CH 1302 – Chemical Thermodynamics Principles / Problems for Coaching

1. A non flow reversible process occurs for whichP=3 V 3+ 1V

where P is in bar and V is in m3. Calculate the

work done when volume changes from 0.4 m3 to 1.2 m3

2. A system undergoes a process 1-2 during which 50 kJ energy is added as heat while it does 30 kJ work. Then the system follows the process 2 – 3 during which 40 kJ is rejected as heat while 50 kJ work is done on it. Then the system returns to the initial state by an adiabatic process. Calculate the net work done by the system.

3. An elevator with a mass of 3 tons rests at a level 15 m above the base of an elevator shaft. It is raised to 125 m above the base of the shaft. The elevator falls freely to the base of the shaft and strikes a spring and comes to rest. Calculate:

a) The potential energy of the elevator in its initial and final position.b) Work required raising the elevator.c) The velocity and kinetic energy of the elevator before it strikes the springd) The potential energy of the compressed springe) If the elevator and spring is considered as a system, calculate the energy of the systems at different

conditions mentioned above.4. A rigid and insulated tank of volume 2 m3 is divided into two equal compartments by a partition. One

compartment contains an ideal gas at 1 MPa and 327°C (600K), while the other compartment contains the same gas at 0.1 MPa and 27°C (300 K). Find the final temperature and pressure of the gas in the tank if the partition gets punctured. (take γ = 1.4 for gases)

5. An inventor claims that a heat engine he has developed has the following specification:Power developed = 70 kWFuel burnt = 4 kg/hrHeating value of fuel = 75,000 kJ/kgTemperature limits = 1000 K and 300 K

State whether his claim is valid.6. Calculate the change in internal energy change in enthalpy, work done and the heat supplied in the following

process.a. An ideal gas is expanded from 5 bar to 4 bar isothermally at 600 Kb. An ideal gas contained in a vessel of 0.1 m3 capacity is initially at 1 bar and 298 K. It is heated at

constant volume to 400 K.Assume CP = 30 J/ mol. K

7. 1 kg mol of CO2 gas having a volume of 0.381 m3 is available at 313 K. Find out the pressure exerted by it using ideal gas law and van der Waals equation.

a = 0.365 Nm4/mol2 and b = 4.28 X 10 -5 m3/ mola. Molar volume = volume/ moles

P = RT/Vb. Van der Waals equation

( P + a/V2) (V – b) = RT8. Calculate the molar volume for methanol vapor at 500 K using Virial equation truncated to 3 terms and

Redlich - Kwong equation. The Virial coefficients are B = - 2.19 X 10-4 m3/ mol and C = - 1.73 X 10 -8 m6/mol2. The critical temperature and pressure are 512.6 K and 81 bar.

9. Calculate the volume of 1 kg mole of methane at 150°C and 100 atm using ideal gas law and van der Waals equation. A = 2.28 atm (m3 kg mol -1)2 ; b = 0.0428 m3/k molR = 0.08206 atm/mol K

10. 220 kg of CO2 gas at 27oC and 1 atm is compressed adiabatically to 1/5th of its volume. It is then cooled to its original temperature at constant volume. Find Q, ΔU and W for each step and for the entire process.

11. A vertical cylinder with a freely floating piston contains 0.1 kg air at 1.2 bar and a small electric resistor. The resistor is wired to an external 12 V battery. When a current of 1.5 Amp is passed through the resistor for 90

CH 1302 – Chemical Thermodynamics Principles / Problems for Coaching              Compiled by R.Arul Kamalakumar Page 1

sec, the piston sweeps a volume of 0.01 m3. Assume (i) piston and the cylinder are insulated and (ii) air behaves as an ideal gas with Cv = 700 J/(kg.K). Find the rise in temperature of air.

12. A 28 liter rigid enclosure contains air at 140 kPa and 20oC. Heat is added to the container until the pressure reaches 345 kPa. Calculate the heat added.

13. A tank having a volume of 0.1 m3 contains air at 14 MPa and 50oC. It is connected through a valve to a larger tank having a volume of 15 m3, which is completely evacuated. The entire assembly is completely insulated. The valve is opened and the gas allowed to come to equilibrium in both tanks. Calculate the final pressure.

