the triple point?

Ttr= 195 K

(a)

Ttr – from the equation

(b)

From the equation for liq. ammonia

The Clausius-Clapeyron eq. gives

(c)

Denote Sg, Sl, and Ss as the entropy for vapor, liquid and solid at triple point.

The latent heat of vaporization -

The latent heat of sublimation -

The latent heat of melting -

### C-C eq., phase transformation)

A cylinder closed with a piston is filled with the saturated water vapor at T = 1000C. The vapor is heated up by 10C, and, at the same time, the piston is moved to prevent condensation and to keep the vapor saturated (the system is “moving” along the coexistence curve). Find the relative change in the vapor volume, V/V. Assume that the vapor is an ideal gas, the latent heat of vaporization Lvap=40.7 kJ/mol, and the vapor density is negligible in comparison with the water density.

The Clausius-Clapeyron equation for the coexistence curve “liquid-gas”:

For an ideal gas:

For a small temperature changes :

The triple point for water corresponds to Ttr=0.010C and Ptr=0.006 bar. At T~Ttr,,the latent heat of melting is 335 kJ/kg, and the slope of the solid-vapor phase coexistence curve is dP/dT=50 Pa/K . Assume that water vapor behaves as an ideal gas near its triple point. Find the latent heat of vaporization.

### phase transformations

Along the solid-gas phase equilibrium curve:

- the latent heat of sublimation at the triple point

### Chem. equilibrium

Consider the following equilibrium at 500 K:

CO(g) + 2 H2 (g)  CH3OH(g)

The equilibrium concentrations are: [CO] = 0.0911 M, [H2] = 0.0822 M, [CH3OH] = 0.00892 M, what is the value of the equilibrium constant? Is G positive or negative when the reactants are transformed into products?

Since the value of the equilibrium constant is greater than one, G <0.

Consider the reaction:

• N2(g) + 3H2 (g)  2NH3 (g) Keq = 0.5 at 400 K

Suppose we mix the following initial concentrations:

In which direction will the reaction go?

Thus, the reaction will go to the left (some ammonia molecules will be transformed into hydrogen and nitrogen).

CO

H2O

CO2

H2

0.1

0.1

0

0

-x

- x

+ x

+ x

0.1-x

0.1 - x

x

x

### Problem (chem. equilibrium)

• Consider the following reaction:

• CO(g) + H2O(g)  CO2 (g) + H2 (g) Keq = 23.8 at 600 K

• If the initial amounts of CO and H2O were both 0.100 M, what will be the amounts of each reactant and product at equilibrium?

start

change

finish

We start withn0CO  and n0H20  moles of the reacting gases and define as the yieldx the number of moles of CO2 and H2 that the reaction will produce at equilibrium:

The mass action law requires:

This is a quadratic equation with respect to x :

Thus, in equilibrium,

A

B

C

D

2

1

1

1

-x

- 2x

+ x

+ 3x

2-x

1 - 2x

1+x

1+3x

### Chem. equilibrium

• Consider the following reaction: A (g) + 2B (g)  C (g) +3D (g), where A, B, C, D are some molecules. In equilibrium,

• (a) Calculate the equilibrium constant and the “standard” Gibbs energy for this reaction at a given temperature.

• (b) To the above equilibrium system we add an extra 1M of A. What are the new equilibrium concentrations of each component?

• (a) The mass action law:

(b)

initial

change

final