14. Liquid water at 25oC flows in a straight horizontal pipe, in which there is no exchange of either heat or work with the surroundings. Its velocity is 12 m/s in a pipe with an i.d. of 2.5 cm until it flows into a section where the pipe diameter increases to 7.5 cm. What is the temperature change?

15. One kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and 1 bar until its volume triples. Calculate W, Q, ΔU, and ΔH for the process. Assume air obeys the relation PV/T = 83.14 bar cm3 mol-1 K-1 and that Cp = 29 J mol-1 K-1.

16. Liquid water at 100oC and 1 bar has an internal energy (on an arbitrary scale) of 419 kJ/kg and a specific volume of 1.044 cm3/g.

(a) What is its enthalpy?(b) The water is brought to the vapor state at 200oC and 800 kPa, where its enthalpy is 2838.6 kJ/kg and its specific volume is 260.79 cm3/g. Calculate ΔU and ΔH for the process.

17. Calculate the change in entropy when 10 kg of air is heated at constant volume from a pressure of 101325 N/m2 and a temperature of 20oC to a pressure of 405300 N/m2. CV = 20.934 kJ/kmol.oC.

18. A power plant operates with a heat source reservoir at 350oC and a heat sink reservoir at 30oC. It has a thermal efficiency of 55% of the Carnot engine thermal efficiency for the same temperature levels. What is the thermal efficiency of the plant?

19. A heat exchanger uses 5000 kg/hr of water to cool hydrocarbon oil from 140oC to 65oC. The oil, flowing at the rate of 2500 kg/hr has an average specific heat of 0.6 kcal/ kg.oC. The water enters at 20oC. Determine

i. The entropy change of the oil ii. Entropy change of water

iii. The total entropy change as a result of this heat exchange process.20. From the following vapor pressure data, construct the temperature - composition diagram at 1 atm, for the

system benzene-toluene, assuming ideal solution behavior. Temperature oC

Vapor pressure, mm HgBenzene Toluene

80 760 300 92 1078 432 100 1344 559 110.4 1748 760

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UNIT – IV-THERMODYNAMIC FORMULATIONS

1. How are the thermodynamic properties classified? Explain it with examples. Thermodynamic properties of fluids can classify into 3 broad groups.

Reference properties Energy Properties Derived Properties

2. Reference Properties: It is also known as primary properties. It is used to define the state of the system. They have absolute values, as against energy properties, which are measured relative to some arbitrary

reference state. Ex: Temperature, Pressure, Volume and Entropy

3. Energy Properties: There are Four energy properties Internal energy (U), enthalpy (H), Helmholtz free energy (A) and Gibbs free energy (G) All the extensive thermodynamic properties and are known relative to reference state.

4. Derived Properties: These are partial derivatives of energy properties (or) the reference properties. Ex: Specific heat (C), Coefficient of expansion (β) , Joule–Thomson Coefficient (µ) and Coefficient of

compressibility (Қ)

5. Define work function:- The Helmholtz free energy (or) work function of a system is defined as

A=U-TS Where U- internal energy, A – Helmholtz free energy, T – Temperature, S- Entropy Work – Function is a state function. Work function is also an extensive property.

6. Define free energy; Free energy is the difference between the enthalpy of a system and the product of its entropy and absolute

temperature.G=H−TS Where G- Gibbs free energy, H – Enthalpy, T – Temperature, S – Entropy

It is a state function It is also an extensive property.

7. Clapeyron Equation (Pl refer previous notes)

8. Write the fundamental property relations.

dU =TdS−PdVdH =TdS+VdP

dA=−SdT−PdVdG=−SdT +VdP

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9. Maxwell’s Relations from fundamentals.

UNIT-V

THERMODYNAMIC PROPERTIES OF REAL GASES AND COMPRESSION OF FLUIDS

10. Give any two equations used to calculate fugacity coefficient of a real gas?Effect of temperature:

(δlnf/δT)p=(Ho-H/RT2)Effect of pressure

(δlnf/δT)T=(V/RT)

11. Write a note on acentric factor?(Definition is already given)

The acentric factor ω is a concept that was introduced by Pitzer in 1955 I is very useful in the characterization of substances. It has become a standard for the proper characterization of any single pure component, along with

other common properties, such as molecular weight, critical temperature, critical pressure, and critical volume.

The acentric factor is said to be a measure of the non- sphericity (acentricity) of the molecules. It is defined as:

12. State the laws of corresponding states?Z = f (PR,TR)

The equation is the mathematical statement of the law of corresponding state. The law says that all gases under the reduced conditions of temperature and pressure have the same

compressibility factor. Compressibility factor.

Z º PV/RT = 1 + BP + CP2 + DP3 + …or

Z = 1 + B/V + C/V2 + D/V3 + … The Virial equation of state is the only one which has a firm basis in theory. \ It follows from statistical mechanics. It can be used to represent both liquids and gases. The term B/V arises due to pair-wise interactions of molecules. The term C/V2 arises due to interactions among three molecules, etc.

13. Define equation of state?

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Any equation which relates pressure volume and temperature is called on equation of state f (P,V,T) = 0

14. What are the basic assumptions involved in ideal gas equation? There is no interaction between the molecules of gas. The volume of molecule is compared to the total volume of molecule is negligible.

15. Write a note on Compressor and their types. Compressors are similar to pumps Both increase the pressure on a fluid Both can transport the fluid through a pipe As gases are compressible, the compressor also reduces the volume of a gas Liquids are relatively incompressible, while some can be compressed, The main action of a pump is to pressurize and transport liquids.

Types of Air Compressors There are many different types of air compressors available in the market today. They come in a wide range of sizes and types. Here are some of the most common air compressors that you will come across today.

a. Piston compressors or reciprocating compressors – These are the most common types available in the market today. Basically, they are positive displacement compressors and are generally found in wide ranges that vary from fractional to very high horsepower. They work by pumping air into an air chamber and then reducing this chamber’s volume. The manner in which they work is very similar to that of an internal combustion engine but more or less in a reverse manner. They have pistons, valves, cylinders, housing blocks and crankshafts.

b. Rotary Screw Compressors – These types work on the principle of air filling into a void that is present between two helical mated screws. As these screws are turned, the volume will be reduced, which results in increased air pressure. Most of these compressors inject oil into the compression area and the bearing. This is done for lubrication, cooling and even creating a kind of seal to reduce any leakage.

c. Rotary Sliding Vane Compressors – like the piston and rotary compressors, these are also positive displacement air compressors. The pump is made up of a stator, rotor and 8 blades. The rotor is arranged in an eccentric manner within the stator and forms a crescent shaped swept area between the two ports. As the rotor makes its revolutions, compression will take place and the volume will go directly from maximum at the intake port to minimum at the exhaust port. Oil will be injected into the intake for cooling, lubrications and sealing the vanes.

d. Centrifugal Compressors – These are not positive displacement air compressors like the others mentioned above. They make use of extremely high speed spinning impellers to accelerate the air, and then a diffuser in order to decelerate the air.

16. Ejectors A vacuum producing device, without any moving part It uses high velocity steam as a motive fluid The design of the unit is extremely critical in view of the large steam consumption and requirement

of high levels of vacuum. Advantages

Extremely reliable and stable operation Ejectors will handle highly corrosive vapors and fluids Complete absence of moving parts. Available in all sizes Can operate safely under the most hazardous conditions Can be fabricated in any materials of construction

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17. SOME DEFINITIONS in Compressors Bore = Cylinder diameter. Stroke = Distance through which the piston moves. The two extreme positions of the piston are known as head-end and crank-end dead centres. Clearance Volume (Cl) : Volume occupied by the fluid when the piston is at head-end dead centre. Piston Displacement (PD) : Volume, a piston sweeps through. Compression Ratio (rv) : Ratio of cylinder volume with the piston at crank-end dead centre to the

cylinder volume with the piston at head-end dead centre. Single-acting : Where only one side of the piston is used. Double-acting : Where both sides of the piston are used.

Mechanical Efficiency :Break Work

Indicated Work , which gives an indication of the losses occurring between

the piston and driving shaft. Volumetric Efficiency : Is a measure of the effectiveness of the machine with regard to gas handling.

ηVol= Vol. of gasactually compressed∧delivered asmeasured at inlet pressure∧temperature

Piston displacement

= Mass of gas actually compressed∧delivered

Mass of gas occupyingthe pistondisplacement at inlet pressure∧temperature

18. What is the effect of pressure ratio on capacity of the compressor?The pressure ratio is increased the volumetric efficiency of a compressor of fixed

clearance decreases, eventually becoming zero.

19. Various compression processes. Adiabatic - This model assumes that no energy (heat) is transferred to or from the gas during the

compression, and all supplied work is added to the internal energy of the gas, resulting in increases of temperature and pressure.

Theoretical temperature rise is,

with T1 and T2 in degrees Rankine or kelvins, p2 and p1 being absolute pressures and k = ratio of specific heats (approximately 1.4 for air).

The rise in air and temperature ratio means compression does not follow a simple pressure to volume ratio. This is less efficient, but quick.

Adiabatic compression or expansion more closely model real life when a compressor has good insulation, a large gas volume, or a short time scale (i.e., a high power level). In practice there will always be a certain amount of heat flow out of the compressed gas. Thus, making a perfect adiabatic compressor would require perfect heat insulation of all parts of the machine. For example, even a bicycle tire pump's metal tube becomes hot as you compress the air to fill a tire.

The relation between temperature and compression ratio described above means that the value of “n” for an adiabatic process is k (the ratio of specific heats).

Isothermal - This model assumes that the compressed gas remains at a constant temperature throughout the compression or expansion process.

In this cycle, internal energy is removed from the system as heat at the same rate that it is added by the mechanical work of compression.

Isothermal compression or expansion more closely models real life when the compressor has a large heat exchanging surface, a small gas volume, or a long time scale (i.e., a small power level).

Compressors that utilize inter-stage cooling between compression stages come closest to achieving perfect isothermal compression. However, with practical devices perfect isothermal

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compression is not attainable. For example, unless you have an infinite number of compression stages with corresponding intercoolers, you will never achieve perfect isothermal compression.

For an isothermal process, n is 1, so the value of the work integral for an isothermal process is:

When evaluated, the isothermal work is found to be lower than the adiabatic work. Polytropic - This model takes into account both a rise in temperature in the gas as well as some loss of

energy (heat) to the compressor's components. This assumes that heat may enter or leave the system, and that input shaft work can appear as

both increased pressure (usually useful work) and increased temperature above adiabatic (usually losses due to cycle efficiency).

Compression efficiency is then the ratio of temperature rise at theoretical 100 percent (adiabatic) vs. actual (polytropic).

Polytropic compression will use a value of n between 0 (a constant-pressure process) and infinity (a constant volume process).

For the typical case where an effort is made to cool the gas compressed by an approximately adiabatic process, the value of n will be between 1 and k.

20. What is the physical significance of compressibility factor For an ideal gas the compressibility factor is defined as Z=1. In real gases this is seldom the case. Z generally increases with pressure and decreases with

temperature. At high pressures or low temperatures, molecules are moving less rapidly and are colliding more often.

This allows attractive forces between molecules to dominate, making the volume of the real gas less than the volume of an ideal gas which causes to drop below one.

When pressures are lower or temperatures higher, the molecules are more free to move. In this case repulsive forces dominate, making Z

. The closer the gas is to its critical point or its boiling point, the more Z deviates from the ideal case.

21. Define fugacity. G=RT ln f +C Fugacity is the escaping tendency of a fluid at high pressures Fugacity is mainly applied to mixtures , but also studied for gases Fugacity is always equal to pressure for ideal gases Fugacity coefficient is the ratio of fugacity to pressure and denoted by φ

For real gases the ratio fP

is not a constant and if the pressure is reduced, the behavior of real gases

approaches ideal gases. So at very low pressures fugacity approaches the pressure.

limP →0

fP

=1∨ fP

→ 1 as P=0

22. Define Activity. Activity is the relative fugacity and defined as fugacity to fugacity at the standard condition.

a= f

f ° , but the temperature at standard state is same as the temperature at the given condition.

For gases the standard state fugacity is chosen by convenience to be unity and therefore fugacity and activity are numerically equal.

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